PARAMETRIC EQUATIONS,
11 POLAR COORDINATES,
AND VECTOR FUNCTIONS
11.1 Parametric Equations
Preliminary Questions
1. Describe the shape of the curve x D 3 cos t; y D 3 sin t.
SOLUTION
For all t,
x 2 C y 2 D .3 cos t/2 C .3 sin t/2 D 9.cos2 t C sin2 t/ D 9 ! 1 D 9;
therefore the curve is on the circle x 2 C y 2 D 9. Also, each point on the circle x 2 C y 2 D 9 can be represented in the form
.3 cos t; 3 sin t/ for some value of t. We conclude that the curve x D 3 cos t, y D 3 sin t is the circle of radius 3 centered at the
origin.
2. How does x D 4 C 3 cos t; y D 5 C 3 sin t differ from the curve in the previous question?
SOLUTION
In this case we have
.x " 4/2 C .y " 5/2 D .3 cos t/2 C .3 sin t/2 D 9.cos2 t C sin2 t/ D 9 ! 1 D 9
Therefore, the given equations parametrize the circle of radius 3 centered at the point .4; 5/.
3. What is the maximum height of a particle whose path has parametric equations x D t 9 , y D 4 " t 2 ?
SOLUTION
The particle’s height is y D 4 " t 2 . To find the maximum height we set the derivative equal to zero and solve:
dy
d
D
.4 " t 2 / D "2t D 0 or
dt
dt
t D0
The maximum height is y.0/ D 4 " 02 D 4.
4. Can the parametric curve .t; sin t/ be represented as a graph y D f .x/? What about .sin t; t/?
SOLUTION In the parametric curve .t; sin t/ we have x D t and y D sin t, therefore, y D sin x. That is, the curve can be
represented as a graph of a function. In the parametric curve .sin t; t/ we have x D sin t, y D t, therefore x D sin y. This
equation does not define y as a function of x, therefore the parametric curve .sin t; t/ cannot be represented as a graph of a function
y D f .x/.
5. Match the derivatives with a verbal description:
dx
dy
(a)
(b)
dt
dt
(i) Slope of the tangent line to the curve
(ii) Vertical rate of change with respect to time
(iii) Horizontal rate of change with respect to time
SOLUTION
dx
is the horizontal rate of change with respect to time.
dt
dy
(b) The derivative
is the vertical rate of change with respect to time.
dt
dy
(c) The derivative
is the slope of the tangent line to the curve.
dx
Hence, (a) $ (iii), (b) $ (ii), (c) $ (i)
(a) The derivative
(c)
dy
dx
1348
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
Exercises
1. Find the coordinates at times t D 0, 2, 4 of a particle following the path x D 1 C t 3 , y D 9 " 3t 2 .
Substituting t D 0, t D 2, and t D 4 into x D 1 C t 3 , y D 9 " 3t 2 gives the coordinates of the particle at these
times respectively. That is,
SOLUTION
.t D 0/
.t D 2/
.t D 4/
2. Find the coordinates at t D 0;
SOLUTION
Setting t D 0, t D
!
4;!
!
4,
x D 1 C 03 D 1; y D 9 " 3 ! 02 D 9
x D 1 C 23 D 9; y D 9 " 3 ! 22 D "3
x D 1 C 43 D 65; y D 9 " 3 ! 42 D "39
) .1; 9/
) .9; "3/
) .65; "39/:
of a particle moving along the path c.t/ D .cos 2t; sin2 t/.
and t D ! in c.t/ D .cos 2t; sin2 t/ we obtain the following coordinates of the particle:
t D 0W
tD
!
4W
.cos 2 ! 0; sin2 0/ D .1; 0/
2!
2 !
4 ; sin 4 /
.cos 2!; sin2 !/
.cos
t D !W
D .0; 12 /
D .1; 0/
3. Show that the path traced by the bullet in Example 3 is a parabola by eliminating the parameter.
SOLUTION
The path traced by the bullet is given by the following parametric equations:
x D 80t; y D 200t " 4:9t 2
We eliminate the parameter. Since x D 80t, we have t D
x
80 .
y D 200t " 4:9t 2 D 200 !
The equation y D 25 x "
4:9 2
6400 x
Substituting into the equation for y we obtain:
! x "2
x
5
4:9 2
" 4:9
D x"
x
80
80
2
6400
is the equation of a parabola.
4. Use the table of values to sketch the parametric curve .x.t/; y.t//, indicating the direction of motion.
t
"3
"2
"1
0
1
2
3
x
"15
0
3
0
"3
0
15
y
5
0
"3
"4
"3
0
5
SOLUTION We mark the given points on the xy-plane and connect the points corresponding to successive values of t in the
direction of increasing t. We get the following trajectory (there are other correct answers):
y
6
t = −3
t=3
4
2
t = −2
−15 −10 −5
t=1
t=2
5
−2
x
10
15
t = −1
−4 t = 0
−6
5. Graph the parametric curves. Include arrows indicating the direction of motion.
(a) .t; t/, "1 < t < 1
(b) .sin t; sin t/, 0 # t # 2!
(c) .e t ; e t /, "1 < t < 1
(d) .t 3 ; t 3 /, "1 # t # 1
SOLUTION
(a) For the trajectory c.t/ D .t; t/, "1 < t < 1 we have y D x. Also the two coordinates tend to 1 and "1 as t ! 1 and
t ! "1 respectively. The graph is shown next:
y
x
S E C T I O N 11.1
Parametric Equations
1349
(b) For the curve c.t/ D .sin t; sin t/, 0 # t # 2!, we have y D x. sin t is increasing for 0 # t # !2 , decreasing for !2 # t # 3!
2
!
and increasing again for 3!
2 # t # 2!. Hence the particle moves from c.0/ D .0; 0/ to c. 2 / D .1; 1/, then moves back to
c. 3!
2 / D ."1; "1/ and then returns to c.2!/ D .0; 0/. We obtain the following trajectory:
y
y
t=
π
(1,1)
2
t=
π
(1,1)
2
x
x
t=0
t=
0<t #
y
!
2
3π
(−1,−1)
2
!
2
x
t=
#t #
3!
2
3π
(−1,−1)
2
3!
2
# t < 2!
These three parts of the trajectory are shown together in the next figure:
y
t=
t=0
t = 2π
t=
π
(1,1)
2
x
3π
(−1,−1)
2
(c) For the trajectory c.t/ D .e t ; e t /, "1 < t < 1, we have y D x. However since
trajectory is the part of the line y D x, 0 < x.
lim e t D 0 and lim e t D 1, the
t !!1
t !1
y
x
(d) For the trajectory c.t/ D .t 3 ; t 3 /, "1 # t # 1, we have again y D x. Since the function t 3 is increasing the particle moves in
one direction starting at .."1/3 ; ."1/3 / D ."1; "1/ and ending at .13 ; 13 / D .1; 1/. The trajectory is shown next:
y
t = 1(1,1)
x
t = −1 (−1,−1)
6. Give two different parametrizations of the line through .4; 1/ with slope 2.
SOLUTION The equation of the line through .4; 1/ with slope 2 is y " 1 D 2.x " 4/ or y D 2x " 7. One parametrization is
obtained by choosing the x coordinate as the parameter. That is, x D t. Hence y D 2t " 7 and we get x D t, y D 2t " 7,
"1 < t < 1. Another parametrization is given by x D 2t , y D t " 7, "1 < t < 1.
In Exercises 7–14, express in the form y D f .x/ by eliminating the parameter.
7. x D t C 3,
SOLUTION
y D 4t
We eliminate the parameter. Since x D t C 3, we have t D x " 3. Substituting into y D 4t we obtain
y D 4t D 4.x " 3/ ) y D 4x " 12
1350
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
8. x D t !1 ,
SOLUTION
y D t !2
From x D t !1 , we have t D x !1 . Substituting in y D t !2 we obtain
9. x D t, y D
SOLUTION
tan!1 .t 3
C et /
y D t !2 D .x !1 /!2 D x 2 ) y D x 2 ;
x ¤ 0:
Replacing t by x in the equation for y we obtain y D tan!1 .x 3 C e x /.
10. x D t 2 , y D t 3 C 1
SOLUTION
p
From x D t 2 we get t D ˙ x. Substituting into y D t 3 C 1 we obtain
p
p
y D t 3 C 1 D .˙ x/3 C 1 D ˙ x 3 C 1;
x $ 0:
Since we must have y a function of x, we should probably choose either the positive or negative root.
11. x D e !2t , y D 6e 4t
SOLUTION
We eliminate the parameter. Since x D e !2t , we have "2t D ln x or t D " 12 ln x. Substituting in y D 6e 4t we get
D 6x !2 ) y D
6
;
x2
We now substitute t D
1
x!1
1
y D 6e 4t D 6e 4".! 2 ln x/ D 6e !2 ln x D 6e ln x
12. x D 1 C t !1 ,
SOLUTION
y D t2
From x D 1 C t !1 , we get t !1 D x " 1 or t D
y D t2 D
13. x D ln t,
SOLUTION
#
y D2"t
1
x"1
$2
1
x!1 .
)yD
!2
1
;
.x " 1/2
x > 0:
in y D t 2 to obtain
x ¤ 1:
Since x D ln t we have t D e x . Substituting in y D 2 " t we obtain y D 2 " e x .
14. x D cos t, y D tan t
SOLUTION
p
We use the trigonometric identity sin t D ˙ 1 " cos2 t to write
sin t
y D tan t D
D˙
cos t
p
1 " cos2 t
:
cos t
We now express y in terms of x:
y D tan t D ˙
p
1 " x2
)yD˙
x
p
1 " x2
;
x
x ¤ 0:
Since we must have y a function of x, we should probably choose either the positive or negative root.
In Exercises 15–18, graph the curve and draw an arrow specifying the direction corresponding to motion.
15. x D
1
2 t,
y D 2t 2
Let c.t/ D .x.t/; y.t// D . 12 t; 2t 2 /. Then c."t/ D ."x.t/; y.t// so the curve is symmetric with respect to the
y-axis. Also, the function 12 t is increasing. Hence there is only one direction of motion on the curve. The corresponding function
is the parabola y D 2 ! .2x/2 D 8x 2 . We obtain the following trajectory:
SOLUTION
y
t=0
16. x D 2 C 4t,
y D 3 C 2t
x
Parametric Equations
S E C T I O N 11.1
1351
x!2
x!2
SOLUTION We find the function by eliminating the parameter. Since x D 2 C 4t we have t D 4 , hence y D 3 C 2. 4 / or
y D x2 C 2. Also, since 2 C 4t and 3 C 2t are increasing functions, the direction of motion is the direction of increasing t. We
obtain the following curve:
y
6
4
2
t=1
(6, 5)
t=0
(2, 3)
x
−6
−4
2
−2
4
6
−2
−4
−6
17. x D ! t, y D sin t
SOLUTION We find the function by eliminating
x
! . We obtain the following curve:
y D sin
t. Since x D ! t, we have t D
x
!.
Substituting t D
x
!
into y D sin t we get
y
(4π2,0)
(−2π2,0)
x
18. x D t 2 , y D t 3
From x D t 2 we have t D ˙x 1=2 . Hence, y D ˙x 3=2 . Since the functions t 2 and t 3 are increasing, there is only
one direction of motion, which is the direction of increasing t. Notice that for c.t/ D .t 2 ; t 3 / we have c."t/ D .t 2 ; "t 3 / D
.x.t/; "y.t//. Hence the curve is symmetric with respect to the x axis. We obtain the following curve:
SOLUTION
y
x
19. Match the parametrizations (a)–(d) below with their plots in Figure 1, and draw an arrow indicating the direction of motion.
y
y
y
5
10
20
x
x
5
(I)
y
2π
5
5
(II)
(III)
x
−1
1
x
(IV)
FIGURE 1
(a) c.t/ D .sin t; "t/
(c) c.t/ D .1 " t; t 2 " 9/
(b) c.t/ D .t 2 " 9; 8t " t 3 /
(d) c.t/ D .4t C 2; 5 " 3t/
SOLUTION
(a) In the curve c.t/ D .sin t; "t/ the x-coordinate is varying between "1 and 1 so this curve corresponds to plot IV. As t increases,
the y-coordinate y D "t is decreasing so the direction of motion is downward.
1352
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
2π
1
x
−1
−2π
(IV) c.t/ D .sin t; "t/
(b) The curve
c.t/ D .t 2 " 9; 8t " t 3 /
p
p
intersects the x-axis where y D 8t " t 3 D 0; or t D "2 2; 0, and 2 2. These are the points ."1; 0/ and ."9; 0/. The x-intercepts
are obtained where x D t 2 " 9; or t D ˙3: The y-intercepts are .0; 3/ and .0; "3/. As t increases from "1 to 0, x decreases, and
as t increases from 0 to 1, x increases. We obtain the following trajectory:
(c) The curve c.t/ D .1 " t; t 2 " 9/ intersects the y-axis where x D 1 " t D 0, or t D 1. The y-intercept is .0; "8/. The
x-intercepts are obtained where t 2 " 9 D 0 or t D ˙3. These are the points ."2; 0/ and .4; 0/. Setting t D 1 " x we get
y D t 2 " 9 D .1 " x/2 " 9 D x 2 " 2x " 8:
As t increases the x coordinate decreases and we obtain the following trajectory:
y
10
4 5
−2
x
(III)
(d) The curve c.t/ D .4t C 2; 5 " 3t/ is a straight line, since eliminating t in x D 4t C 2 and substituting in y D 5 " 3t gives
3
13
y D 5 " 3 ! x!2
4 D " 4 x C 2 which is the equation of a line. As t increases, the x coordinate x D 4t C 2 increases and the
y-coordinate y D 5 " 3t decreases. We obtain the following trajectory:
y
5
5
(I)
x
Parametric Equations
S E C T I O N 11.1
1353
20. A particle follows the trajectory
x.t/ D
1 3
t C 2t;
4
y.t/ D 20t " t 2
with t in seconds and distance in centimeters.
(a) What is the particle’s maximum height?
(b) When does the particle hit the ground and how far from the origin does it land?
SOLUTION
(a) To find the maximum height y.t/, we set the derivative of y.t/ equal to zero and solve:
dy
d
D
.20t " t 2 / D 20 " 2t D 0 ) t D 10:
dt
dt
The maximum height is y.10/ D 20 ! 10 " 102 D 100 cm.
(b) The object hits the ground when its height is zero. That is, when y.t/ D 0. Solving for t we get
20t " t 2 D t.20 " t/ D 0 ) t D 0; t D 20:
t D 0 is the initial time, so the solution is t D 20. At that time, the object’s x coordinate is x.20/ D
when it hits the ground, the object is 2040 cm away from the origin.
1
4
! 203 C 2 ! 20 D 2040. Thus,
21. Find an interval of t-values such that c.t/ D .cos t; sin t/ traces the lower half of the unit circle.
SOLUTION For t D !, we have c.!/ D ."1; 0/. As t increases from ! to 2!, the x-coordinate of c.t/ increases from "1 to 1,
and the y-coordinate decreases from 0 to "1 (at t D 3!=2) and then returns to 0. Thus, for t in Œ!; 2!", the equation traces the
lower part of the circle.
22. Find an interval of t-values such that c.t/ D .2t C 1; 4t " 5/ parametrizes the segment from .0; "7/ to .7; 7/.
SOLUTION Note that 2t C 1 D 0 at t D "1=2, and 2t C 1 D 7 at t D 3. Also, 4t " 5 takes on the values of "7 and 7 at
t D "1=2 and t D 3. Thus, the interval is Œ"1=2; 3".
In Exercises 23–38, find parametric equations for the given curve.
23. y D 9 " 4x
SOLUTION This is a line through P D .0; 9/ with slope m D "4. Using the parametric representation of a line, as given in
Example 3, we obtain c.t/ D .t; 9 " 4t/.
24. y D 8x 2 " 3x
SOLUTION
Letting t D x yields the parametric representation c.t/ D .t; 8t 2 " 3t/.
25. 4x " y 2 D 5
SOLUTION
We define the parameter t D y. Then, x D
26. x 2 C y 2 D 49
! 5 C t2 "
5 C y2
5 C t2
D
, giving us the parametrization c.t/ D
;t .
4
4
4
The curve x 2 C y 2 D 49 is a circle of radius 7 centered at the origin. We use the parametric representation of a circle
to obtain the representation c.t/ D .7 cos t; 7 sin t/.
SOLUTION
27. .x C 9/2 C .y " 4/2 D 49
SOLUTION This is a circle of radius 7 centered at ."9; 4/. Using the parametric representation of a circle we get c.t/ D ."9 C
7 cos t; 4 C 7 sin t/.
! x "2 ! y "2
28.
C
D1
5
12
SOLUTION This is an ellipse centered at the origin with a D 5 and b D 12. Using the parametric representation of an ellipse we
get c.t/ D .5 cos t; 12 sin t/ for "! # t # !.
29. Line of slope 8 through ."4; 9/
SOLUTION
Using the parametric representation of a line given in Equation 3, we get the parametrization c.t/ D ."4 C t; 9 C 8t/.
30. Line through .2; 5/ perpendicular to y D 3x
The line perpendicular to y D 3x has slope m D " 13 . We use the parametric representation of a line given in Equation
3 to obtain the parametrization c.t/ D .2 C t; 5 " 31 t/.
SOLUTION
1354
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
31. Line through .3; 1/ and ."5; 4/
SOLUTION We use the two-point parametrization of a line with P D .a; b/ D .3; 1/ and Q D .c; d / D ."5; 4/. Then c.t/ D
.3 " 8t; 1 C 3t/ for "1 < t < 1.
%
&
%
&
32. Line through 13 ; 16 and " 76 ; 53
!
"
!
"
1 1
7 5
SOLUTION We use the two-point parametrization of a line with P D .a; b/ D 3 ; 6 and Q D .c; d / D " 6 ; 3 . Then
c.t/ D
#
1 3 1
3
" t; C t
3 2 6
2
$
for "1 < t < 1.
33. Segment joining .1; 1/ and .2; 3/
SOLUTION We use the two-point parametrization of a line with P D .a; b/ D .1; 1/ and Q D .c; d / D .2; 3/. Then c.t/ D
.1 C t; 1 C 2t/; since we want only the segment joining the two points, we want 0 # t # 1.
34. Segment joining ."3; 0/ and .0; 4/
SOLUTION We use the two-point parametrization of a line with P D .a; b/ D ."3; 0/ and Q D .c; d / D .0; 4/. Then c.t/ D
."3 C 3t; 4t/; since we want only the segment joining the two points, we want 0 # t # 1.
35. Circle of radius 4 with center .3; 9/
SOLUTION
Substituting .a; b/ D .3; 9/ and R D 4 in the parametric equation of the circle we get c.t/ D .3 C 4 cos t; 9 C 4 sin t/.
36. Ellipse of Exercise 28, with its center translated to .7; 4/
SOLUTION Since the center is translated by .7; 4/, so is every point. Thus the original parametrization becomes c.t/ D .7 C
5 cos t; 4 C 12 sin t/ for "! # t # !.
37. y D x 2 , translated so that the minimum occurs at ."4; "8/
We may parametrize y D x 2 by .t; t 2 / for "1 < t < 1. The minimum of y D x 2 occurs at .0; 0/, so the desired
curve is translated by ."4; "8/ from y D x 2 . Thus a parametrization of the desired curve is c.t/ D ."4 C t; "8 C t 2 /.
SOLUTION
38. y D cos x translated so that a maximum occurs at .3; 5/
SOLUTION A maximum value 1 of y D cos x occurs at x D 0. Hence, the curve y " 4 D cos.x " 3/, or y D 4 C cos.x " 3/
has a maximum at the point .3; 5/. We let t D x " 3, then x D t C 3 and y D 4 C cos t. We obtain the representation c.t/ D
.t C 3; 4 C cos t/.
In Exercises 39–42, find a parametrization c.t/ of the curve satisfying the given condition.
39. y D 3x " 4,
c.0/ D .2; 2/
SOLUTION Let x.t/ D t C a and y.t/ D 3x " 4 D 3.t C a/ " 4. We want x.0/ D 2, thus we must use a D 2. Our line is
c.t/ D .x.t/; y.t// D .t C 2; 3.t C 2/ " 4/ D .t C 2; 3t C 2/.
40. y D 3x " 4,
c.3/ D .2; 2/
SOLUTION Let x.t/ D t C a; since x.3/ D 2 we have 2 D 3 C a so that a D "1. Then y D 3x " 4 D 3.t " 1/ " 4 D 3t " 7,
so that the line is c.t/ D .t " 1; 3t " 7/ for "1 < t < 1.
41. y D x 2 , c.0/ D .3; 9/
Let x.t/ D t C a and y.t/ D x 2 D .t C a/2 . We want x.0/ D 3, thus we must use a D 3. Our curve is
c.t/ D .x.t/; y.t// D .t C 3; .t C 3/2 / D .t C 3; t 2 C 6t C 9/.
p
42. x 2 C y 2 D 4, c.0/ D .1; 3/
SOLUTION
SOLUTION This is a circle of radius 2 centered at the origin, so we are looking for a parametrization of that circle that starts at a
different point. Thus instead of the standard parametrization .2 cos #; 2 sin #/, # D 0 must correspond to some other angle !. We
choose the parametrization .2 cos.# C !/; 2 sin.# C !// and must determine the value of !. Now,
x.0/ D 1;
so 1 D 2 cos.0 C !/ D 2 cos !
1
!
5!
D
or
2
3
3
and ! D cos!1
Since
y.0/ D
p
3;
we have
p
3 D 2 sin.0 C !/ D 2 sin !
and ! D sin
!
so that the parametrization is
3
!
!
!
!"
! ""
c.t/ D 2 cos # C
; 2 sin # C
3
3
Comparing these results we see that we must have ! D
!1
p
2!
3
!
D
or
2
3
3
S E C T I O N 11.1
43. Describe c.t/ D .sec t; tan t/ for 0 # t <
!
2
Parametric Equations
1355
in the form y D f .x/. Specify the domain of x.
The function x D sec t has period 2! and y D tan t has period !. The graphs of these functions in the interval
"! # t # !, are shown below:
SOLUTION
y
y
−
x
x
−
−
2
−
2
2
x D sec t
2
y D tan t
x D sec t ) x 2 D sec2 t
y D tan t ) y 2 D tan2 t D
sin2 t
1 " cos2 t
D
D sec2 t " 1 D x 2 " 1
2
cos t
cos2 t
Hence the graph of the curve is the hyperbola x 2 " y 2 D 1. The function x D sec t is an even function while y D tan t is odd.
!
!
Also x has period 2! and y has period !. It follows that the intervals "! # t < " !2 , !!
2 < t < 2 and 2 < t < ! trace the curve
exactly once. The corresponding curve is shown next:
y
t=−
2
t=
−
t=−
−
2
t=0
x
(−1, 0) (1, 0)
t=
2
t=−
+
2
+
c.t/ D .sec t; tan t/
44. Find a parametrization of the right branch (x > 0) of the hyperbola
! x "2 ! y "2
"
D1
a
b
using the functions cosh t and sinh t. How can you parametrize the branch x < 0?
SOLUTION
We show first that x D cosh t, y D sinh t parametrizes the hyperbola when a D b D 1: then
x 2 " y 2 D .cosh t/2 " .sinh t/2 D 1:
using the identity cosh2 " sinh2 D 1. Generalize this parametrization to get a parametrization for the general hyperbola . xa /2 "
. yb /2 D 1:
x D a cosh t; y D b sinh t:
We must of course check that this parametrization indeed parametrizes the curve, i.e. that x D a cosh t and y D b sin t satisfy the
equation . xa /2 " . yb /2 D 1:
! x "2
a
"
! y "2
b
D
#
a cosh t
a
$2
"
#
b sinh t
b
$2
D .cosh t/2 " .sinh t/2 D 1:
The left branch of the hyperbola is the reflection of the right branch around the line x D 0, so it clearly has the parametrization
x D "a cosh t; y D b sinh t:
45. The graphs of x.t/ and y.t/ as functions of t are shown in Figure 2(A). Which of (I)–(III) is the plot of c.t/ D .x.t/; y.t//?
Explain.
y
y
y
y
x(t)
y(t)
t
(A)
x
(I)
FIGURE 2
x
x
(II)
(III)
1356
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
SOLUTION As seen in Figure 2(A), the x-coordinate is an increasing function of t, while y.t/ is first increasing and then decreasing. In Figure I, x and y are both increasing or both decreasing (depending on the direction on the curve). In Figure II, x does not
maintain one tendency, rather, it is decreasing and increasing for certain values of t. The plot c.t/ D .x.t/; y.t// is plot III.
46. Which graph, (I) or (II), is the graph of x.t/ and which is the graph of y.t/ for the parametric curve in Figure 3(A)?
y
y
y
x
t
(A)
t
(I)
(II)
FIGURE 3
SOLUTION As indicated by Figure 3(A), the y-coordinate is decreasing and then increasing, so plot I is the graph of y. Figure
3(A) also shows that the x-coordinate is increasing, decreasing and then increasing, so plot II is the graph for x.
47. Sketch c.t/ D .t 3 " 4t; t 2 / following the steps in Example 7.
SOLUTION We note that x.t/ D t 3 " 4t is odd and y.t/ D t 2 is even, hence c."t/ D .x."t/; y."t// D ."x.t/; y.t//. It
follows that c."t/ is the reflection of c.t/ across y-axis. That is, c."t/ and c.t/ are symmetric with respect to the y-axis; thus, it
suffices to graph the curve for t $ 0. For t D 0, we have c.0/ D .0; 0/ and the y-coordinate y.t/ D t 2 tends to 1 as t ! 1. To
analyze the x-coordinate, we graph x.t/ D t 3 " 4t for t $ 0:
y
8
6
4
2
1
−2
2
3
x
4
−4
x D t 3 " 4t
p
p
We see that x.t/ < 0 and decreasing for 0 < t < 2= 3, x.t/ < 0 and increasing for 2= 3 < t < 2 and x.t/ > 0 and increasing
for t >p
2. Also x.t/ tends to 1 as t ! 1. Therefore, starting at the origin, the curve first directs to the left of the y-axis, then at
t D 2= 3 it turns to the right, always keeping an upward direction. The part of the path for t # 0 is obtained by reflecting across
the y-axis. We also use the points c.0/ D .0; 0/, c.1/ D ."3; 1/, c.2/ D .0; 4/ to obtain the following graph for c.t/:
y
y
t=2
(0, 4)
t=2
t = −2
t = 1 (−3, 1)
t=1
x
t=0
t = −1
x
t=0
2
2
Graph of c.t/ for t $ 0.
Graph of c.t/ for all t.
48. Sketch c.t/ D .t " 4t; 9 " t / for "4 # t # 10.
SOLUTION
The graphs of x.t/ D t 2 " 4t and y.t/ D 9 " t 2 for "4 # t # 10 are shown in the following figures:
y
y
20
30
10
20
10
x
2
−4 −2
4
6
−10
x.t/ D t 2 " 4t
8
3
−3
6
10
−10
y.t/ D 9 " t 2
9
x
S E C T I O N 11.1
Parametric Equations
1357
The curve starts at c."4/ D .32; "7/. For "4 < t < 0, x.t/ is decreasing and y.t/ is increasing, so the graph turns to the left and
upwards to c.0/ D .0; 9/. Then for 0 < t < 2, x.t/ is decreasing and so is y.t/, hence the graph turns to the left and downwards
towards c.2/ D ."4; 5/.
For 2 < t < 10, x.t/ is increasing and y.t/ is decreasing, hence the graph turns to the right and downwards, ending at
c.10/ D .60; "91/. The intercept are the points where t 2 " 4t D t.t " 4/ D 0 or 9 " t 2 D 0, that is t D 0; 4; ˙3. These are the
points c.0/ D .0; 9/, c.4/ D .0; "7/, c.3/ D ."3; 0/, c."3/ D .21; 0/. These properties lead to the following path:
y
t = 0 (0, 9)
t = 2, (−4, 5)
(21, 0)
x
t = −3
t = 3, (−3, 0)
t = 4 (0, −7)
t = −4
(32, −7)
In Exercises 49–52, use Eq. (7) to find dy=dx at the given point.
49. .t 3 ; t 2 " 1/,
SOLUTION
t D "4
By Eq. (7) we have
0
dy
y 0 .t/
.t 2 " 1/
2t
2
D 0
D
D 2 D
0
dx
x .t/
3t
3t
.t 3 /
Substituting t D "4 we get
50. .2t C 9; 7t " 9/,
SOLUTION
We find
t D1
dy
:
dx
51. .s !1 " 3s; s 3 /, s D "1
SOLUTION
ˇ
dy
2ˇ
2
1
D ˇˇ
D
D" :
dx
3t t D!4
3 ! ."4/
6
ˇ
dy
.7t " 9/0
7
dy ˇˇ
7
D
D
)
D :
dx
2
dx ˇt D1
2
.2t C 9/0
Using Eq. (7) we get
0
dy
y 0 .s/
.s 3 /
3s 2
3s 4
D 0
D
D
D
0
!2
!1
dx
x .s/
"s " 3
"1 " 3s 2
.s " 3s/
Substituting s D "1 we obtain
52. .sin 2#; cos 3#/,
SOLUTION
#D
!
6
ˇ
dy
3s 4 ˇˇ
3 ! ."1/4
3
D
D
D" :
2
dx
4
"1 " 3s 2 ˇsD!1
"1 " 3 ! ."1/
Using Eq. (7) we get
dy
y 0 .#/
"3 sin 3#
D 0
D
dx
x .#/
2 cos 2#
Substituting # D
!
we get
6
ˇ
"3 sin !2
dy
"3 sin 3# ˇˇ
"3
D
D
D
D "3
ˇ
dx
2 cos 2# " D!=6
2 cos !3
2 ! 12
1358
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
In Exercises 53–56, find an equation y D f .x/ for the parametric curve and compute dy=dx in two ways: using Eq. (7) and by
differentiating f .x/.
53. c.t/ D .2t C 1; 1 " 9t/
Since x D 2t C 1, we have t D
SOLUTION
x "1
. Substituting in y D 1 " 9t we have
2
#
$
x"1
9
11
y D1"9
D" xC
2
2
2
9
11
dy
9
dy
Differentiating y D " x C
gives
D " . We now find
using Eq. (7):
2
2
dx
2
dx
54. c.t/ D
SOLUTION
%1
1 2
2 t; 4 t
"t
dy
y 0 .t/
.1 " 9t/0
9
D 0
D
D"
dx
x .t/
2
.2t C 1/0
&
Since x D 21 t we have t D 2x. Substituting in y D
yD
1 2
4t
" t yields
1
.2x/2 " 2x D x 2 " 2x:
4
We differentiate y D x 2 " 2x:
dy
D 2x " 2
dx
Now, we find
dy
dx
using Eq. (7). Thus,
y 0 .t/
dy
D 0
D
dx
x .t/
Since t D 2x, then this t " 2 is the same as 2x " 2.
55. x D s 3 , y D s 6 C s !3
SOLUTION We find y as a function of x:
!
1 2
4t
"t
! "0
1
2t
"0
D
1
2t
"1
1
2
D t " 2:
! "2 ! "!1
y D s 6 C s !3 D s 3 C s 3
D x 2 C x !1 :
We now differentiate y D x 2 C x !1 . This gives
dy
D 2x " x !2 :
dx
Alternatively, we can use Eq. (7) to obtain the following derivative:
% 6
&0
s C s !3
dy
y 0 .s/
6s 5 " 3s !4
D 0
D
D
D 2s 3 " s !6 :
% &0
dx
x .s/
3s 2
s3
Hence, since x D s 3 ,
dy
D 2x " x !2 :
dx
56. x D cos #, y D cos # C sin2 #
SOLUTION To find y as a function of x, we first use the trigonometric identity sin2 # D 1 " cos2 # to write
y D cos # C 1 " cos2 #:
We substitute x D cos # to obtain y D x C 1 " x 2 . Differentiating this function yields
dy
D 1 " 2x:
dx
Alternatively, we can compute
dy
dx
using Eq. (7). That is,
%
&0
cos # C sin2 #
" sin # C 2 sin # cos #
dy
y 0 .#/
D
D 0
D
D 1 " 2 cos #:
dx
x .#/
" sin #
.cos #/0
Hence, since x D cos #,
dy
D 1 " 2x:
dx
Parametric Equations
S E C T I O N 11.1
1359
57. Find the points on the curve c.t/ D .3t 2 " 2t; t 3 " 6t/ where the tangent line has slope 3.
SOLUTION
We solve
dy
3t 2 " 6
D
D3
dx
6t " 2
or 3t 2 " 6 D 18t " 6, or t 2 " 6t D 0, so the slope is 3 at t D 0; 6 and the points are .0; 0/ and .96; 180/
!
2.
58. Find the equation of the tangent line to the cycloid generated by a circle of radius 4 at t D
SOLUTION
The cycloid generated by a circle of radius 4 can be parametrized by
c.t/ D .4t " 4 sin t; 4 " 4 cos t/
Then we compute
ˇ
ˇ
4 sin t ˇˇ
dy ˇˇ
4
D
D D1
dx ˇt D!=2
4 " 4 cos t ˇt D!=2
4
so that the slope of the tangent line is 1 and the equation of the tangent line is
!
!
! !
!"
! ""
y " 4 " 4 cos
D 1 ! x " 4 ! " 4 sin
2
2
2
or
y D x C 8 " 2!
In Exercises 59–62, let c.t/ D .t 2 " 9; t 2 " 8t/ (see Figure 4).
y
60
40
20
20
x
60
40
FIGURE 4 Plot of c.t/ D .t 2 " 9; t 2 " 8t/.
59. Draw an arrow indicating the direction of motion, and determine the interval of t-values corresponding to the portion of the
curve in each of the four quadrants.
SOLUTION
We plot the functions x.t/ D t 2 " 9 and y.t/ D t 2 " 8t :
y
x
3
−3
t
−3 −2 −1
x D t2 " 9
1 2 3 4 5 6 7 8 9
t
y D t 2 " 8t
We note carefully where each of these graphs are positive or negative, increasing or decreasing. In particular, x.t/ is decreasing for
t < 0, increasing for t > 0, positive for jtj > 3, and negative for jtj < 3. Likewise, y.t/ is decreasing for t < 4, increasing for
t > 4, positive for t > 8 or t < 0, and negative for 0 < t < 8. We now draw arrows on the path following the decreasing/increasing
behavior of the coordinates as indicated above. We obtain:
y
60
40
t = −3 (0,33)
20
t=0
(−9,0)
−20
t=3
(0,−15)
−20
t=8
(55,0)
20
t = 4 (7,−16)
40
60
x
1360
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
This plot also shows that:
#
#
#
#
The graph is in the first quadrant for t < "3 or t > 8.
The graph is in the second quadrant for "3 < t < 0.
The graph is in the third quadrant for 0 < t < 3.
The graph is in the fourth quadrant for 3 < t < 8.
60. Find the equation of the tangent line at t D 4.
SOLUTION
Using the formula for the slope m of the tangent line we have:
%2
&0 ˇ
ˇ
t " 8t ˇ
dy ˇˇ
2t " 8
4
mD
D %
D
jt D4 D 1 " jt D4 D 0:
&0 ˇˇ
2
dx ˇt D4
2t
t
t " 9 t D4
Since the slope is zero, the tangent line is horizontal. The y-coordinate corresponding to t D 4 is y D 42 " 8 ! 4 D "16. Hence the
equation of the tangent line is y D "16.
61. Find the points where the tangent has slope 12 .
SOLUTION
The slope of the tangent at t is
The point where the tangent has slope
1
2
%2
&0
t " 8t
dy
2t " 8
4
D %
D1"
&0 D
2
dx
2t
t
t "9
corresponds to the value of t that satisfies
dy
4
1
4
1
D 1 " D ) D ) t D 8:
dx
t
2
t
2
We substitute t D 8 in x.t/ D t 2 " 9 and y.t/ D t 2 " 8t to obtain the following point:
x.8/ D 82 " 9 D 55
)
y.8/ D 82 " 8 ! 8 D 0
.55; 0/
62. Find the points where the tangent is horizontal or vertical.
SOLUTION
In Exercise 61 we found that the slope of the tangent at t is
dy
4
t "4
D1" D
dx
t
t
The tangent is horizontal where its slope is zero. We set the slope equal to zero and solve for t. This gives
t "4
D 0 ) t D 4:
t
The corresponding point is
.x.4/; y.4// D .42 " 9; 42 " 8 ! 4/ D .7; "16/:
The tangent is vertical where it has infinite slope; that is, at t D 0. The corresponding point is
.x.0/; y.0// D .02 " 9; 02 " 8 ! 0/ D ."9; 0/:
63. Let A and B be the points where the ray of angle # intersects the two concentric circles of radii r < R centered at the
origin (Figure 5). Let P be the point of intersection of the horizontal line through A and the vertical line through B. Express the
coordinates of P as a function of # and describe the curve traced by P for 0 # # # 2!.
y
B
A
P
r
FIGURE 5
R
x
S E C T I O N 11.1
SOLUTION
Parametric Equations
1361
We use the parametric representation of a circle to determine the coordinates of the points A and B. That is,
A D .r cos #; r sin #/; B D .R cos #; R sin #/
The coordinates of P are therefore
P D .R cos #; r sin #/
In order to identify the curve traced by P , we notice that the x and y coordinates of P satisfy
! x "2
R
The equation
C
! y "2
D cos2 # C sin2 # D 1:
! x "2
C
r
R
! y "2
r
x
R
D cos # and
y
r
D sin #. Hence
D1
is the equation of ellipse. Hence, the coordinates of P , .R cos #; r sin #/ describe an ellipse for 0 # # # 2!.
64. A 10-ft ladder slides down a wall as its bottom B is pulled away from the wall (Figure 6). Using the angle # as parameter, find
the parametric equations for the path followed by (a) the top of the ladder A, (b) the bottom of the ladder B, and (c) the point P
located 4 ft from the top of the ladder. Show that P describes an ellipse.
y
A
4
P = (x, y)
6
x
B
FIGURE 6
SOLUTION
(a) We define the xy-coordinate system as shown in the figure:
y
A
10
0
B
x
As the ladder slides down the wall, the x-coordinate of A is always zero and the y-coordinate is y D 10 sin #. The parametric
equations for the path followed by A are thus
x D 0; y D 10 sin #;
# is between
The path described by A is the segment Œ0; 10" on the y-axis.
y
10
0
x
!
2
and 0.
1362
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
(b) As the ladder slides down the wall, the y-coordinate of B is always zero and the x-coordinate is x D 10 cos #. The parametric
equations for the path followed by B are therefore
x D 10 cos #; y D 0;
# is between
!
2
and 0.
The path is the segment Œ0; 10" on the x-axis.
y
0
10
x
(c) The x and y coordinates of P are x D 4 cos #, y D 6 sin #. The path followed by P has the following parametrization:
c.#/ D .4 cos #; 6 sin #/;
# is between
!
2
and 0.
y
4
x
P(x, y)
y
6
x
0
As shown in Example 4, the corresponding path is a part of an ellipse. Since # is varying between
the ellipse in the first quadrant.
!
2
and 0, we obtain the part of
y
6
0
4
x
In Exercises 65–68, refer to the Bézier curve defined by Eqs. (8) and (9).
65. Show that the Bézier curve with control points
P0 D .1; 4/;
P1 D .3; 12/;
P2 D .6; 15/;
P3 D .7; 4/
has parametrization
c.t/ D .1 C 6t C 3t 2 " 3t 3 ; 4 C 24t " 15t 2 " 9t 3 /
Verify that the slope at t D 0 is equal to the slope of the segment P0 P1 .
SOLUTION For the given Bézier curve we have a0 D 1, a1 D 3, a2 D 6, a3 D 7, and b0 D 4, b1 D 12, b2 D 15, b3 D 4.
Substituting these values in Eq. (8)–(9) and simplifying gives
x.t/ D .1 " t/3 C 9t.1 " t/2 C 18t 2 .1 " t/ C 7t 3
D 1 " 3t C 3t 2 " t 3 C 9t.1 " 2t C t 2 / C 18t 2 " 18t 3 C 7t 3
D 1 " 3t C 3t 2 " t 3 C 9t " 18t 2 C 9t 3 C 18t 2 " 18t 3 C 7t 3
D "3t 3 C 3t 2 C 6t C 1
S E C T I O N 11.1
Parametric Equations
1363
y.t/ D 4.1 " t/3 C 36t.1 " t/2 C 45t 2 .1 " t/ C 4t 3
D 4.1 " 3t C 3t 2 " t 3 / C 36t.1 " 2t C t 2 / C 45t 2 " 45t 3 C 4t 3
D 4 " 12t C 12t 2 " 4t 3 C 36t " 72t 2 C 36t 3 C 45t 2 " 45t 3 C 4t 3
D 4 C 24t " 15t 2 " 9t 3
Then
c.t/ D .1 C 6t C 3t 2 " 3t 3 ; 4 C 24t " 15t 2 " 9t 3 /;
0 # t # 1:
We find the slope at t D 0. Using the formula for slope of the tangent line we get
ˇ
dy
.4 C 24t " 15t 2 " 9t 3 /0
24 " 30t " 27t 2
dy ˇˇ
24
D
D
)
D
D 4:
dx
dx ˇt D0
6
.1 C 6t C 3t 2 " 3t 3 /0
6 C 6t " 9t 2
The slope of the segment P0 P1 is the slope of the line determined by the points P0 D .1; 4/ and P1 D .3; 12/. That is,
12!4
8
3!1 D 2 D 4. We see that the slope of the tangent line at t D 0 is equal to the slope of the segment P0 P1 , as expected.
66. Find an equation of the tangent line to the Bézier curve in Exercise 65 at t D 13 .
SOLUTION We have
dy
y.t/0
24 " 30t " 27t 2
D
D
0
dx
x.t/
6 C 6t " 9t 2
so that at t D 13 ,
ˇ
ˇ
dy ˇˇ
24 " 30t " 27t 2 ˇˇ
11
D
D
dx ˇt D1=3
7
6 C 6t " 9t 2 ˇt D1=3
and
# $
1
29
D
;
3
9
y
#
$
11
29
x"
7
9
or
x
# $
1
D 10
3
Thus the tangent line is
y " 10 D
67.
SOLUTION
11
311
xC
7
63
Find and plot the Bézier curve c.t/ passing through the control points
P0 D .3; 2/;
gives
yD
P1 D .0; 2/;
P2 D .5; 4/;
P3 D .2; 4/
Setting a0 D 3, a1 D 0, a2 D 5, a3 D 2, and b0 D 2, b1 D 2, b2 D 4, b3 D 4 into Eq. (8)–(9) and simplifying
x.t/ D 3.1 " t/3 C 0 C 15t 2 .1 " t/ C 2t 3
D 3.1 " 3t C 3t 2 " t 3 / C 15t 2 " 15t 3 C 2t 3 D 3 " 9t C 24t 2 " 16t 3
y.t/ D 2.1 " t/3 C 6t.1 " t/2 C 12t 2 .1 " t/ C 4t 3
D 2.1 " 3t C 3t 2 " t 3 / C 6t.1 " 2t C t 2 / C 12t 2 " 12t 3 C 4t 3
D 2 " 6t C 6t 2 " 2t 3 C 6t " 12t 2 C 6t 3 C 12t 2 " 12t 3 C 4t 3 D 2 C 6t 2 " 4t 3
We obtain the following equation
c.t/ D .3 " 9t C 24t 2 " 16t 3 ; 2 C 6t 2 " 4t 3 /;
The graph of the Bézier curve is shown in the following figure:
y
4
3
2
1
1
2
3
x
0 # t # 1:
1364
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
68. Show that a cubic Bézier curve is tangent to the segment P2 P3 at P3 .
SOLUTION
The equations of the cubic Bézier curve are
x.t/ D a0 .1 " t/3 C 3a1 t.1 " t/2 C 3a2 t 2 .1 " t/ C a3 t 3
y.t/ D b0 .1 " t/3 C 3b1 t.1 " t/2 C 3b2 t 2 .1 " t/ C b3 t 3
We use the formula for the slope of the tangent line to find the slope of the tangent line at P3 . We obtain
dy
y 0 .t/
"3b0 .1 " t/2 C 3b1 ..1 " t/2 " 2t.1 " t// C 3b2 .2t.1 " t/ " t 2 / C 3b3 t 2
D 0
D
dx
x .t/
"3a0 .1 " t/2 C 3a1 ..1 " t/2 " 2t.1 " t// C 3a2 .2t.1 " t/ " t 2 / C 3a3 t 2
(1)
The slope of the tangent line at P3 is obtained by setting t D 1 in (1). That is,
0 C 0 " 3b2 C 3b3
b3 " b2
D
0 C 0 " 3a2 C 3a3
a3 " a2
m1 D
(2)
We compute the slope of the segment P2 P3 for P2 D .a2 ; b2 / and P3 D .a3 ; b3 /. We get
m2 D
b3 " b2
a3 " a2
Since the two slopes are equal, we conclude that the tangent line to the curve at the point P3 is the segment P2 P3 .
69. A bullet fired from a gun follows the trajectory
y D bt " 16t 2 .a; b > 0/
% &
Show that the bullet leaves the gun at an angle # D tan!1 ba and lands at a distance ab=16 from the origin.
x D at;
SOLUTION The height of the bullet equals the value of the y-coordinate. When the bullet leaves the gun, y.t/ D
b
t.b " 16t/ D 0. The solutions to this equation are t D 0 and t D 16
, with t D 0 corresponding to the moment the bullet
leaves the gun. We find the slope m of the tangent line at t D 0:
ˇ
dy
y 0 .t/
b " 32t
b " 32t ˇˇ
b
D 0
D
)mD
D
ˇ
dx
x .t/
a
a
a
t D0
! "
b
It follows that tan # D ba or # D tan!1 ba . The bullet lands at t D 16
. We find the distance of the bullet from the origin at this
time, by substituting t D
b
16
in x.t/ D at. This gives
x
70.
SOLUTION
#
b
16
$
D
ab
16
Plot c.t/ D .t 3 " 4t; t 4 " 12t 2 C 48/ for "3 # t # 3. Find the points where the tangent line is horizontal or vertical.
The graph of c.t/ D .t 3 " 4t; t 4 " 12t 2 C 48/, "3 # t # 3 is shown in the following figure:
y
60
t=0
50 (0, 48)
t = 1.15 40
(−3.1, 33.8)
30
t = −3
(−15, 21) 20
t = −2.45 10
(−4.9, 12)
−15 −10 −5
t = −1.15
(3.1, 33.8)
t = 3, (15, 21)
t = 2.45
(4.9, 12)
5
x
10
15
We find the slope of the tangent line at t:
0
.t 4 " 12t 2 C 48/
4t 3 " 24t
dy
y 0 .t/
D
D
D 0
0
dx
x .x/
3t 2 " 4
.t 3 " 4t/
The tangent line is horizontal where
dy
dx
D 0. That is,
p
p
4t 3 " 24t
D 0 ) 4t.t 2 " 6/ D 0 ) t D 0; t D " 6; t D 6:
2
3t " 4
(1)
S E C T I O N 11.1
Parametric Equations
1365
We find the corresponding points by substituting these values of t in c.t/. We obtain:
p
p
c.0/ D .0; 48/; c." 6/ % ."4:9; 12/; c. 6/ % .4:9; 12/:
The tangent line is vertical where the slope in (1) is infinite, that is, where 3t 2 " 4 D 0 or t D ˙ p2 % ˙1:15. We find the points
3
by setting t D ˙ p2 in c.t/. We get
3
c
#
2
p
3
$
#
$
2
c "p
% .3:1; 33:8/:
3
% ."3:1; 33:8/;
Plot the astroid x D cos3 #, y D sin3 # and find the equation of the tangent line at # D
71.
SOLUTION
!
3.
The graph of the astroid x D cos3 #, y D sin3 # is shown in the following figure:
y
=
π
(0, 1)
2
=0
(1, 0)
x
=π
(−1, 0)
=
The slope of the tangent line at # D
!
3
3π
(0, −1)
2
is
ˇ
ˇ
ˇ
ˇ
ˇ
p
ˇ
dy ˇˇ
.sin3 #/0 ˇˇ
3 sin2 # cos # ˇˇ
sin # ˇˇ
ˇ
mD
D
D
D
"
D
"
tan
#
D" 3
ˇ
ˇ
ˇ
ˇ
ˇ
3
0
2
dx " D!=3
cos # " D!=3
.cos #/ " D!=3
3 cos #." sin #/ " D!=3
!=3
We find the point of tangency:
! ! ! " ! ! "" !
!
!"
x
;y
D cos3 ; sin3
D
3
3
3
3
The equation of the tangent line at # D
!
3
is, thus,
p
p
#
$
p
p
3 3
1
3
y"
D" 3 x"
) y D " 3x C
8
8
2
72. Find the equation of the tangent line at t D
SOLUTION
dy
tive dx
:
p !
1 3 3
;
8 8
!
4
to the cycloid generated by the unit circle with parametric equation (5).
We find the equation of the tangent line at t D
!
4
to the cycloid x D t " sin t , y D 1 " cos t. We first find the deriva-
dy
y 0 .t/
.1 " cos t/0
sin t
D 0
D
D
dx
x .t/
1 " cos t
.t " sin t/0
The slope of the tangent line at t D
We find the point of tangency:
!
4
is therefore:
ˇ
sin !4
dy ˇˇ
mD
D
ˇ
dx t D!=4
1 " cos
p
!
4
D
2
2
p
1"
2
2
! ! ! " ! ! "" ! !
!
!"
x
;y
D
" sin ; 1 " cos
D
4
4
4
4
4
The equation of the tangent line is, thus,
p !
2
1
y" 1"
D p
x"
2
2"1
D p
1
2"1
p
p !
!
2
2
"
;1 "
4
2
2
p !!
#
$
!
!
2
1
"
)yD p
xC 2" p 4
4
2
2"1
2"1
73. Find the points with horizontal tangent line on the cycloid with parametric equation (5).
1366
C H A P T E R 11
SOLUTION
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
The parametric equations of the cycloid are
x D t " sin t;
y D 1 " cos t
We find the slope of the tangent line at t:
dy
.1 " cos t/0
sin t
D
0 D 1 " cos t
dx
.t " sin t/
The tangent line is horizontal where it has slope zero. That is,
dy
sin t
D
D0
dx
1 " cos t
sin t D 0
)
)
cos t ¤ 1
t D .2k " 1/!;
k D 0; ˙1; ˙2; : : :
We find the coordinates of the points with horizontal tangent line, by substituting t D .2k " 1/! in x.t/ and y.t/. This gives
x D .2k " 1/! " sin.2k " 1/! D .2k " 1/!
y D 1 " cos..2k " 1/!/ D 1 " ."1/ D 2
The required points are
..2k " 1/!; 2/;
k D 0; ˙1; ˙2; : : :
74. Property of the Cycloid Prove that the tangent line at a point P on the cycloid always passes through the top point on the
rolling circle as indicated in Figure 7. Assume the generating circle of the cycloid has radius 1.
y
Tangent line
Cycloid
x
FIGURE 7
SOLUTION The definition of the cycloid is such that at time t, the top of the circle has coordinates Q D .t; 2/ (since at time
t D 2! the circle has rotated exactly once, and its circumference is 2!). Let L be the line through P and Q. To show that L is
tangent to the cycloid at P it suffices to show that the slope of L equals the slope of the tangent at P . Recall that the cycloid is
parametrized by c.t/ D .t " sin t; 1 " cos t/. Then the slope of L is
2 " .1 " cos t/
1 C cos t
D
t " .t " sin t/
sin t
and the slope of the tangent line is
y 0 .t/
.1 " cos t/0
sin t
sin t.1 C cos t/
sin t.1 C cos t/
1 C cos t
D
D
D
D
D
0
x .t/
.t " sin t/0
1 " cos t
sin t
1 " cos2 t
sin2 t
and the two are equal.
75. A curtate cycloid (Figure 8) is the curve traced by a point at a distance h from the center of a circle of radius R rolling along
the x-axis where h < R. Show that this curve has parametric equations x D Rt " h sin t, y D R " h cos t.
y
h
R
2π
4π
x
FIGURE 8 Curtate cycloid.
SOLUTION Let P be a point at a distance h from the center C of the circle. Assume that at t D 0, the line of CP is passing
c (see figure) is also Rt and the
through the origin. When the circle rolls a distance Rt along the x-axis, the length of the arc SQ
angle †SCQ has radian measure t. We compute the coordinates x and y of P .
S
P
A
h
C
C
t
R
0
Rt
Q
S E C T I O N 11.1
Parametric Equations
1367
x D Rt " PA D Rt " h sin.! " t/ D Rt " h sin t
y D R C AC D R C h cos.! " t/ D R " h cos t
We obtain the following parametrization:
x D Rt " h sin t; y D R " h cos t:
76.
Use a computer algebra system to explore what happens when h > R in the parametric equations of Exercise 75.
Describe the result.
SOLUTION Look first at the parametric equations x D "h sin t, y D "h cos t. These describe a circle of radius h. See for instance
the graphs below obtained for h D 3 and h D 5.
y
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
c(t) = (−h*sin(t), −h*cos(t)) h = 3, 5
Adding R to the y coordinate to obtain the parametric equations x D "h sin t, y D R " h cos t, yields a circle with its center
moved up by R units:
y
10
8
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
c(t) = (−h*sin(t), R−h*cos(t)) R = 1, 5 h = 5
Now, we add Rt to the x coordinate to obtain the given parametric equation; the curve becomes a spring. The figure below shows
the graphs obtained for R D 1 and various values of h. We see the inner loop formed for h > R.
y
8
6
4
2
x
−10 −8 −6 −4 −2
−2
2
4
6
8 10
−4
−6
77.
Show that the line of slope t through ."1; 0/ intersects the unit circle in the point with coordinates
xD
1 " t2
;
t2 C 1
yD
2t
t2 C 1
10
Conclude that these equations parametrize the unit circle with the point ."1; 0/ excluded (Figure 9). Show further that t D
y=.x C 1/.
1368
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
(x, y)
Slope t
(−1, 0)
x
FIGURE 9 Unit circle.
SOLUTION The equation of the line of slope t through ."1; 0/ is y D t.x C 1/. The equation of the unit circle is x 2 C y 2 D 1.
Hence, the line intersects the unit circle at the points .x; y/ that satisfy the equations:
y D t.x C 1/
2
2
(1)
(2)
x Cy D1
Substituting y from equation (1) into equation (2) and solving for x we obtain
x 2 C t 2 .x C 1/2 D 1
x 2 C t 2 x 2 C 2t 2 x C t 2 D 1
.1 C t 2 /x 2 C 2t 2 x C .t 2 " 1/ D 0
This gives
x1;2 D
"2t 2 ˙
p
4t 4 " 4.t 2 C 1/.t 2 " 1/
"2t 2 ˙ 2
˙1 " t 2
D
D
2
2
2.1 C t /
2.1 C t /
1 C t2
1 " t2
. The solution x D "1 corresponds to the point ."1; 0/. We are interested in the second point of
t2 C 1
intersection that is varying as t varies. Hence the appropriate solution is
So x1 D "1 and x2 D
xD
1 " t2
t2 C 1
We find the y-coordinate by substituting x in equation (1). This gives
!
1 " t2
1 " t2 C t2 C 1
2t
y D t.x C 1/ D t 2
C1 Dt !
D 2
t C1
t2 C 1
t C1
We conclude that the line and the unit circle intersect, besides at ."1; 0/, at the point with the following coordinates:
1 " t2
2t
; yD 2
(3)
t2 C 1
t C1
Since these points determine all the points on the unit circle except for ."1; 0/ and no other points, the equations in (3) parametrize
the unit circle with the point ."1; 0/ excluded.
y
We show that t D
. Using (3) we have
xC1
xD
y
D
xC1
2t
t 2 C1
1!t 2
C1
t 2 C1
2t
2t
2t
t 2 C1
t 2 C1
D
D
D t:
2
1!t 2 Ct 2 C1
2
2 C1
2
t
t C1
x 3 C y 3 D 3axy, where a ¤ 0 is a constant
D
78. The folium of Descartes is the curve with equation
(Figure 10).
(a) Show that the line y D tx intersects the folium at the origin and at one other point P for all t ¤ "1; 0. Express the coordinates
of P in terms of t to obtain a parametrization of the folium. Indicate the direction of the parametrization on the graph.
(b) Describe the interval of t-values parametrizing the parts of the curve in quadrants I, II, and IV. Note that t D "1 is a point of
discontinuity of the parametrization.
(c) Calculate dy=dx as a function of t and find the points with horizontal or vertical tangent.
y
II
I
2
2
−2
III
−2
FIGURE 10 Folium
x
IV
x3
C y 3 D 3axy.
S E C T I O N 11.1
Parametric Equations
1369
SOLUTION
(a) We find the points where the line y D tx (t ¤ "1; 0) and the folium intersect, by solving the following equations:
(1)
y D tx
3
3
(2)
x C y D 3axy
Substituting y from (1) in (2) and solving for x we get
x 3 C t 3 x 3 D 3axtx
.1 C t 3 /x 3 " 3atx 2 D 0
x 2 .x.1 C t 3 / " 3at/ D 0 ) x1 D 0; x2 D
3at
1 C t3
Substituting in (1) we find the corresponding y-coordinates. That is,
y1 D t ! 0 D 0; y2 D t !
3at
3at 2
D
3
1Ct
1 C t3
We conclude that the line y D tx, t ¤ 0; "1 intersects the folium in a unique point P besides the origin. The coordinates of P are:
xD
3at
3at 2
; yD
;
3
1Ct
1 C t3
t ¤ 0; "1
The coordinates of P determine a parametrization for the folium. We add the origin so t D 0 must be included in the interval of t.
We get
!
3at
3at 2
c.t/ D
;
; t ¤ "1
1 C t3 1 C t3
To indicate the direction on the curve (for a > 0), we first consider the following limits:
lim x.t/ D 1
lim y.t/ D "1
t !!1!
t !!1!
lim x.t/ D lim x.t/ D 0
t !!1
lim
t !!1C
lim y.t/ D lim y.t/ D 0
t !1
t !!1
lim
x.t/ D "1
t !!1C
lim x.t/ D 0
t !1
y.t/ D 1
lim y.t/ D 0
t !0
t !0
These limits determine the directions of the two parts of the folium in the second and fourth quadrant. The loop in the first quadrant,
3a
2a 4a
corresponds to the values 0 # t < 1, and it is directed from c.1/ D . 3a
2 ; 2 / to c.2/ D . 3 ; 3 / where t D 1 and t D 2 are two
chosen values in the interval 0 # t < 1. The following graph shows the directed folium:
y
−1
<
t<
t = −1+
0≤t<∞
0
t=0
t=∞
t = −∞
−∞
<
t<
−1
t = −1−
x
(b) The limits computed in part (a) indicate that the parts of the curve in the second and fourth quadrants correspond to the values
"1 < t < 0 and "1 < t < "1 respectively. The loop in the first quadrant corresponds to the remaining interval 0 # t < 1.
(c) We find the derivative
dy
,
dx
using the Formula for the Slope of the Tangent Line. We get
!
"0
3at 2
dy
y 0 .t/
1Ct 3
D 0
D !
"0 D
dx
x .t/
3at
1Ct 3
Horizontal tangent occurs when
dy
dx
6at .1Ct 3 /!3at 2 "3t 2
2
.1Ct 3 /
3
3a.1Ct /!3at "3t 2
2
.1Ct 3 /
D
6at " 3at 4
t.2 " t 3 /
D
3a " 6at 3
1 " 2t 3
D 0. That is,
p
t.2 " t 3 /
3
D 0 ) t.2 " t 3 / D 0; 1 " 2t 3 ¤ 0 ) t D 0; t D 2:
3
1 " 2t
1370
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
The corresponding points are:
c.0/ D .x.0/; y.0// D .0; 0/
! p " ! ! p " ! p ""
3
3
3
c
2 D x
2 ;y
2 D
Vertical tangent line occurs when
dy
dx
p
p ! !
"
p
p
3a 3 2 3a 3 4
3
3
;
D a 2; a 4
1C2 1C2
is infinite. That is,
1
1 " 2t 3 D 0 ) t D p
3
2
The corresponding point is
c
#
1
p
3
2
$
0 3a
1
3a
# #
$ #
$$
p
p
!p
"
3
3
p
1
1
3
3
2
4 A
@
D x p
;
y
D
;
D
4a;
2a
:
p
3
3
1 C 12 1 C 12
2
2
79. Use the results of Exercise 78 to show that the asymptote of the folium is the line x C y D "a. Hint: Show that lim .x C y/ D
t !!1
"a.
SOLUTION We must show that as x ! 1 or x ! "1 the graph of the folium is getting arbitrarily close to the line x C y D "a,
dy
and the derivative dx
is approaching the slope "1 of the line.
In Exercise 78 we showed that x ! 1 when t ! ."1! / and x ! "1 when t ! ."1C /. We first show that the graph is
approaching the line x C y D "a as x ! 1 or x ! "1, by showing that lim x C y D lim x C y D "a.
t !!1!
t !!1C
3at
3at 2
For x.t/ D
, y.t/ D
, a > 0, calculated in Exercise 78, we obtain using L’Hôpital’s Rule:
3
1Ct
1 C t3
lim .x C y/ D
t !!1!
lim .x C y/ D
t !!1C
We now show that
obtain
3at C 3at 2
3a C 6at
3a " 6a
D lim
D
D "a
t !!1!
t !!1!
3
1 C t3
3t 2
lim
3at C 3at 2
3a C 6at
3a " 6a
D lim
D
D "a
t !!1C
t !!1C
3
1 C t3
3t 2
lim
dy
dy
6at " 3at 4
is approaching "1 as t ! "1" and as t ! "1C. We use
D
computed in Exercise 78 to
dx
dx
3a " 6at 3
dy
6at " 3at 4
"9a
D lim
D
D "1
3
t !!1! dx
t !!1! 3a " 6at
9a
lim
dy
6at " 3at 4
"9a
D lim
D
D "1
t !!1C dx
t !!1C 3a " 6at 3
9a
lim
We conclude that the line x C y D "a is an asymptote of the folium as x ! 1 and as x ! "1.
80. Find a parametrization of x 2nC1 C y 2nC1 D ax n y n , where a and n are constants.
SOLUTION Following the method in Exercise 78, we first find the coordinates of the point P where the curve and the line y D tx
intersect. We solve the following equations:
y D tx
x 2nC1 C y 2nC1 D ax n y n
Substituting y D tx in the second equation and solving for x yields
x 2nC1 C t 2nC1 x 2nC1 D ax n ! t n x n
.1 C t 2nC1 /x 2nC1 " at n x 2n D 0
x 2n ..1 C t 2nC1 /x " at n / D 0 ) x D 0; x D
at n
1 C t 2nC1
We assume that t ¤ "1 (so 1 C t 2nC1 ¤ 0) and obtain one solution besides the origin. The corresponding y coordinates are
y D tx D t !
Hence, the points x D
parametrization:
at n
at nC1
D
1 C t 2nC1
1 C t 2nC1
at n
at nC1
, yD
, t ¤ "1, are exactly the points on the curve. We obtain the following
2nC1
1Ct
1 C t 2nC1
xD
at nC1
at n
;
y
D
;
1 C t 2nC1
1 C t 2nC1
t ¤ "1:
S E C T I O N 11.1
Parametric Equations
1371
81. Second Derivative for a Parametrized Curve Given a parametrized curve c.t/ D .x.t/; y.t//, show that
d ! dy "
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
D
dt dx
x 0 .t/2
Use this to prove the formula
d 2y
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
D
2
dx
x 0 .t/3
11
By the formula for the slope of the tangent line we have
SOLUTION
dy
y 0 .t/
D 0
dx
x .t/
Differentiating with respect to t, using the Quotient Rule, gives
# $
#
$
d dy
d y 0 .t/
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
D
D
dt dx
dt x 0 .t/
x 0 .t/2
By the Chain Rule we have
d 2y
d
D
dx
dx 2
#
dy
dx
$
D
d
dt
#
dy
dx
$
!
dt
dx
#
$
dt
1
1
Substituting into the above equation and using
D
D 0
gives
dx
dx=dt
x .t/
d 2y
x 0 .t/y 0 0 .t/ " y 0 .t/x 0 0 .t/
1
x 0 .t/y 0 0 .t/ " y 0 .t/x 0 0 .t/
D
! 0
D
2
2
x .t/
dx
x 0 .t/
x 0 .t/3
82. The second derivative of y D x 2 is dy 2 =d 2 x D 2. Verify that Eq. (11) applied to c.t/ D .t; t 2 / yields dy 2 =d 2 x D 2. In fact,
any parametrization may be used. Check that c.t/ D .t 3 ; t 6 / and c.t/ D .tan t; tan2 t/ also yield dy 2 =d 2 x D 2.
For the parametrization c.t/ D .t; t 2 /, we have
SOLUTION
x 0 .t/ D 1;
x 00 .t/ D 0;
y 0 .t/ D 2t;
y 00 .t/ D 2
so that indeed
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
1 ! 2 " 2t ! 0
D
D2
x 0 .t/3
13
For c.t/ D .t 3 ; t 6 /, we have
x 0 .t/ D 3t 2 ;
x 00 .t/ D 6t;
y 0 .t/ D 6t 5 ;
y 00 .t/ D 30t 4
so that again
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
3t 2 ! 30t 4 " 6t 5 ! 6t
54t 6
D
D
D2
0
3
2
3
x .t/
.3t /
27t 6
Finally, for c.t/ D .tan t; tan2 t/,
x 0 .t/ D sec2 t;
x 00 .t/ D 2 tan t sec2 t;
y 0 .t/ D 2 tan t sec2 t;
y 00 .t/ D 6 sec4 t " 4 sec2 t
and
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
sec2 t.6 sec4 t " 4 sec2 t/ " 2 tan t sec2 t.2 tan t sec2 t/
D
0
3
x .t/
sec6 t
D
6 sec6 t " 4 sec4 t " 4 sec4 t tan2 t
6 sec6 t " 4 sec4 t.1 " .1 C sec2 t///
D
6
sec t
sec6 t
D
2 sec6 t
D2
sec6 t
1372
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
In Exercises 83–86, use Eq. (11) to find d 2 y=dx 2 .
83. x D t 3 C t 2 ,
SOLUTION
y D 7t 2 " 4,
t D2
We find the first and second derivatives of x.t/ and y.t/:
x 0 .t/ D 3t 2 C 2t ) x 0 .2/ D 3 ! 22 C 2 ! 2 D 16
x 00 .t/ D 6t C 2
y 0 .t/ D 14t
y 00 .t/ D 14
) x 00 .2/ D 6 ! 2 C 2 D 14
) y 0 .2/ D 14 ! 2 D 28
) y 00 .2/ D 14
Using Eq. (11) we get
ˇ
ˇ
d 2 y ˇˇ
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/ ˇˇ
16 ! 14 " 28 ! 14
"21
D
D
D
ˇ
3
512
dx 2 ˇt D2
16
x 0 .t/3
t D2
84. x D s !1 C s, y D 4 " s !2 , s D 1
SOLUTION
Since x 0 .s/ D "s !2 C 1 D 1 "
85. x D 8t C 9,
SOLUTION
y D 1 " 4t,
2
1
0 .1/ D 0. Hence, Eq. (11) cannot be used to compute d y at s D 1.
,
we
have
x
s2
dx 2
t D "3
We compute the first and second derivatives of x.t/ and y.t/:
x 0 .t/ D 8 ) x 0 ."3/ D 8
x 00 .t/ D 0 ) x 00 ."3/ D 0
y 0 .t/ D "4 ) y 0 ."3/ D "4
y 00 .t/ D 0 ) y 00 ."3/ D 0
Using Eq. (11) we get
86. x D cos #, y D sin #,
SOLUTION
ˇ
d 2 y ˇˇ
x 0 ."3/y 00 ."3/ " y 0 ."3/x 00 ."3/
8 ! 0 " ."4/ ! 0
D
D
D0
ˇ
2
dx t D!3
83
x 0 ."3/3
#D
!
4
We find the first and second derivatives of x.#/ and y.#/:
x .#/ D " sin # ) x
Using Eq. (11) we get
!! "
p
2
2
p
! "
2
00
00 !
x .#/ D " cos # ) x
D"
4
2
p
!! "
2
y 0 .#/ D cos # ) y 0
D
4
2
p
!! "
2
y 00 .#/ D " sin # ) y 00
D"
4
2
0
0
4
D"
! p "! p " p ! p "
% & % &
% & % &
ˇ
" 22 " 22 " 22 ! " 22
p
x 0 !4 y 00 !4 " y 0 !4 x 00 !4
d 2 y ˇˇ
D
D
D "2 2
%
%
&&
p
!
"
ˇ
2
3
3
dx " D !
x 0 !4
4
" 22
87. Use Eq. (11) to find the t-intervals on which c.t/ D .t 2 ; t 3 " 4t/ is concave up.
SOLUTION
The curve is concave up where
d 2y
> 0. Thus,
dx 2
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
x 0 .t/3
>0
We compute the first and second derivatives:
x 0 .t/ D 2t;
x 00 .t/ D 2
y 0 .t/ D 3t 2 " 4;
y 00 .t/ D 6t
(1)
S E C T I O N 11.1
Parametric Equations
1373
Substituting in (1) and solving for t gives
12t 2 " .6t 2 " 8/
6t 2 C 8
D
3
8t
8t 3
Since 6t 2 C 8 > 0 for all t, the quotient is positive if 8t 3 > 0. We conclude that the curve is concave up for t > 0.
88. Use Eq. (11) to find the t-intervals on which c.t/ D .t 2 ; t 4 " 4t/ is concave up.
SOLUTION
The curve is concave up where
d 2y
> 0. That is,
dx 2
x 0 .t/y 00 .t/ " y 0 .t/x 00 .t/
x 0 .t/3
>0
(1)
We compute the first and second derivatives:
x 0 .t/ D 2t;
x 00 .t/ D 2
y 0 .t/ D 4t 3 " 4;
y 00 .t/ D 12t 2
Substituting in (1) and solving for t gives
24t 3 " .8t 3 " 8/
16t 3 C 8
1
D
D1C 3
3
3
8t
8t
2t
1
1
> 0, which means 3 > "1, so 2t 3 < "1 (by taking the reciprocal
3
2t
2t
1
1
of both sides), so t < " p
.
Thus,
we
see
that
our
curve
is
concave
up
for
t
<
"
p
and for t > 0.
3
3
2
2
This is clearly positive for t > 0. For t < 0, we want 1 C
Let c.t/ D .x.t/; y.t//, where y.t/ > 0 and x 0 .t/ > 0 (Figure 11). Show that the area
89. Area Under a Parametrized Curve
A under c.t/ for t0 # t # t1 is
AD
Z
t1
t0
y.t/x 0 .t/ dt
12
Hint: Because it is increasing, the function x.t/ has an inverse t D g.x/ and c.t/ is the graph of y D y.g.x//. Apply the
R x.t /
change-of-variables formula to A D x.t 1/ y.g.x// dx.
0
y
c(t)
x(t 0)
x(t 1)
x
FIGURE 11
Let x0 D x.t0 / and x1 D x.t1 /. We are given that x 0 .t/ > 0, hence x D x.t/ is an increasing function of t, so it has
Rx
an inverse function t D g.x/. The area A is given by x01 y.g.x// dx. Recall that y is a function of t and t D g.x/, so the height
SOLUTION
y at any point x is given by y D y.g.x//. We find the new limits of integration. Since x0 D x.t0 / and x1 D x.t1 /, the limits for t
are t0 and t1 , respectively. Also since x 0 .t/ D
AD
Since g.x/ D t, we have A D
Z
t1
t0
dx
,
dt
Z
we have dx D x 0 .t/dt . Performing this substitution gives
x1
x0
y.g.x// dx D
Z
t1
t0
y.g.x//x 0.t/ dt:
y.t/x 0 .t/ dt.
90. Calculate the area under y D x 2 over Œ0; 1" using Eq. (12) with the parametrizations .t 3 ; t 6 / and .t 2 ; t 4 /.
SOLUTION
The area A under y D x 2 on Œ0; 1" is given by the integral
AD
Z
t1
t0
y.t/x 0 .t/ dt
where x.t0 / D 0 and x.t1 / D 1. We first use the parametrization .t 3 ; t 6 /. We have x.t/ D t 3 , y.t/ D t 6 . Hence,
0 D x.t0 / D t03 ) t0 D 0
1374
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
1 D x.t1 / D t13 ) t1 D 1
Also x 0 .t/ D 3t 2 . Substituting these values in Eq. (12) we obtain
AD
Z
1
0
t 6 ! 3t 2 dt D
Z
1
0
3t 8 dt D
ˇ
3 9 ˇˇ1
3
1
t ˇ D D
9 0
9
3
Using the parametrization x.t/ D t 2 , y.t/ D t 4 , we have x 0 .t/ D 2t. We find t0 and t1 :
0 D x.t0 / D t02 ) t0 D 0
1 D x.t1 / D t12 ) t1 D 1 or
t1 D "1:
Equation (12) is valid if x 0 .t/ > 0, that is if t > 0. Hence we choose the positive value, t1 D 1. We now use Eq. (12) to obtain
ˇ
Z 1
Z 1
2 ˇ1
2
1
AD
t 4 ! 2t dt D
2t 5 dt D t 6 ˇˇ D D
6 0
6
3
0
0
Both answers agree, as expected.
91. What does Eq. (12) say if c.t/ D .t; f .t//?
SOLUTION
In the parametrization x.t/ D t, y.t/ D f .t/ we have x 0 .t/ D 1, t0 D x.t0 /, t1 D x.t1 /. Hence Eq. (12) becomes
AD
Z
t1
t0
y.t/x 0 .t/ dt D
Z
x.t1 /
f .t/ dt
x.t0 /
We see that in this parametrization Eq. (12) is the familiar formula for the area under the graph of a positive function.
92. Sketch the graph of c.t/ D .ln t; 2 " t/ for 1 # t # 2 and compute the area under the graph using Eq. (12).
SOLUTION
We use the following graphs of x.t/ D ln t and y.t/ D 2 " t for 1 # t # 2:
y
y
1
1
t
1
t
2
1
x.t/ D ln t
2
y.t/ D 2 " t
We see that for 1 < t < 2, x.t/ is positive and increasing and y.t/ is positive and decreasing. Also c.1/ D .ln 1; 2 " 1/ D
.0; 1/ and c.2/ D .ln 2; 2 " 2/ D .ln 2; 0/. Additional information is obtained from the derivative
dy
.2 " t/0
1
D
D"
D "t;
dx
1=t
.ln t/0
yielding
We obtain the following graph:
ˇ
dy ˇˇ
" 1 and
dx ˇt D1
ˇ
dy ˇˇ
" 2:
dx ˇt D2
y
t=1
(0, 1)
t=2
(ln 2, 0)
t
1
S E C T I O N 11.1
Parametric Equations
1375
We now use Eq. (12) to compute the area A under the graph. We have x.t/ D ln t, x 0 .t/ D 1t , y.t/ D 2 " t, t0 D 1, t1 D 2. Hence,
$
Z t1
Z 2
Z 2#
1
2
AD
y.t/x 0 .t/ dt D
.2 " t/ ! dt D
" 1 dt
t
t
t0
1
1
ˇ2
ˇ
D 2 ln t " t ˇˇ D .2 ln 2 " 2/ " .2 ln 1 " 1/ D 2 ln 2 " 1 % 0:386
1
93. Galileo tried unsuccessfully to find the area under a cycloid. Around 1630, Gilles de Roberval proved that the area under one
arch of the cycloid c.t/ D .Rt " R sin t; R " R cos t/ generated by a circle of radius R is equal to three times the area of the circle
(Figure 12). Verify Roberval’s result using Eq. (12).
y
R
2πR
πR
x
FIGURE 12 The area of one arch of the cycloid equals three times the area of the generating circle.
SOLUTION
This reduces to
Z
2!
0
.R " R cos t/.Rt " R sin t/0 dt D
Z
2!
0
R2 .1 " cos t/2 dt D 3!R2 :
Further Insights and Challenges
94. Prove the following generalization of Exercise 93: For all t > 0, the area of the cycloidal sector OP C is equal to three times
the area of the circular segment cut by the chord P C in Figure 13.
y
y
P
O
t
P
t
R
C = (Rt, 0)
x
O
(A) Cycloidal sector OPC
R
x
C = (Rt, 0)
(B) Circular segment cut
by the chord PC
FIGURE 13
SOLUTION Drop a perpendicular from point P to the x-axis and label the point of intersection T , and denote by D the center
of the circle. Then the area of the cycloidal sector is equal to the area of OP T plus the area of P T C . The latter is a triangle with
1
height y.t/ D R " R cos t and base Rt " .Rt " R sin t/ D R sin t, so its area is R2 sin t.1 " cos t/. The area of OP T , using
2
Eq. (12), is
Z t
Z t
Z t
y.u/x 0 .u/ du D
.R " R cos u/.Ru " R sin u/0 du D R2
.1 " cos u/2 du
0
0
D R2
#
3
1
t " 2 sin t C sin t cos t
2
2
$
0
so that the total area of the cycloidal sector is
#
$
#
$
3
1
1
1 2
1
1
R2
t " 2 sin t C sin t cos t C R2 sin t.1 " cos t/ D 3
R t " R2 sin t D 3 ! R2 .t " sin t/
2
2
2
2
2
2
The area of the circular segment is the area of the circular sector DP C subtended by the angle t less the area of the triangle DP C .
t
t
t
1
The triangle DP C has height R cos and base 2R sin 2t so that its area is R2 cos sin D R2 sin t, and the area of the circular
2
2
2
2
t
1
sector is !R2 !
D R2 t. Thus the area of the circular segment is
2!
2
1 2
R .t " sin t/
2
which is one third the area of the cycloidal sector.
1376
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
95.
Derive the formula for the slope of the tangent line to a parametric curve c.t/ D .x.t/; y.t// using a method different
from that presented in the text. Assume that x 0 .t0 / and y 0 .t0 / exist and that x 0 .t0 / ¤ 0. Show that
y.t0 C h/ " y.t0 /
y 0 .t0 /
D 0
x .t0 /
h!0 x.t0 C h/ " x.t0 /
lim
Then explain why this limit is equal to the slope dy=dx. Draw a diagram showing that the ratio in the limit is the slope of a secant
line.
SOLUTION
Since y 0 .t0 / and x 0 .t0 / exist, we have the following limits:
y.t0 C h/ " y.t0 /
x.t0 C h/ " x.t0 /
D y 0 .t0 /; lim
D x 0 .t0 /
h
h
h!0
h!0
lim
(1)
We use Basic Limit Laws, the limits in (1) and the given data x 0 .t0 / ¤ 0, to write
y.t0 C h/ " y.t0 /
D lim
h!0 x.t0 C h/ " x.t0 /
h!0
lim
y.t0 Ch/!y.t0 /
h
x.t0 Ch/!x.t0 /
h
D
y.t0 Ch/!y.t0 /
h
0/
limh!0 x.t0 Ch/!x.t
h
limh!0
D
y 0 .t0 /
x 0 .t0 /
y.t0 C h/ " y.t0 /
is the slope of the secant line determined by the points P D .x.t0 /; y.t0 // and
x.t0 C h/ " x.t0 /
Q D .x.t0 C h/; y.t0 C h//. Hence, the limit of the quotient as h ! 0 is the slope of the tangent line at P , that is the derivative
dy
.
dx
Notice that the quotient
y
Q
y(t0, h)
P
y(t0 )
x(t0 )
x(t0 + h)
x
96. Verify that the tractrix curve (` > 0)
#
$
t
t
c.t/ D t " ` tanh ; ` sech
`
`
has the following property: For all t, the segment from c.t/ to .t; 0/ is tangent to the curve and has length ` (Figure 14).
y
c(t)
t
x
#
$
t
t
FIGURE 14 The tractrix c.t/ D t " ` tanh ; ` sech
.
`
`
SOLUTION
Let P D c.t/ and Q D .t; 0/.
y
P(x(t), y(t))
Q = (t, 0)
x
The slope of the segment PQ is
% &
` sech `t
1
y.t/ " 0
%t & D "
% &
m1 D
D
x.t/ " t
"` tanh `
sinh `t
S E C T I O N 11.1
Parametric Equations
1377
We compute the slope of the tangent line at P :
%
% &&0
%
% &
% &&
` sech `t
` ! 1` " sech `t tanh `t
dy
y 0 .t/
D 0
D %
D
%
&
% &&0
dx
x .t/
1 " ` ! 1` sech2 `t
t " ` tanh `t
% &
% &
% &
% &
% &
" sech `t tanh `t
" sech `t tanh `t
" sech `t
1
%t & D "
% &
D"
%t &
D
%t &
D
2
2
tanh `
sinh `t
1 " sech `
tanh `
m2 D
Since m1 D m2 , we conclude that the segment from c.t/ to .t; 0/ is tangent to the curve.
We now show that jPQj D `:
s
#
$
#
# $$2
q
t 2
t
2
2
jPQj D .x.t/ " t/ C .y.t/ " 0/ D
"` tanh
C ` sech
`
`
s #
s
# $
# $$
# $
# $
# $
t
t
t
t
t
2
2
2
D ` tanh
C sech
D ` sech2
sinh2
C sech2
`
`
`
`
`
# $s
# $
# $
# $
t
t
t
t
D ` sech
sinh2
C 1 D ` sech
cosh
D`!1D`
`
`
`
`
97. In Exercise 58 of Section 9.1, we described the tractrix by the differential equation
dy
y
D "p
2
dx
` " y2
Show that the curve c.t/ identified as the tractrix in Exercise 96 satisfies this differential equation. Note that the derivative on the
left is taken with respect to x, not t.
SOLUTION
Note that dx=dt D 1 " sech2 .t=`/ D tanh2 .t=`/ and dy=dt D " sech.t=`/ tanh.t=`/. Thus,
Multiplying top and bottom by `=` gives
dy
dy=dt
" sech.t=`/
"y=`
D
D
D p
dx
dx=dt
tanh.t=`/
1 " y 2 =`2
dy
"y
D p
dx
`2 " y 2
In Exercises 98 and 99, refer to Figure 15.
98. In the parametrization c.t/ D .a cos t; b sin t/ of an ellipse, t is not an angular parameter unless a D b (in which case the
ellipse is a circle). However, t can be interpreted in terms of area: Show that if c.t/ D .x; y/, then t D .2=ab/A, where A is the
area of the shaded region in Figure 15. Hint: Use Eq. (12).
y
(x, y)
x
FIGURE 15 The parameter # on the ellipse
! x "2
a
C
! y "2
b
D 1.
SOLUTION We compute the area A of the shaded region as the sum of the area S1 of the triangle and the area S2 of the region
under the curve. The area of the triangle is
S1 D
xy
.a cos t/.b sin t/
ab sin 2t
D
D
2
2
4
y
(x, y)
S1
S2
x
(1)
1378
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
The area S2 under the curve can be computed using Eq. (12). The lower limit of the integration is t0 D 0 (corresponds to .a; 0/)
and the upper limit is t (corresponds to .x.t/; y.t//). Also y.t/ D b sin t and x 0 .t/ D "a sin t . Since x 0 .t/ < 0 on the interval
0 < t < !2 (which represents the ellipse on the first quadrant), we use the positive value a sin t to obtain a positive value for the
area. This gives
Z t
Z t
S2 D
b sin u ! a sin u du D ab
sin2 u du
0
0
'
(ˇ
1 1
u sin 2u ˇˇt
" cos 2u du D ab
"
ˇ
2 2
2
4
0
0
'
(
t
sin 2t
abt
ab sin 2t
D ab
"
"0 D
"
2
4
2
4
D ab
Z
t
#
$
(2)
Combining (1) and (2) we obtain
A D S1 C S2 D
Hence, t D
ab sin 2t
abt
ab sin 2t
abt
C
"
D
4
2
4
2
2A
.
ab
99. Show that the parametrization of the ellipse by the angle # is
xD p
SOLUTION
yD p
We consider the ellipse
ab cos #
a2 sin2 #
C b 2 cos2 #
ab sin #
a2 sin2 #
C b 2 cos2 #
x2
y2
C 2 D 1:
2
a
b
y
x,
For the angle # we have tan # D
hence,
y D x tan #
(1)
Substituting in the equation of the ellipse and solving for x we obtain
x2
x 2 tan2 #
C
D1
2
a
b2
b 2 x 2 C a2 x 2 tan2 # D a2 b 2
.a2 tan2 # C b 2 /x 2 D a2 b 2
x2 D
a2 b 2
a2 b 2 cos2 #
D
a2 tan2 # C b 2
a2 sin2 # C b 2 cos2 #
We now take the square root. Since the sign of the x-coordinate is the same as the sign of cos #, we take the positive root, obtaining
ab cos #
xD p
2
a sin2 # C b 2 cos2 #
Hence by (1), the y-coordinate is
(2)
ab cos # tan #
ab sin #
y D x tan # D p
D p
a2 sin2 # C b 2 cos2 #
a2 sin2 # C b 2 cos2 #
Equalities (2) and (3) give the following parametrization for the ellipse:
c1 .#/ D
ab sin #
ab cos #
;p
p
2
2
2
2
2
a sin # C b cos #
a sin2 # C b 2 cos2 #
!
(3)
Arc Length and Speed
S E C T I O N 11.2
1379
11.2 Arc Length and Speed
Preliminary Questions
1. What is the definition of arc length?
SOLUTION
A curve can be approximated by a polygonal path obtained by connecting points
p0 D c.t0 /; p1 D c.t1 /; : : : ; pN D c.tN /
on the path with segments. One gets an approximation by summing the lengths of the segments. The definition of arc length is the
limit of that approximation when increasing the number of points so that the lengths of the segments approach zero. In doing so,
we obtain the following theorem for the arc length:
Z
SD
b
a
q
x 0 .t/2 C y 0 .t/2 dt;
which is the length of the curve c.t/ D .x.t/; y.t// for a # t # b.
p
2. What is the interpretation of x 0 .t/2 C y 0 .t/2 for a particle following the trajectory .x.t/; y.t//?
q
SOLUTION The expression
x 0 .t/2 C y 0 .t/2 denotes the speed at time t of a particle following the trajectory .x.t/; y.t//.
3. A particle travels along a path from .0; 0/ to .3; 4/. What is the displacement? Can the distance traveled be determined from
the information given?
SOLUTION
The net displacement is the distance between the initial point .0; 0/ and the endpoint .3; 4/. That is
q
p
.3 " 0/2 C .4 " 0/2 D 25 D 5:
The distance traveled can be determined only if the trajectory c.t/ D .x.t/; y.t// of the particle is known.
4. A particle traverses the parabola y D x 2 with constant speed 3 cm/s. What is the distance traveled during the first minute?
Hint: No computation is necessary.
SOLUTION
Since the speed is constant, the distance traveled is the following product: L D st D 3 ! 60 D 180 cm.
Exercises
In Exercises 1–10, use Eq. (3) to find the length of the path over the given interval.
1. .3t C 1; 9 " 4t/,
SOLUTION
0#t #2
Since x D 3t C 1 and y D 9 " 4t we have x 0 D 3 and y 0 D "4. Hence, the length of the path is
SD
2. .1 C 2t; 2 C 4t/,
SOLUTION
1#t #4
3. .2t 2 ; 3t 2 " 1/,
32 C ."4/2 dt D 5
0
Z
2
dt D 10:
0
1
4p
22 C 42 dt D
Z
4p
1
20 dt D
p
p
20.4 " 1/ D 6 5
Since x D 2t 2 and y D 3t 2 " 1, we have x 0 D 4t and y 0 D 6t . By the formula for the arc length we get
.3t; 4t 3=2 /,
SOLUTION
Z
0#t #4
SD
4.
2q
We have x D 1 C 2t and y D 2 C 4t , hence x 0 D 2 and y 0 D 4. Using the formula for arc length we obtain
SD
SOLUTION
Z
Z
4q
x 0 .t/2 C y 0 .t/2 dt D
0
Z
4p
16t 2 C 36t 2 dt D
0
0#t #1
p Z
52
4
0
t dt D
p
52 !
ˇ
p
t 2 ˇˇ4
D 16 13
ˇ
2 0
We have x D 3t and y D 4t 3=2 , hence x 0 D 3 and y 0 D 6t 1=2 . Using the formula for the arc length we obtain
SD
Z
1q
0
x 0 .t/2 C y 0 .t/2 dt D
Setting u D 1 C 4t we get
SD
3
4
Z
1
Z
5p
0
1q
&2
%
32 C 6t 1=2 dt D
u du D
Z
1p
0
9 C 36t dt D 3
ˇ
1
3 2 3=2 ˇˇ5
! u ˇ D .53=2 " 1/ % 5:09
4 3
2
1
Z
1p
0
1 C 4t dt
1380
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
5. .3t 2 ; 4t 3 /,
SOLUTION
1#t #4
We have x D 3t 2 and y D 4t 3 . Hence x 0 D 6t and y 0 D 12t 2 . By the formula for the arc length we get
SD
Z
4q
x 0 .t/2 C y 0 .t/2 dt D
1
Z
1
Using the substitution u D 1 C 4t 2 , du D 8t dt we obtain
SD
6.
.t 3
C 1; t 2
SOLUTION
" 3/,
6
8
Z
65 p
u du D
5
0#t #1
4p
36t 2 C 144t 4 dt D 6
Z
4p
1 C 4t 2 t dt:
1
ˇ
3 2 3=2 ˇˇ65
1
! u ˇ D .653=2 " 53=2 / % 256:43
4 3
2
5
We have x D t 3 C 1; y D t 2 " 3, hence, x 0 D 3t 2 and y 0 D 2t. By the formula for the arc length we get
Z
SD
1q
x 0 .t/2 C y 0 .t/2 dt D
0
Z
1p
0
9t 4 C 4t 2 dt D
Z
1
t
0
p
9t 2 C 4 dt
We compute the integral using the substitution u D 4 C 9t 2 . This gives
ˇ
Z 13
p
1
1 2 3=2 ˇˇ13
1
1
SD
u du D
! u ˇ D
.133=2 " 43=2 / D
.133=2 " 8/ % 1:44:
18 4
18 3
27
27
4
7. .sin 3t; cos 3t/,
SOLUTION
0#t #!
We have x D sin 3t, y D cos 3t, hence x 0 D 3 cos 3t and y 0 D "3 sin 3t. By the formula for the arc length we obtain:
SD
Z
!
0
Z
q
x 0 .t/2 C y 0 .t/2 dt D
8. .sin # " # cos #; cos # C # sin #/,
0## #2
!
0
p
9 cos2 3t C 9 sin2 3t dt D
Z
0
!
p
9 dt D 3!
SOLUTION We have x D sin # " # cos # and y D cos # C # sin #. Hence, x 0 D cos # " .cos # " # sin #/ D # sin # and
y 0 D " sin # C sin # C # cos # D # cos #: Using the formula for the arc length we obtain:
Z 2q
Z 2q
SD
x 0 .#/2 C y 0 .#/2 d# D
.# sin #/2 C .# cos #/2 d#
0
D
Z
0
2q
0
# 2 .sin2 # C cos2 #/ d# D
Z
2
0
# d# D
In Exercises 9 and 10, use the identity
ˇ
# 2 ˇˇ2
D2
2 ˇ0
1 " cos t
t
D sin2
2
2
9. .2 cos t " cos 2t; 2 sin t " sin 2t/, 0 # t #
SOLUTION
get
!
2
We have x D 2 cos t " cos 2t, y D 2 sin t " sin 2t. Thus, x 0 D "2 sin t C 2 sin 2t and y 0 D 2 cos t " 2 cos 2t. We
x 0 .t/2 C y 0 .t/2 D ."2 sin t C 2 sin 2t/2 C .2 cos t " 2 cos 2t/2
D 4 sin2 t " 8 sin t sin 2t C 4 sin2 2t C 4 cos2 t " 8 cos t cos 2t C 4 cos2 2t
D 4.sin2 t C cos2 t/ C 4.sin2 2t C cos2 2t/ " 8.sin t sin 2t C cos t cos 2t/
D 4 C 4 " 8 cos.2t " t/ D 8 " 8 cos t D 8.1 " cos t/
We now use the formula for the arc length to obtain
SD
Z
!=2 q
0
x 0 .t/2
C y 0 .t/2
D
Z
!=2 p
0
8.1 " cos t/ dt D
Z
0
!=2
r
!
p
ˇ
!
"
!
t ˇˇ!=2
2
D
"8
cos
0
D
"8
D "8 cos ˇ
" cos
" 1 % 2:34
2 0
4
2
10. .5.# " sin #/; 5.1 " cos #//, 0 # # # 2!
16 sin2
t
dt D 4
2
Z
0
!=2
sin
t
dt
2
Arc Length and Speed
S E C T I O N 11.2
1381
SOLUTION Since x D 5.# " sin #/ and y D 5.1 " cos #/, we have x 0 D 5.1 " cos #/ and y 0 D 5 sin #. Using the formula for
the arc length we obtain:
Z 2! q
Z 2! q
2
2
0
0
SD
x .#/ C y .#/ d# D
25.1 " cos #/2 C 25 sin2 # d#
0
D5
D5
0
Z
2!
0
Z
2!
0
Z
p
1 " 2 cos # C cos2 # C sin2 # d# D 5
r
2!
0
4 sin2
#
d# D 10
2
Z
2!
sin
0
#
d# D 20
2
ˇ!
ˇ
D 20." cos u/ˇˇ D "20."1 " 1/ D 40:
Z
!
p
2.1 " cos #/ d#
sin u du
0
0
11. Show that one arch of a cycloid generated by a circle of radius R has length 8R.
Recall from earlier that the cycloid generated by a circle of radius R has parametric equations x D Rt " R sin t,
t
1 " cos t
y D R " R cos t. Hence, x 0 D R " R cos t, y 0 D R sin t. Using the identity sin2 D
, we get
2
2
SOLUTION
x 0 .t/2 C y 0 .t/2 D R2 .1 " cos t/2 C R2 sin2 t D R2 .1 " 2 cos t C cos2 t C sin2 t/
D R2 .1 " 2 cos t C 1/ D 2R2 .1 " cos t/ D 4R2 sin2
t
2
One arch of the cycloid is traced as t varies from 0 to 2!. Hence, using the formula for the arc length we obtain:
Z 2! q
Z 2! r
Z 2!
Z !
t
t
SD
x 0 .t/2 C y 0 .t/2 dt D
4R2 sin2 dt D 2R
sin dt D 4R
sin u du
2
2
0
0
0
0
ˇ!
ˇ
D "4R cos uˇˇ D "4R.cos ! " cos 0/ D 8R
0
12. Find the length of the spiral c.t/ D .t cos t; t sin t/ for 0 # t # 2! to three decimal places (Figure 1). Hint: Use the formula
Z p
p
&
1 p
1 %
1 C t 2 dt D t 1 C t 2 C ln t C 1 C t 2
2
2
y
5
t=2
t=0
−10
10
x
−10
FIGURE 1 The spiral c.t/ D .t cos t; t sin t/.
SOLUTION
We use the formula for the arc length:
SD
Z
2!
0
q
x 0 .t/2 C y 0 .t/2 dt
Differentiating x D t cos t and y D t sin t yields
d
.t cos t/ D cos t " t sin t
dt
d
y 0 .t/ D
.t sin t/ D sin t C t cos t
dt
x 0 .t/ D
Thus,
q
q
x 0 .t/2 C y 0 .t/2 D .cos t " t sin t/2 C .sin t C t cos t/2
p
D cos2 t " 2t cos t sin t C t 2 sin2 t C sin2 t C 2t sin t cos t C t 2 cos2 t
q
p
D .cos2 t C sin2 t/.1 C t 2 / D 1 C t 2
We substitute into (1) and use the integral given in the hint to obtain the following arc length:
Z 2! p
" ˇˇ2!
p
1 p
1 !
SD
1 C t 2 dt D t 1 C t 2 C ln t C 1 C t 2 ˇˇ
2
2
0
0
(1)
1382
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
#
$ #
$
q
q
1
1
1
! 2! 1 C .2!/2 C ln 2! C 1 C .2!/2 " 0 C ln 1
2
2
2
!
"
p
p
1
D ! 1 C 4! 2 C ln 2! C 1 C 4! 2 % 21:256
2
13. Find the length of the tractrix (see Figure 6)
D
c.t/ D .t " tanh.t/; sech.t//;
SOLUTION
0#t #A
Since x D t " tanh.t/ and y D sech.t/ we have x 0 D 1 " sech2 .t/ and y 0 D "sech.t/ tanh.t/. Hence,
2
x 0 .t/2 C y 0 .t/2 D .1 " sech2 .t// C sech2 .t/tanh2 .t/
D 1 " 2 sech2 .t/ C sech4 .t/ C sech2 .t/tanh2 .t/
D 1 " 2 sech2 .t/ C sech2 .t/.sech2 .t/ C tanh2 .t//
D 1 " 2 sech2 .t/ C sech2 .t/ D 1 " sech2 .t/ D tanh2 .t/
Hence, using the formula for the arc length we get:
SD
Z
0
Aq
x 0 .t/2 C y 0 .t/2 dt D
Z
Aq
tanh2 .t/ dt D
0
Z
A
0
ˇA
ˇ
tanh.t/ dt D ln.cosh.t//ˇˇ
0
D ln.cosh.A// " ln.cosh.0// D ln.cosh.A// " ln 1 D ln.cosh.A//
14.
Find a numerical approximation to the length of c.t/ D .cos 5t; sin 3t/ for 0 # t # 2! (Figure 2).
y
1
x
1
FIGURE 2
SOLUTION
Since x D cos 5t and y D sin 3t, we have
x 0 .t/ D "5 sin 5t;
y 0 .t/ D 3 cos 3t
so that
x 0 .t/2 C y 0 .t/2 D 25 sin2 5t C 9 cos2 3t
Then the arc length is
Z
2!
0
Z
q
0
2
0
2
x .t/ C y .t/ dt D
0
2!
p
25 sin2 5t C 9 cos2 3t dt % 24:60296
In Exercises 15–18, determine the speed s at time t (assume units of meters and seconds).
15. .t 3 ; t 2 /, t D 2
We have x.t/ D t 3 ; y.t/ D t 2 hence x 0 .t/ D 3t 2 ; y 0 .t/ D 2t . The speed of the particle at time t is thus,
At time t D 2 the speed is
ˇ
p
p
p
ds ˇˇ
D 2 9 ! 22 C 4 D 2 40 D 4 10 % 12:65 m=s:
ˇ
dt t D2
SOLUTION
16. .3 sin 5t; 8 cos 5t/,
tD
!
4
ds
dt
D
q
p
x 0 .t/2 C y 0 .t/2 D 9t 4 C 4t 2 D
S E C T I O N 11.2
SOLUTION
is
1383
We have x D 3 sin 5t, y D 8 cos 5t, hence x 0 D 15 cos 5t, y 0 D "40 sin 5t. Thus, the speed of the particle at time t
ds
D
dt
D
Thus,
The speed at time t D
17. .5t C 1; 4t " 3/,
SOLUTION
Arc Length and Speed
!
4
q
p
x 0 .t/2 C y 0 .t/2 D 225 cos2 5t C 1600 sin2 5t
q
p
225.cos2 5t C sin2 5t/ C 1375 sin2 5t D 5 9 C 55 sin2 5t
p
ds
D 5 9 C 55 sin2 5t:
dt
is thus
r
ˇ
! !"
ds ˇˇ
D
5
9 C 55 sin2 5 !
Š 30:21 m=s
ˇ
dt t D!=4
4
t D9
Since x D 5t C 1, y D 4t " 3, we have x 0 D 5 and y 0 D 4. The speed of the particle at time t is
p
p
p
ds
D x 0 .t/ C y 0 .t/ D 52 C 42 D 41 % 6:4 m=s:
dt
We conclude that the particle has constant speed of 6:4 m=s.
18. .ln.t 2 C 1/; t 3 /, t D 1
SOLUTION
2t
and y 0 D 3t 2 . The speed of the particle at time t is thus
C1
s
s
q
ds
4t 2
4
2
2
4 Dt
D x 0 .t/ C y 0 .t/ D
C
9t
C 9t 2 :
2
2
dt
.t 2 C 1/
.t 2 C 1/
We have x D ln.t 2 C 1/, y D t 3 , so x 0 D
The speed at time t D 1 is
t2
r
ˇ
p
ds ˇˇ
4
D
C 9 D 10 % 3:16 m=s:
dt ˇt D1
22
19. Find the minimum speed of a particle with trajectory c.t/ D .t 3 " 4t; t 2 C 1/ for t $ 0. Hint: It is easier to find the minimum
of the square of the speed.
We first find the speed of the particle. We have x.t/ D t 3 " 4t, y.t/ D t 2 C 1, hence x 0 .t/ D 3t 2 " 4 and y 0 .t/ D 2t .
The speed is thus
q
p
p
ds
2
D .3t 2 " 4/ C .2t/2 D 9t 4 " 24t 2 C 16 C 4t 2 D 9t 4 " 20t 2 C 16:
dt
SOLUTION
The square root function is an increasing function, hence the minimum speed occurs at the value of t where the function
f .t/ D 9t 4 " 20t 2 C 16 has minimum value. Since lim f .t/ D 1, f has a minimum value on the interval 0 # t < 1, and it
t !1
occurs at a critical point or at the endpoint t D 0. We find the critical point of f on t $ 0:
f 0 .t/ D 36t 3 " 40t D 4t.9t 2 " 10/ D 0 ) t D 0; t D
r
10
:
9
We compute the values of f at these points:
f
f .0/ D 9 ! 04 " 20 ! 02 C 16 D 16
r !
r !4
r !2
10
10
10
44
D9
" 20
C 16 D
% 4:89
9
9
9
9
We conclude that the minimum value of f on t $ 0 is 4:89. The minimum speed is therefore
# $
p
ds
% 4:89 % 2:21:
dt min
20. Find the minimum speed of a particle with trajectory c.t/ D .t 3 ; t !2 / for t $ 0:5.
1384
C H A P T E R 11
SOLUTION
The speed is
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
We first compute the speed of the particle. Since x.t/ D t 3 and y.t/ D t !2 , we have x 0 .t/ D 3t 2 and y 0 .t/ D "2t !3 .
ds
D
dt
q
x 0 .t/2 C y 0 .t/2 D
p
9t 4 C 4t !6 :
The square root function is an increasing function, hence the minimum value of ds
occurs at the point where the function
dt
4
!6
f .t/ D 9t C 4t attains its minimum value. We find the critical points of f on the interval t $ 0:5:
f 0 .t/ D 36t 3 " 24t !7 D 0
r
10 2
3t 10 " 2 D 0 ) t D
% 0:96
3
Since lim f .t/ D 1, the minimum value on 0:5 # t < 1 exists, and it occurs at the critical point t D 0:96 or at the endpoint
t !1
t D 0:5. We compute the values of f at these points:
f .0:96/ D 9 ! .0:96/4 C 4 ! .0:96/!6 D 12:75
f .0:5/ D 9.0:5/4 C 4.0:5/!6 D 256:56
We conclude that the minimum value of f on the interval t $ 0:5 is 12:75. The minimum speed for t $ 0:5 is therefore
# $
p
ds
D 12:75 % 3:57
dt min
21. Find the speed of the cycloid c.t/ D .4t " 4 sin t; 4 " 4 cos t/ at points where the tangent line is horizontal.
SOLUTION
We first find the points where the tangent line is horizontal. The slope of the tangent line is the following quotient:
dy
dy=dt
4 sin t
sin t
D
D
D
:
dx
dx=dt
4 " 4 cos t
1 " cos t
To find the points where the tangent line is horizontal we solve the following equation for t $ 0:
dy
sin t
D 0;
D 0 ) sin t D 0
dx
1 " cos t
and
cos t ¤ 1:
Now, sin t D 0 and t $ 0 at the points t D !k, k D 0; 1; 2; : : : : Since cos !k D ."1/k , the points where cos t ¤ 1 are t D !k for
k odd. The points where the tangent line is horizontal are, therefore:
t D !.2k " 1/;
k D 1; 2; 3; : : :
The speed at time t is given by the following expression:
q
q
ds
D x 0 .t/2 C y 0 .t/2 D .4 " 4 cos t/2 C .4 sin t/2
dt
p
p
D 16 " 32 cos t C 16 cos2 t C 16 sin2 t D 16 " 32 cos t C 16
r
ˇ
ˇ
p
ˇ
t
t ˇˇ
2
ˇ
D 32.1 " cos t/ D 32 ! 2 sin
D 8 ˇsin ˇ
2
2
That is, the speed of the cycloid at time t is
We now substitute
ˇ
ˇ
ˇ
ds
tˇ
D 8 ˇˇsin ˇˇ :
dt
2
t D !.2k " 1/;
k D 1; 2; 3; : : :
to obtain
ˇ
ˇ
ˇ !.2k " 1/ ˇ
ds
ˇ D 8j."1/kC1 j D 8
D 8 ˇˇsin
ˇ
dt
2
22. Calculate the arc length integral s.t/ for the logarithmic spiral c.t/ D .e t cos t; e t sin t/.
SOLUTION
We have x 0 .t/ D e t .cos t " sin t/, y 0 .t/ D e t .cos t C sin t/ so that
x 0 .t/2 C y 0 .t/2 D e 2t .cos2 t " 2 cos t sin t C sin2 t C cos2 t C 2 cos t sin t C sin2 t/ D 2e 2t .cos2 t C sin2 t/ D 2e 2t
S E C T I O N 11.2
Arc Length and Speed
1385
so that the arc length integral is
Z
b
a
If neither a nor b is ˙1, then this equals
p
q
p Z
x 0 .t/2 C y 0 .t/2 dt D 2
b
e t dt
a
2.e b " e a /. Note that the origin corresponds to t D "1.
In Exercises 23–26, plot the curve and use the Midpoint Rule with N D 10, 20, 30, and 50 to approximate its length.
23. c.t/ D .cos t; e sin t / for 0 # t # 2!
SOLUTION
The curve of c.t/ D .cos t; e sin t / for 0 # t # 2! is shown in the figure below:
y
π
(0, e)
2
t=
t = π, (−1, 1)
t = 0, t = 2π, (1, 1)
x
t=
3π
1
(0, )
2
e
c.t/ D .cos t; e sin t /, 0 # t # 2!:
The length of the curve is given by the following integral:
Z 2! q
Z
SD
x 0 .t/2 C y 0 .t/2 dt D
0
2!
0
q
2
." sin t/2 C .cos t e sin t / dt:
R 2! p 2
That is, S D 0
sin t C cos2 t e 2 sin t dt . We approximate the integral using the Mid-Point Rule with N D 10; 20;
p
30; 50. For f .t/ D sin2 t C cos2 t e 2 sin t we obtain
#
$
2!
!
1
!
.N D 10/W x D
D ; ci D i "
!
10
5
2
5
10
!X
f .ci / D 6:903734
5
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
20
10
2
10
M10 D
.N D 20/W
20
! X
f .ci / D 6:915035
10
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
30
15
2
15
M20 D
.N D 30/W
30
! X
f .ci / D 6:914949
15
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
50
25
2
25
M30 D
.N D 50/W
50
M50 D
24. c.t/ D .t " sin 2t; 1 " cos 2t/
SOLUTION
for 0 # t # 2!
! X
f .ci / D 6:914951
25
i D1
The curve is shown in the figure below:
y
2
1
x
2
4
6
c.t/ D .t " sin 2t; 1 " cos 2t/, 0 # t # 2!:
1386
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
The length of the curve is given by the following integral:
Z 2! q
Z 2! p
Z
SD
.1 " 2 cos 2t/2 C .2 sin 2t/2 dt D
1 " 4 cos 2t C 4 cos2 2t C 4 sin2 2t dt D
0
0
2!
0
p
5 " 4 cos 2t dt:
That is,
SD
Z
2!
p
0
5 " 4 cos 2t dt:
p
Approximating the length using the Mid-Point Rule with N D 10; 20; 30; 50 for f .t/ D 5 " 4 cos 2t we obtain
#
$
2!
!
1
!
.N D 10/W $x D
D ; ci D i "
!
10
5
2
5
10
!X
f .ci / D 13:384047
5
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
20
10
2
10
M10 D
.N D 20/W
20
! X
f .ci / D 13:365095
10
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
30
15
2
15
M20 D
.N D 30/W
30
! X
f .ci / D 13:364897
15
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
50
25
2
25
M30 D
.N D 50/W
50
M50 D
25. The ellipse
! x "2
5
C
! y "2
3
! X
f .ci / D 13:364893
25
i D1
D1
SOLUTION We use the parametrization given in Example 5, section 11.1, that is, c.t/ D .5 cos t; 3 sin t/, 0 # t # 2!. The curve
is shown in the figure below:
y
t=0
x
t = 2π
c.t/ D .5 cos t; 3 sin t/, 0 # t # 2!:
The length of the curve is given by the following integral:
Z 2! q
Z 2! q
SD
x 0 .t/2 C y 0 .t/2 dt D
."5 sin t/2 C .3 cos t/2 dt
0
D
Z
0
0
2!
p
25 sin2 t C 9 cos2 t dt D
Z
2!
0
Z
q
9.sin2 t C cos2 t/ C 16 sin2 t dt D
0
2!
p
9 C 16 sin2 t dt:
That is,
SD
Z
0
2!
p
9 C 16 sin2 t dt:
We approximate the integral using the Mid-Point Rule with N D 10; 20; 30; 50, for f .t/ D
$
#
2!
!
!
1
.N D 10/W $x D
D ; ci D i "
!
10
5
2
5
p
9 C 16 sin2 t. We obtain
S E C T I O N 11.2
Arc Length and Speed
10
!X
f .ci / D 25:528309
5
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
20
10
2
10
M10 D
.N D 20/W
20
! X
f .ci / D 25:526999
10
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
30
15
2
15
M20 D
.N D 30/W
30
! X
f .ci / D 25:526999
15
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
50
25
2
25
M30 D
.N D 50/W
M50 D
26. x D sin 2t,
SOLUTION
y D sin 3t
for 0 # t # 2!
50
! X
f .ci / D 25:526999
25
i D1
The curve is shown in the figure below:
y
x
c.t/ D .sin 2t; sin 3t/, 0 # t # 2!:
The length of the curve is given by the following integral:
Z 2! q
Z 2! q
Z
SD
x 0 .t/2 C y 0 .t/2 dt D
.2 cos 2t/2 C .3 cos 3t/2 dt D
0
0
2!
0
We approximate the length using the Mid-Point Rule with N D 10; 20; 30; 50 for f .t/ D
#
$
2!
!
1
!
.N D 10/W $x D
D ; ci D i "
!
10
5
2
5
p
10
!X
f .ci / D 15:865169
5
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
20
10
2
10
M10 D
.N D 20/W
20
! X
f .ci / D 15:324697
10
i D1
#
$
2!
!
1
!
$x D
D
; ci D i "
!
30
15
2
15
M20 D
.N D 30/W
30
! X
f .ci / D 15:279322
15
i D1
$
#
2!
!
!
1
$x D
D
; ci D i "
!
50
25
2
25
M30 D
.N D 50/W
M50 D
50
! X
f .ci / D 15:287976
25
i D1
p
4 cos2 2t C 9 cos2 3t dt:
4 cos2 2t C 9 cos2 3t. We obtain
1387
1388
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
27. If you unwind thread from a stationary circular spool, keeping the thread taut at all times, then the endpoint traces a curve C
called the involute of the circle (Figure 3). Observe that PQ has length R#. Show that C is parametrized by
%
&
c.#/ D R.cos # C # sin #/; R.sin # " # cos #/
Then find the length of the involute for 0 # # # 2!.
y
Q
R
P = (x, y)
x
FIGURE 3 Involute of a circle.
b
Suppose that the arc QT corresponding to the angle # is unwound. Then the length of the segment QP equals the
length of this arc. That is, QP D R#. With the help of the figure we can see that
SOLUTION
x D OA C AB D OA C EP D R cos # C QP sin # D R cos # C R# sin # D R.cos # C # sin #/:
Furthermore,
y D QA " QE D R sin # " QP cos # D R sin # " R# cos # D R.sin # " # cos #/
The coordinates of P with respect to the parameter # form the following parametrization of the curve:
c.#/ D .R.cos # C # sin #/; R.sin # " # cos #//;
0 # # # 2!:
We find the length of the involute for 0 # # # 2!, using the formula for the arc length:
Z 2! q
SD
x 0 .#/2 C y 0 .#/2 d#:
0
We compute the integrand:
d
.R.cos # C # sin #// D R." sin # C sin # C # cos #/ D R# cos #
d#
d
y 0 .#/ D
.R.sin # " # cos #// D R.cos # " .cos # " # sin #// D R# sin #
d#
q
q
q
p
x 0 .#/2 C y 0 .#/2 D .R# cos #/2 C .R# sin #/2 D R2 # 2 .cos2 # C sin2 #/ D R2 # 2 D R#
x 0 .#/ D
We now compute the arc length:
Z
SD
2!
0
R# d# D
28. Let a > b and set
ˇ
R# 2 ˇˇ2!
R ! .2!/2
D
D 2! 2 R:
ˇ
2 0
2
s
kD
Use a parametric representation to show that the ellipse
% x &2
a
G.#; k/ D
Z
C
"
0
1"
b2
a2
% y &2
D 1 has length L D 4aG
b
p
1 " k 2 sin2 t dt
%!
2 ;k
&
, where
is the elliptic integral of the second kind.
SOLUTION Since the ellipse is symmetric with respect to the x and y axis, its length L is four times the length of the part of
the ellipse which is in the first quadrant. This part is represented by the following parametrization: x.t/ D a sin t, y.t/ D b cos t,
0 # t # !2 : Using the formula for the arc length we get:
LD4
Z
0
!=2 q
x 0 .t/2 C y 0 .t/2 dt D 4
Z
!=2 q
0
.a cos t/2 C ."b sin t/2 dt
Arc Length and Speed
S E C T I O N 11.2
D4
Z
0
1389
!=2 p
a2 cos2 t C b 2 sin2 t dt
We rewrite the integrand as follows:
Z !=2 q
LD4
a2 cos2 t C a2 sin2 t C .b 2 " a2 / sin2 t dt
0
D4
Z
D4
Z
a2 .cos2 t C sin2 t/ C .b 2 " a2 / sin2 t dt
0
!=2 q
a2 C .b 2 " a2 / sin2 t
0
D 4a
q
where k D 1 "
!=2 q
Z
!=2
0
s
#
1" 1"
$
2
b
a2
dt D 4a
0
s
Z
!=2 p
Z
sin2 t dt D 4a
!=2
0
a2
b 2 " a2
C
sin2 t dt
2
a
a2
1 " k 2 sin2 t dt D 4aG
!!
2
;k
"
b2
.
a2
In Exercises 29–32, use Eq. (4) to compute the surface area of the given surface.
29. The cone generated by revolving c.t/ D .t; mt/ about the x-axis for 0 # t # A
SOLUTION
Substituting y.t/ D mt, y 0 .t/ D m, x 0 .t/ D 1, a D 0, and b D 0 in the formula for the surface area, we get
S D 2!
Z
A
0
30. A sphere of radius R
mt
p
1 C m2 dt D 2!
p
1 C m2 m
Z
A
0
ˇ
p
p
t 2 ˇA
t dt D 2! m 1 C m2 ! ˇˇ D m 1 C m2 !A2
2
0
SOLUTION The sphere of radius R is generated by revolving the half circle c.t/ D .R cos t; R sin t/, 0 # t # ! about the x-axis.
We have x.t/ D R cos t, x 0 .t/ D "R sin t, y.t/ D R sin t, y 0 .t/ D R cos t. Using the formula for the surface area, we get
Z !
Z !
q
p
S D 2!
y.t/ x 0 .t/2 C y 0 .t/2 dt D 2!
R sin t R2 sin2 t C R2 cos2 t dt
0
D 2!R 2
0
Z
!
0
ˇ!
ˇ
2
sin t dt D "2!R cos t ˇˇ D "2!R2 ."1 " 1/ D 4!R2
0
31. The surface generated by revolving one arch of the cycloid c.t/ D .t " sin t; 1 " cos t/ about the x-axis
SOLUTION One arch of the cycloid is traced as t varies from 0 to 2!. Since x.t/ D t " sin t and y.t/ D 1 " cos t, we have
x 0 .t/ D 1 " cos t and y 0 .t/ D sin t . Hence, using the identity 1 " cos t D 2 sin2 2t , we get
x 0 .t/2 C y 0 .t/2 D .1 " cos t/2 C sin2 t D 1 " 2 cos t C cos2 t C sin2 t D 2 " 2 cos t D 4 sin2
By the formula for the surface area we obtain:
Z 2!
Z
q
S D 2!
y.t/ x 0 .t/2 C y 0 .t/2 dt D 2!
0
D 2!
Z
0
2!
t
t
2 sin2 ! 2 sin dt D 8!
2
2
Z
0
2!
2!
0
sin3
.1 " cos t/ ! 2 sin
t
dt D 16!
2
Z
!
t
dt
2
sin3 u du
0
We use a reduction formula to compute this integral, obtaining
'
( ˇ!
' (
ˇ
1
4
64!
S D 16!
cos3 u " cos u ˇˇ D 16!
D
3
3
3
0
32. The surface generated by revolving the astroid c.t/ D .cos3 t; sin3 t/ about the x-axis for 0 # t #
SOLUTION
t
2
!
2
We have x.t/ D cos3 t, y.t/ D sin3 t, x 0 .t/ D "3 cos2 t sin t, y 0 .t/ D 3 sin2 t cos t. Hence,
x 0 .t/2 C y 0 .t/2 D 9 cos4 t sin2 t C 9 sin4 t cos2 t D 9 cos2 t sin2 t.cos2 t C sin2 t/ D 9 cos2 t sin2 t
Using the formula for the surface area we get
Z !=2
Z
q
S D 2!
y.t/ x 0 .t/2 C y 0 .t/2 dt D 2!
0
!=2
0
sin3 t ! 3 cos t sin t dt D 6!
We compute the integral using the substitution u D sin t du D cos t dt. We obtain
ˇ
Z 1
6!
u5 ˇ 1
:
S D 6!
u4 du D 6! ˇˇ D
5 0
5
0
Z
!=2
0
sin4 t cos t dt
1390
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
Further Insights and Challenges
33.
Let b.t/ be the “Butterfly Curve”:
x.t/ D sin t e
#
t
" 2 cos 4t " sin
12
cos t
y.t/ D cos t e
cos t
$5 !
#
t
" 2 cos 4t " sin
12
$5 !
(a) Use a computer algebra system to plot b.t/ and the speed s 0 .t/ for 0 # t # 12!.
(b) Approximate the length b.t/ for 0 # t # 10!.
SOLUTION
(a) Let f .t/ D e cos t " 2 cos 4t " sin
%
&
t 5
12 ,
then
x.t/ D sin tf .t/
y.t/ D cos tf .t/
and so
2
2
2
2
.x 0 .t// C .y 0 .t// D Œsin tf 0 .t/ C cos tf .t/" C Œcos tf 0 .t/ " sin tf .t/"
Using the identity sin2 t C cos2 t D 1, we get
2
2
2
.x 0 .t// C .y 0 .t// D .f 0 .t// C .f .t//2 :
Thus, s 0 .t/ is the following:
v
u"
# $5 #2 "
# $4
# $5 #2
u
5
t
t
t e cos t " 2 cos 4t " sin t
C " sin te cos t C 8 sin 4t "
cos
:
12
12 12
12
The following figures show the curves of b.t/ and the speed s 0 .t/ for 0 # t # 10! :
y
20
y
15
10
t=0
t = 10
x
5
x
10
The “Butterfly Curve” b.t/, 0 # t # 10!
20
30
s 0 .t/, 0 # t # 10!
Looking at the graph, we see it would be difficult to compute the length using numeric integration; due to the high frequency
oscillations, very small steps would be needed.
R 10!
(b) The length of b.t/ for 0 # t # 10! is given by the integral: L D 0 s 0 .t/ dt where s 0 .t/ is given in part (a). We approximate
the length using the Midpoint Rule with N D 30. The numerical methods in Mathematica approximate the answer by 211:952.
Using the Midpoint Rule with N D 50, we get 204:48; with N D 500, we get 211:6; and with N D 5000, we get 212:09.
p
2 ab
34.
Let a $ b > 0 and set k D
. Show that the trochoid
a"b
has length 2.a " b/G
SOLUTION
%T
We have
&
2 ;k
x 0 .t/
x D at " b sin t;
y D a " b cos t;
0#t #T
with G.#; k/ as in Exercise 28.
D a " b cos t, y 0 .t/ D b sin t. Hence,
x 0 .t/2 C y 0 .t/2 D .a " b cos t/2 C .b sin t/2 D a2 " 2ab cos t C b 2 cos2 t C b 2 sin2 t
D a2 C b 2 " 2ab cos t
Arc Length and Speed
S E C T I O N 11.2
The length of the trochoid for 0 # t # T is
LD
Z
p
a2 C b 2 " 2ab cos t dt
T
0
We rewrite the integrand as follows to bring it to the required form. We use the identity 1 " cos t D 2 sin2
LD
D
Z
T
0
Z
T
0
q
r
.a " b/2 C 2ab " 2ab cos t dt D
D .a " b/
p
ab
(where k D 2a!b
).
Substituting u D 2t , du D
1
2
t
dt D
2
.a " b/2 C 4ab sin2
Z
T
0
r
1 C k 2 sin2
Z
0
T
Z
T
0
s
t
2
1391
to obtain
q
.a " b/2 C 2ab.1 " cos t/ dt
#
.a " b/2 1 C
4ab
.a " b/2
sin2
$
t
dt
2
t
dt
2
dt, we get
L D 2.a " b/
Z
T =2 p
1 C k 2 sin2 u du D 2.a " b/G.T =2; k/
0
35. A satellite orbiting at a distance R from the center of the earth follows the circular path x D R cos !t, y D R sin !t.
(a) Show that the period T (the time of one revolution) is T D 2!=!.
(b) According to Newton’s laws of motion and gravity,
x 00 .t/ D "Gme
x
;
R3
y 00 .t/ D "Gme
y
R3
where G is the universal gravitational constant and me is the mass of the earth. Prove that R3 =T 2 D Gme =4! 2 . Thus, R3 =T 2 has
the same value for all orbits (a special case of Kepler’s Third Law).
SOLUTION
(a) As shown in Example 4, the circular path has constant speed of
T is
T D
ds
dt
D !R. Since the length of one revolution is 2!R, the period
2!R
2!
D
:
!R
!
(b) Differentiating x D R cos !t twice with respect to t gives
x 0 .t/ D "R! sin !t
x 00 .t/ D "R! 2 cos !t
Substituting x.t/ and x 00 .t/ in the equation x 00 .t/ D "Gme
x
and simplifying, we obtain
R3
R cos !t
R3
Gme
Gme
"R! 2 D " 2 ) R3 D
R
!2
"R! 2 cos !t D "Gme !
By part (a), T D
2!
2!
. Hence, ! D
. Substituting yields
!
T
R3 D
Gme
4! 2
T2
D
T 2 Gme
R3
Gme
)
D
4! 2
T2
4! 2
36. The acceleration due to gravity on the surface of the earth is
gD
Gme
Re2
D 9:8 m/s2 ;
where Re D 6378 km
p
Use Exercise 35(b) to show that a satellite orbiting at the earth’s surface would have period Te D 2! Re =g % 84:5 min.
Then estimate the distance Rm from the moon to the center of the earth. Assume that the period of the moon (sidereal month) is
Tm % 27:43 days.
1392
C H A P T E R 11
SOLUTION
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
By part (b) of Exercise 35, it follows that
Re3
Gme
4! 2 Re3
4! 2 Re
4! 2 Re
D
) Te2 D
D Gm D
2
2
e
Gme
g
4!
Te
2
Re
Hence,
Te D 2!
s
Re
D 2!
g
s
6378 ! 103
% 5068:8 s % 84:5 min:
9:8
R3
is the same for all orbits. It follows that this quotient is the same for the satellite
T2
orbiting at the earth’s surface and for the moon orbiting around the earth. Thus,
In part (b) of Exercise 35 we showed that
#
$
3
Rm
Re3
Tm 2=3
D
)
R
D
R
:
m
e
2
Te
Tm
Te2
Setting Tm D 27:43 ! 1440 D 39;499:2 minutes, Te D 84:5 minutes, and Re D 6378 km we get
#
$
39;499:2 2=3
Rm D 6378
% 384;154 km:
84:5
11.3 Polar Coordinates
Preliminary Questions
1. Points P and Q with the same radial coordinate (choose the correct answer):
(a) Lie on the same circle with the center at the origin.
(b) Lie on the same ray based at the origin.
SOLUTION Two points with the same radial coordinate are equidistant from the origin, therefore they lie on the same circle
centered at the origin. The angular coordinate defines a ray based at the origin. Therefore, if the two points have the same angular
coordinate, they lie on the same ray based at the origin.
2. Give two polar representations for the point .x; y/ D .0; 1/, one with negative r and one with positive r.
SOLUTION The point .0; 1/ is on the y-axis, distant one unit from the origin, hence the polar representation with positive r is
%
&
%
&
%
&
.r; #/ D 1; !2 . The point .r; #/ D "1; !2 is the reflection of .r; #/ D 1; !2 through the origin, hence we must add ! to return
to the original point.
We obtain the following polar representation of .0; 1/ with negative r:
$
!
" #
!
3!
.r; #/ D "1; C ! D "1;
:
2
2
3. Describe each of the following curves:
(a) r D 2
(b) r 2 D 2
(c) r cos # D 2
SOLUTION
(a) Converting to rectangular coordinates we get
q
x 2 C y 2 D 2 or
x 2 C y 2 D 22 :
This is the equation of the circle of radius 2 centered at the origin.
p
(b) We convert to rectangular coordinates, obtaining x 2 C y 2 D 2. This is the equation of the circle of radius 2, centered at the
origin.
(c) We convert to rectangular coordinates. Since x D r cos # we obtain the following equation: x D 2. This is the equation of the
vertical line through the point .2; 0/.
4. If f ."#/ D f .#/, then the curve r D f .#/ is symmetric with respect to the (choose the correct answer):
(a) x-axis
(b) y-axis
(c) origin
SOLUTION The equality f ."#/ D f .#/ for all # implies that whenever a point .r; #/ is on the curve, also the point .r; "#/ is on
the curve. Since the point .r; "#/ is the reflection of .r; #/ with respect to the x-axis, we conclude that the curve is symmetric with
respect to the x-axis.
S E C T I O N 11.3
Polar Coordinates
Exercises
1. Find polar coordinates for each of the seven points plotted in Figure 1.
y
4
A
(x, y) = (2!3, 2)
F
E
B
x
4
D
C
G
FIGURE 1
SOLUTION
We mark the points as shown in the figure.
y
A
F(2 3, 2)
E
x
B
C
D
G(2 3, −2)
Using the data given in the figure for the x and y coordinates and the quadrants in which the point are located, we obtain:
q
p
!
"
r
D
."3/2 C 32 D 18 ) .r; #/ D 3p2; 3!
(A), with rectangular coordinates ."3; 3/:
4
# D ! " !4 D 3!
4
y
A
3 2
(B), with rectangular coordinates ."3; 0/:
3π
4
x
r D3
) .r; #/ D .3; !/
# D!
y
π
B
x
3
(C), with rectangular coordinates ."2; "1/:
p
p
!p
"
r D 22 C! 12 D
" 5 % 2:2
! "
) .r; #/ %
5; 3:6
!1
1
!1
!1
# D tan
!2 D tan
2 D ! C 0:46 % 3:6
y
3.6
x
C
2.2
p
p
!p
"
r D 12 C 12 D 2 % 1:4
(D), with rectangular coordinates ."1; "1/:
) .r; #/ %
2; 5!
5!
!
4
# D!C 4 D 4
1393
1394
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
y
5π
4
x
1.4
D
(E), with rectangular coordinates .1; 1/:
rD
p
12 C! 12" D
# D tan!1
1
1
D
p
!
4
2 % 1:4
) .r; #/ %
!p
2; !4
"
y
E π
1.4
4
x
r!
p "2
p
p
rD
2 3 C 22 D 16 D 4
(F), with rectangular coordinates .2 3; 2/:
!
"
! "
2
# D tan!1 p
D tan!1 p1 D
2 3
3
%
) .r; #/ D 4;
!
6
!
6
&
y
F(2 3, 2)
4
π
6
x
p
(G), with rectangular coordinates .2 3; "2/: G is the reflection of F about the x axis, hence the two points have equal radial
coordinates, and the angular coordinate of G is obtained from the angular coordinate of F : # D 2! " !6 D 11!
6 . Hence, the
!
"
11!
polar coordinates of G are 4; 6 .
2. Plot the points with polar coordinates:
%
&
%
&
(a) 2; !6
(b) 4; 3!
4
%
&
(c) 3; " !2
%
&
(d) 0; !6
SOLUTION We first plot the ray # D #0 for the given angle #0 , and then mark the point on this line distanced r D r0 from the
origin. We obtain the following points:
y
y
6
x
3π
4
4
x
x
3
(3, − π2 )
(a)
y
(4, 3π4 )
(2, π6 ) π
2
y
(b)
−
π
2
(c)
R D 0 is the point .0; 0/ in rect. coords.
3. Convert from rectangular to polar coordinates.
p
(a) .1; 0/
(b) .3; 3/
(c) ."2; 2/
π
6
(0, π6 )
x
(d)
(d) ."1;
p
3/
SOLUTION
(a) The point .1; 0/ is on the positive x axis distanced one unit from the origin. Hence, r D 1 andr# D 0. Thus, .r; #/ D .1; 0/.
! p "
!p "
!p "2 p
(b) The point 3; 3 is in the first quadrant so # D tan!1 33 D !6 . Also, r D 32 C
3 D 12: Hence,
!p
"
.r; #/ D
12; !6 .
(c) The point ."2; 2/ is in the second quadrant. Hence,
# $
2
!
3!
# D tan!1
D tan!1 ."1/ D ! " D
:
"2
4
4
q
"
!p
p
Also, r D ."2/2 C 22 D 8: Hence, .r; #/ D
8; 3!
4 :
S E C T I O N 11.3
Polar Coordinates
1395
!
p "
(d) The point "1; 3 is in the second quadrant, hence,
# D tan!1
Also, r D
r
."1/2 C
p !
! p "
3
!
2!
D tan!1 " 3 D ! " D
:
"1
3
3
!p "2 p
!
3 D 4 D 2. Hence, .r; #/ D 2;
2!
3
"
.
4. Convert from rectangular to polar coordinates using a calculator (make sure your choice of # gives the correct quadrant).
(a) .2; 3/
(b) .4; "7/
(c) ."3; "8/
(d) ."5; 2/
SOLUTION
(a) The point .2; 3/ is in the first quadrant, with x D 2 and y D 3. Hence
# $
3
# D tan!1
% 0:98
2
) .r; #/ % .3:6; 0:98/ :
p
p
r D 22 C 32 D 13 % 3:6
(b) The point .4; "7/ is in the fourth quadrant with x D 4 and y D "7. We have
# $
"7
tan!1
% "1:05
4
q
p
r D ."7/2 C 42 D 65 % 8:1
Note that tan!1 an angle less that zero in the fourth quadrant; since we want an angle between 0 and 2!, we add 2! to get
# % 2! " 1:05 % 5:232. Thus .r; #/ % .8:1; 5:2/.
(c) The point ."3; "8/ is in the third quadrant, with x D "3 and y D "8. We have
# $
# $
!1 "8
!1 8
tan
D tan
% 1:212
"3
3
q
p
r D ."3/2 C ."8/2 D 73 % 8:54
Note that tan!1 produced an angle in the first quadrant; we want the third quadrant angle with the same tangent, so we add ! to get
# % ! C 1:212 % 4:35. Thus .r; #/ % .8:54; 4:35/
(d) The point ."5; 2/ is in the second quadrant, with x D "5 and y D 2. We have
# $
2
tan!1
% "0:38
"5
q
p
r D 22 C ."5/2 D 29 % 5:39
Note that the angle is in the fourth quadrant; to get the second quadrant angle with the same tangent and in the range Œ0; 2!/, we
add ! to get # % ! " 0:38 % 2:76. Thus .r; #/ % .5:39; 2:76/.
5. Convert from polar to rectangular coordinates:
%
&
%
&
(a) 3; !6
(b) 6; 3!
4
SOLUTION
(a) Since r D 3 and # D
!
6,
%
(c) 0;
3!
4
"
p
!
3
D3!
% 2:6
6
2
!
1
y D r sin # D 3 sin D 3 ! D 1:5
6
2
we have r D 6 and # D
3!
4 .
!
5
%
&
(d) 5; " !2
)
.x; y/ % .2:6; 1:5/ :
Hence,
3!
% "4:24
4
3!
% 4:24
y D r sin # D 6 sin
4
x D r cos # D 6 cos
%
(c) For 0;
&
we have:
x D r cos # D 3 cos
!
(b) For 6;
!
5
)
.x; y/ % ."4:24; 4:24/ :
&
, we have r D 0, so that the rectangular coordinates are .x; y/ D .0; 0/.
1396
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
(d) Since r D 5 and # D " !2 we have
6.
(a)
(c)
(e)
! !"
x D r cos # D 5 cos "
D5!0 D0
2
) .x; y/ D .0; "5/
! !"
y D r sin # D 5 sin "
D 5 ! ."1/ D "5
2
Which of the following are possible polar coordinates for the point P with rectangular coordinates .0; "2/?
#
$
! !"
7!
2;
(b) 2;
# 2
$
# 2 $
3!
7!
"2; "
(d) "2;
2
2 $
#
!
!"
7!
"2; "
(f) 2; "
2
2
The point P has distance 2 from the origin and the angle between OP and the positive x-axis in the positive direction
!
"
Hence, .r; #/ D 2; 3!
is one choice for the polar coordinates for P .
2
SOLUTION
is
3!
2 .
y
3π
2
x
0
P
The polar coordinates .2; #/ are possible for P if # " 3!
2 is a multiple of 2!. The polar coordinate ."2; #/ are possible for P if
# " 3!
is
an
odd
multiple
of
!.
These
considerations
lead
to the following conclusions:
%2 &
% !&
(a) 2; !2 !2 " 3!
D
"!
)
2;
does
not
represent
P:
2
!
"
!2
"
7!
3!
(b) 2; 7!
"
D
2!
)
2; 7!
represents P:
2
2
2
2
!
"
!
"
3!
3!
3!
(c) "2; " 2 " 2 " 2 D "3! ) "2; " 3!
represents P:
2
!
"
!
"
7!
3!
7!
(d) "2; 7!
does not represent P:
2
2 " 2 D 2! ) "2; 2
%
&
%
&
!
(e) "2; " !2 " !2 " 3!
D
"2!
)
"2;
"
2
2 "does not represent P:
!
"
!
3!
7!
(f) 2; " 7!
" 7!
does not represent P:
2
2 " 2 D "5! ) 2; " 2
7. Describe each shaded sector in Figure 2 by inequalities in r and #.
y
y
y
45°
3
(A)
5
x
3
5
x
(B)
3
5
x
(C)
FIGURE 2
SOLUTION
(a) In the sector shown below r is varying between 0 and 3 and # is varying between ! and 2!. Hence the following inequalities
describe the sector:
0#r #3
! # # # 2!
(b) In the sector shown below r is varying between 0 and 3 and # is varying between
sector are:
!
4
and
!
2.
Hence, the inequalities for the
0#r#3
!
!
###
4
2
(c) In the sector shown below r is varying between 3 and 5 and # is varying between
3#r #5
3!
## #!
4
3!
4
and !. Hence, the inequalities are:
S E C T I O N 11.3
Polar Coordinates
1397
8. Find the equation in polar coordinates of the line through the origin with slope 12 .
A line of slope m D
# % 0:46, while r is arbitrary.
SOLUTION
1
2
9. What is the slope of the line # D
SOLUTION
makes an angle #0 D tan!1 12 % 0:46 with the positive x-axis. The equation of the line is
3!
5 ?
This line makes an angle #0 D
3!
5
with the positive x-axis, hence the slope of the line is m D tan
3!
5
% "3:1.
10. Which of r D 2 sec # and r D 2 csc # defines a horizontal line?
SOLUTION The equation r D 2 csc # is the polar equation of a horizontal line, as it can be written as r D 2= sin #, so r sin # D 2,
which becomes y D 2. On the other hand, the equation r D 2 sec # is the polar equation of a vertical line, as it can be written as
r D 2= cos #, so r cos # D 2, which becomes x D 2.
In Exercises 11–16, convert to an equation in rectangular coordinates.
11. r D 7
SOLUTION r D 7 describes the points having distance 7 from the origin, that is, the circle with radius 7 centered at the origin.
The equation of the circle in rectangular coordinates is
x 2 C y 2 D 72 D 49:
12. r D sin #
SOLUTION
Multiplying by r and substituting y D r sin # and r 2 D x 2 C y 2 gives
r 2 D r sin #
x2 C y2 D y
We move the y and then complete the square to obtain
x2 C y2 " y D 0
#
$
# $2
1 2
1
x2 C y "
D
2
2
!
"
Thus, r D sin # is the equation of a circle of radius 12 and center 0; 12 .
13. r D 2 sin #
SOLUTION
We multiply the equation by r and substitute r 2 D x 2 C y 2 ; r sin # D y. This gives
r 2 D 2r sin #
x 2 C y 2 D 2y
Moving the 2y and completing the square yield: x 2 C y 2 " 2y D 0 and x 2 C .y " 1/2 D 1. Thus, r D 2 sin # is the equation of a
circle of radius 1 centered at .0; 1/.
14. r D 2 csc #
SOLUTION
We multiply the equation by sin # and substitute y D r sin # : We get
r sin # D 2
yD2
Thus, r D 2 csc # is the equation of the line y D 2.
15. r D
1
cos # " sin #
SOLUTION
We multiply the equation by cos # " sin # and substitute y D r sin #, x D r cos #. This gives
r .cos # " sin #/ D 1
r cos # " r sin # D 1
x " y D 1 ) y D x " 1. Thus,
rD
is the equation of the line y D x " 1.
1
cos # " sin #
1398
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
1
2 " cos #
p
SOLUTION We multiply the equation by 2 " cos #. Then we substitute x D r cos # and r D x 2 C y 2 , to obtain
16. r D
r .2 " cos #/ D 1
2r " r cos # D 1
q
2 x2 C y2 " x D 1
Moving the x, then squaring and simplifying, we obtain
q
2 x2 C y2 D x C 1
!
"
4 x 2 C y 2 D x 2 C 2x C 1
3x 2 " 2x C 4y 2 D 1
We complete the square:
#
$
2
2
3 x " x C 4y 2 D 1
3
#
$
1 2
4
3 x"
C 4y 2 D
3
3
!
"2
x " 13
y2
C 1 D1
4
9
3
This is the equation of the ellipse shown in the figure:
y
π
2
π
0.5
1 1.5 x
0π
3π
2
In Exercises 17–20, convert to an equation in polar coordinates.
17. x 2 C y 2 D 5
p
SOLUTION We make the substitution x 2 C y 2 D r 2 to obtain; r 2 D 5 or r D 5.
18. x D 5
SOLUTION Substituting x D r cos # gives the polar equation r cos # D 5 or r D 5 sec # .
19. y D x 2
SOLUTION
Substituting y D r sin # and x D r cos # yields
r sin # D r 2 cos2 #:
Then, dividing by r cos2 # we obtain,
sin #
Dr
cos2 #
20. xy D 1
SOLUTION
so
r D tan # sec #
We substitute x D r cos #, y D r sin # to obtain
.r cos #/ .r sin #/ D 1
r 2 cos # sin # D 1
Using the identity cos # sin # D
1
2
sin 2# yields
r2 !
sin 2#
D 1 ) r 2 D 2 csc 2#:
2
S E C T I O N 11.3
Polar Coordinates
1399
21. Match each equation with its description.
(a)
(b)
(c)
(d)
r D2
# D2
r D 2 sec #
r D 2 csc #
(i)
(ii)
(iii)
(iv)
Vertical line
Horizontal line
Circle
Line through origin
SOLUTION
(a) r D 2 describes the points 2 units from the origin. Hence, it is the equation of a circle.
(b) # D 2 describes the points P so that OP makes an angle of #0 D 2 with the positive x-axis. Hence, it is the equation of a line
through the origin.
(c) This is r cos # D 2, which is x D 2, a vertical line.
(d) Converting to rectangular coordinates, we get r D 2 csc #, so r sin # D 2 and y D 2. This is the equation of a horizontal line.
22. Find the values of # in the plot of r D 4 cos # corresponding to points A, B, C , D in Figure 3. Then indicate the portion of the
graph traced out as # varies in the following intervals:
(a) 0 # # # !2
(b) !2 # # # !
(c) ! # # # 3!
2
y
B
2
C
2
−2
A
x
4
D
FIGURE 3 Plot of r D 4 cos #.
The point A is on the x-axis hence # D 0. The point B is in the first quadrant with x D y D 2 hence # D tan!1
The point C is at the origin. Thus,
SOLUTION
r D 0 ) 4 cos # D 0 ) # D
! "
2
2
D tan!1 .1/
! 3!
;
:
2 2
The point D is in the fourth quadrant with x D 2, y D "2, hence
# $
"2
!
7!
# D tan!1
D tan!1 ."1/ D 2! " D
:
2
4
4
0 # # # !2 represents the first quadrant, hence the points .r; #/ where r D 4 cos # and 0 # # #
are in the first quadrant, as shown below:
!
2
are the points on the circle which
y
x
If we insist that r $ 0, then since !2 # # # ! represents the second quadrant and ! # # # 3!
2 represents the third quadrant,
and since the circle r D 4 cos # has no points in the left xy -plane, then there are no points for (b) and (c). However, if we allow
r < 0 then (b) represents the semi-circle
y
y
x
and (c) like (a) represent
x
23. Suppose that P D .x; y/ has polar coordinates .r; #/. Find the polar coordinates for the points:
(a) .x; "y/
(b) ."x; "y/
(c) ."x; y/
(d) .y; x/
1400
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
SOLUTION
(a) .x; "y/ is the symmetric point of .x; y/ with respect to the x-axis, hence the two points have the same radial coordinate, and
the angular coordinate of .x; "y/ is 2! " # . Hence, .x; "y/ D .r; 2! " #/.
y
(x, y)
x
−
2 −
(x, −y)
(b) ."x; "y/ is the symmetric point of .x; y/ with respect to the origin. Hence, ."x; "y/ D .r; # C !/.
y
(x, y)
+
x
(−x, −y)
(c) ."x; y/ is the symmetric point of .x; y/ with respect to the y-axis. Hence the two points have the same radial coordinates and
the angular coordinate of ."x; y/ is ! " #. Hence, ."x; y/ D .r; ! " #/.
y
(−x, y)
(x, y)
−
−
x
(d) Let .r1 ; #1 / denote the polar coordinates of .y; x/. Hence,
q
q
r1 D y 2 C x 2 D x 2 C y 2 D r
tan #1 D
!!
"
x
1
1
D
D
D cot # D tan
"#
y
y=x
tan #
2
Since the points .x; y/% and .y;&x/ are in the same quadrant, the solution for #1 is #1 D
coordinates: .y; x/ D r; !2 " # .
y
(y, x)
−
2
−
(x, y)
x
24. Match each equation in rectangular coordinates with its equation in polar coordinates.
(a)
(b)
(c)
(d)
x2 C y2 D 4
x 2 C .y " 1/2 D 1
x2 " y2 D 4
xCy D4
(i)
(ii)
(iii)
(iv)
r 2 .1 " 2 sin2 #/ D 4
r.cos # C sin #/ D 4
r D 2 sin #
r D2
!
2
" # . We obtain the following polar
S E C T I O N 11.3
Polar Coordinates
1401
SOLUTION
(a) Since x 2 C y 2 D r 2 , we have r 2 D 4 or r D 2.
(b) Using Example 7, the equation of the circle x 2 C .y " 1/2 D 1 has polar equation r D 2 sin #.
(c) Setting x D r cos # , y D r sin # in x 2 " y 2 D 4 gives
!
"
x 2 " y 2 D r 2 cos2 # " r 2 sin2 # D r 2 cos2 # " sin2 # D 4:
We now use the identity cos2 # " sin2 # D 1 " 2 sin2 # to obtain the following equation:
!
"
r 2 1 " 2 sin2 # D 4:
(d) Setting x D r cos # and y D r sin # in x C y D 4 we get:
xCy D4
r cos # C r sin # D 4
so
r .cos # C sin #/ D 4
%
&
25. What are the polar equations of the lines parallel to the line r cos # " !3 D 1?
%
&
%
&
!
!
!
SOLUTION The line r cos # " 3 D 1, or r D sec # " 3 , is perpendicular to the ray # D 3 and at distance d D 1 from the
!
origin. Hence,
the& lines parallel
to& this line are also perpendicular to the ray # D 3 , so the polar equations of these lines are
%
%
r D d sec # " !3 or r cos # " !3 D d .
%
&
26. Show that the circle with center at 12 ; 12 in Figure 4 has polar equation r D sin # C cos # and find the values of # between 0
and ! corresponding to points A; B, C , and D.
y
A
D
( 12 , 12 )
B
C
x
FIGURE 4 Plot of r D sin # C cos #.
SOLUTION
We show that the rectangular equation of r D sin # C cos # is
#
$
#
$
1 2
1 2
1
x"
C y"
D :
2
2
2
We multiply the polar equation by r and substitute r 2 D x 2 C y 2 , r sin # D y, r cos # D x. This gives
r D sin # C cos #
2
2
r D r sin # C r cos #
x C y2 D y C x
Transferring sides and completing the square yields
x2 " x C y2 " y D 0
#
$
#
$
1 2
1 2
1
1
1
x"
C y"
D C D
2
2
4
4
2
Clearly point C corresponds to # D 0 since cos 0 C sin 0 D 1. The circle is traced out counterclockwise as # increases to !, so
pA
corresponds to # D !2 since again cos !2 C sin !2 D 0. Next, D clearly corresponds to # D !4 , and indeed cos !4 C sin !4 D 2,
which is the diameter of the circle. Finally, point A corresponds to # D 3!
4 , since there cos # D " sin #.
27. Sketch the curve r D 12 # (the spiral of Archimedes) for # between 0 and 2! by plotting the points for # D 0; !4 ;
SOLUTION
We first plot the following points .r; #/ on the spiral:
#
$
!! ! "
!! ! "
!! "
3! 3!
;
; BD
;
; C D
;
; DD
;! ;
8 4
4 2
8 4
2
#
#
#
$
$
$
3! 3!
7! 7!
5! 5!
ED
;
; F D
;
; GD
;
; H D .!; 2!/ :
8 4
4 2
8 4
O D .0; 0/ ; A D
!
2 ; : : : ; 2!.
1402
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
2
3
4
4
C
B
D
H 0
2
A
O
E
G
5
4
7
4
F
3
2
Since r.0/ D 02 D 0, the graph begins at the origin and moves toward the points A; B; C; D; E; F; G and H as # varies from # D 0
to the other values stated above. Connecting the points in this direction we obtain the following graph for 0 # # # 2!:
2
3
4
4
C
B
D
H 0
2
A
O
E
G
5
4
7
4
F
3
2
28. Sketch r D 3 cos # " 1 (see Example 8).
SOLUTION We first choose some values of # between 0 and ! and mark the corresponding points on the graph. Then
we use symmetry (due to cos .2! " #/ D cos #) to plot the other half of the graph by reflecting the first half through the
x-axis. Since r D 3 cos # " 1 is periodic, the entire curve is obtained as # varies from 0 to 2!. We start with the values
5!
# D 0; !6 ; !3 ; !2 ; 2!
3 ; 6 ; !, and compute the corresponding values of r:
r D 3 cos 0 " 1 D 3 " 1 D 2 ) A D .2; 0/
p
!
!
3 3
!"
r D 3 cos " 1 D
" 1 % 1:6 ) B D 1:6;
6
2
6
!
!
3
!"
r D 3 cos " 1 D " 1 D 0:5 ) C D 0:5;
3
2
3
!
!
!"
r D 3 cos " 1 D 3 ! 0 " 1 D "1 ) D D "1;
2
2
#
$
2!
2!
r D 3 cos
" 1 D "2:5 ) E D "2:5;
3
3
#
$
5!
5!
r D 3 cos
" 1 D "3:6 ) F D "3:6;
6
6
r D 3 cos ! " 1 D "4 ) G D ."4; !/
The graph begins at the point .r; #/ D .2; 0/ and moves toward the other points in this order, as # varies from 0 to !. Since r is
negative for !2 # # # !, the curve continues into the fourth quadrant, rather than into the second quadrant. We obtain the following
graph:
π
2
2π
3
π
3
5π
6
π
6
C
B
G
A
π
D
E
0
F
Now we have half the curve and we use symmetry to plot the rest. Reflecting the first half through the x axis we obtain the whole
curve:
S E C T I O N 11.3
Polar Coordinates
1403
y
B
1 C
G x
4
A
2
D
F
E
29. Sketch the cardioid curve r D 1 C cos #.
SOLUTION Since cos # is period with period 2!, the entire curve will be traced out as # varies from 0 to 2!. Additionally, since
cos.2! " #/ D cos.#/, we can sketch the curve for # between 0 and ! and reflect the result through the x axis to obtain the whole
3! 5!
curve. Use the values # D 0; !6 ; !4 ; !3 ; !2 ; 2!
3 ; 4 ; 6 , and !:
point
#
r
0
1 C cos 0 D 2
!
6
!
4
!
3
!
2
2!
3
3!
4
5!
6
1 C cos
1 C cos
1 C cos
1 C cos
1 C cos
1 C cos
1 C cos
!
.2; 0/
p
"
2C 3 !
;6
2
! p
"
2C 2 !
;4
2
!
"
3 !
2; 3
% !&
1;
! 2 "
1 2!
;
! 2p 3 "
2! 2 3!
;
! 2p 4 "
2! 3 5!
2 ; 6
p
2C 3
!
6 D
2p
2C 2
!
4 D
2
!
3
3 D 2
!
2 D1
2!
1
3 D 2 p
2! 2
3!
4 D
2p
2! 3
5!
6 D
2
# D 0 corresponds to the point .2; 0/, and the graph moves clockwise as # increases from 0 to !. Thus the graph is
5π
6
π
2
2π
3π 3
4
π
3
π
4
π
6
0
π
Reflecting through the x axis gives the other half of the curve:
y
2
1
x
1
−1
2
−1
−2
30. Show that the cardioid of Exercise 29 has equation
.x 2 C y 2 " x/2 D x 2 C y 2
in rectangular coordinates.
SOLUTION
Multiply through by r and substitute for r, r 2 , and r cos # to get
r D 1 C cos #
r 2 D r C r cos #
q
x2 C y2 D x2 C y2 C x
1404
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
q
x2 C y2
x2 C y2 " x D
.x 2 C y 2 " x/2 D x 2 C y 2
31. Figure 5 displays the graphs of r D sin 2# in rectangular coordinates and in polar coordinates, where it is a “rose with four
petals.” Identify:
(a) The points in (B) corresponding to points A–I in (A).
)
* )
* )
*
) 3!
*
(b) The parts of the curve in (B) corresponding to the angle intervals 0; !2 , !2 ; ! , !; 3!
2 , and 2 ; 2! .
r
y
B
F
A
C
E
π
2
G
I
3π
2
π
D
x
θ
2π
H
(B) Graph of r = sin 2θ
in polar coordinates
(A) Graph of r as a function
of θ, where r = sin 2θ
FIGURE 5
SOLUTION
(a) The graph (A) gives the following polar coordinates of the labeled points:
AW
# D 0;
r D0
!
2!
; r D sin
D1
4
4
!
CW # D ; r D 0
2
3!
2 ! 3!
DW # D
; r D sin
D "1
4
4
EW # D !; r D 0
BW
#D
5!
; r D1
4
3!
GW # D
; r D0
2
7!
HW # D
; r D "1
4
I W # D 2!; r D 0:
FW # D
Since the maximal value of jrj is 1, the points with r D 1 or r D "1 are the furthest points from the origin. The corresponding
quadrant is determined by the value of # and the sign of r. If r0 < 0, the point .r0 ; #0 / is on the ray # D "#0 . These considerations
lead to the following identification of the points in the xy plane. Notice that A, C , G, E, and I are the same point.
3π
4 H
=
y
π
2
π
B 4
7π
4
A,C,E,G,I
r = −1
=
π
4
r=1
x 2π
π
r=1
=
r = −1
5π
4
5π F
4
=
3π
2
3π
4
D 7π
4
(b) We use the graph (A) to find the sign of r D sin 2# : 0 # # # !2 ) r $ 0 ) .r; #/ is in the first quadrant. !2 # # # ! ) r # 0 ) .r; #/ is in the fourth quadrant. ! # # # 3!
2 ) r $ 0 ) .r; #/ is in the third quadrant.
3!
)
r
#
0
)
.r;
#/
is
in
the
second
quadrant.
That
is,
#
#
#
2!
2
S E C T I O N 11.3
y
3π
≤
2
≤ 2π
0≤
≤
Polar Coordinates
1405
π
2
x
π≤
≤
π
≤
2
3π
2
≤π
32. Sketch the curve r D sin 3#. First fill in the table of r-values
points of the curve. Notice that
) below
* ) and plot
* the corresponding
)! *
the three petals of the curve correspond to the angle intervals 0; !3 , !3 ; 2!
3 , and 3 ; ! . Then plot r D sin 3# in rectangular
coordinates and label the points on this graph corresponding to .r; #/ in the table.
#
0
!
12
!
6
!
4
!
3
5!
12
!!!
11!
12
!
r
SOLUTION
We compute the values of r corresponding to the given values of #:
r D sin 0 D 0
# D 0;
#D
#D
#D
#D
#D
#D
#D
#D
#D
#D
#D
!
;
12
!
;
6
!
;
4
!
;
3
5!
;
12
!
;
2
7!
;
12
3!
;
2
3!
;
4
5!
;
6
11!
;
12
# D !;
r D sin
r D sin
r D sin
r D sin
r D sin
r D sin
r D sin
r D sin
r D sin
r D sin
r D sin
3!
% 0:71
12
3!
D1
6
3!
% 0:71
4
3!
D0
3
15!
% "0:71
12
3!
D "1
2
21!
% "0:71
12
9!
D0
2
9!
% 0:71
4
15!
D1
6
33!
% 0:71
12
r D sin 3! D 0
.A/
.B/
.C /
.D/
.E/
.F /
.G/
.H /
.I /
.J /
.K/
.L/
.M /
We plot the points on the xy -plane and join them to obtain the following curve:
r
3
2
3
Using the graph of r D sin 3# we find the sign of r and determine the parts of the graph corresponding to the angle intervals.
We get
1406
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
3
5 4
6
11
12
7
2
12
3
2
≤ ≤
3
5
12
2
0≤
J
K
3
≤
4
3
6
C
D
I E
L
3
≤
≤
12
B
M
2
3
0
A
F
H
G
0## #
!
3
r D sin 3#
## #
2!
3
33.
!
3
2!
3
) r $ 0 ) .r; #/ in the first quadrant.
) r # 0 ) .r; #/ in the third and fourth quadrant.
# # # ! ) r $ 0 ) .r; #/ in the second quadrant.
Plot the cissoid r D 2 sin # tan # and show that its equation in rectangular coordinates is
x3
2"x
y2 D
SOLUTION
Using a CAS we obtain the following curve of the cissoid:
y
π
2
π
1
3 x
2
0π
3π
2
We substitute sin # D
y
r
and tan # D
y
x
in r D 2 sin # tan # to obtain
rD2
y y
! :
r x
Multiplying by rx, setting r 2 D x 2 C y 2 and simplifying, yields
r 2 x D 2y 2
.x 2 C y 2 /x D 2y 2
x 3 C y 2 x D 2y 2
y 2 .2 " x/ D x 3
so
x3
2"x
34. Prove that r D 2a cos # is the equation of the circle in Figure 6 using only the fact that a triangle inscribed in a circle with one
side a diameter is a right triangle.
y2 D
y
r
0
2a
FIGURE 6
x
S E C T I O N 11.3
Polar Coordinates
1407
SOLUTION Since the triangle inscribed in the circle has a diameter as one of its sides, it is a right triangle, so we may use the
definition of cosine for angles in right triangles to write
cos # D
r
) r D 2a cos #:
2a
35. Show that
r D a cos # C b sin #
is the equation of a circle passing through the origin. Express the radius and center (in rectangular coordinates) in terms of a and b.
SOLUTION
gives
We multiply the equation by r and then make the substitution x D r cos #, y D r sin #, and r 2 D x 2 C y 2 . This
r 2 D ar cos # C br sin #
x 2 C y 2 D ax C by
Transferring sides and completing the square yields
x 2 " ax C y 2 " by D 0
#
# $2 ! ! "
# $2
! a "2 $
a
b
b
a 2
b
x2 " 2 ! x C
C y2 " 2 ! y C
D
C
2
2
2
2
2
2
This is the equation of the circle with radius
the circle passes through the origin.
p
a2 Cb 2
2
#
$
!
a "2
b 2
a2 C b 2
x"
C y"
D
2
2
4
!
"
centered at the point a2 ; b2 . By plugging in x D 0 and y D 0 it is clear that
36. Use the previous exercise to write the equation of the circle of radius 5 and center .3; 4/ in the form r D a cos # C b sin #.
In the previous exercise we showed that r D a cos # C b sin # is the equation of the circle with radius
!
"
centered at a2 ; b2 . Thus, we must have
SOLUTION
The radius of the circle is
p
#
a2 Cb 2
2
SOLUTION
p
$
D .3; 4/ )
a2 Cb 2
2
a
b
D 3; D 4 ) a D 6; b D 8:
2
2
62 C82
D 5. Thus, the corresponding equation
2
" sin2 # to find a polar equation of the hyperbola
D
37. Use the identity cos 2# D cos2 #
a b
;
2 2
p
We substitute x D r cos #, y D r sin # in x 2 " y 2 D 1 to obtain
is r D 6 cos # C 8 sin #.
x 2 " y 2 D 1.
r 2 cos2 # " r 2 sin2 # D 1
r 2 .cos2 # " sin2 #/ D 1
Using the identity cos 2# D cos2 # " sin2 # we obtain the following equation of the hyperbola:
r 2 cos 2# D 1 or
r 2 D sec 2#:
38. Find an equation in rectangular coordinates for the curve r 2 D cos 2#.
We first use the identity cos 2# D cos2 # " sin2 # to rewrite the equation of the curve as follows: r 2 D cos2 # " sin2 # .
Multiplying by r 2 and substituting r 2 D x 2 C y 2 , r cos # D x and r sin # D y, we get
SOLUTION
r 4 D .r cos #/2 " .r sin #/2 .x 2 C y 2 /2 D x 2 " y 2 :
2
Thus, the curve has the equation .x 2 C y 2 / D x 2 " y 2 in rectangular coordinates.
39. Show that cos 3# D cos3 # " 3 cos # sin2 # and use this identity to find an equation in rectangular coordinates for the curve
r D cos 3#.
SOLUTION
to write
We use the identities cos.˛ C ˇ/ D cos ˛ cos ˇ " sin ˛ sin ˇ, cos 2˛ D cos2 ˛ " sin2 ˛, and sin 2˛ D 2 sin ˛ cos ˛
cos 3# D cos.2# C #/ D cos 2# cos # " sin 2# sin #
D .cos2 # " sin2 #/ cos # " 2 sin # cos # sin #
D cos3 # " sin2 # cos # " 2 sin2 # cos #
1408
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
D cos3 # " 3 sin2 # cos #
Using this identity we may rewrite the equation r D cos 3# as follows:
r D cos3 # " 3 sin2 # cos #
Since x D r cos # and y D r sin #, we have cos # D
x
r
and sin # D
rD
rD
! x "3
r
y
r.
(1)
Substituting into (1) gives:
! y "2 ! x "
"3
r
r
x 3 3y 2 x
" 3
r3
r
We now multiply by r 3 and make the substitution r 2 D x 2 C y 2 to obtain the following equation for the curve:
r 4 D x 3 " 3y 2 x
2
.x 2 C y 2 / D x 3 " 3y 2 x
40. Use the addition formula for the cosine to show that the line L with polar equation r cos.# " ˛/ D d has the equation in
rectangular coordinates .cos ˛/x C .sin ˛/y D d . Show that L has slope m D " cot ˛ and y-intercept d=sin ˛.
SOLUTION
We use the identity cos .a " b/ D cos a cos b C sin a sin b to rewrite the equation r cos .# " ˛/ D d as follows:
r .cos # cos ˛ C sin # sin ˛/ D d
r cos # cos ˛ C r sin # sin ˛ D d
We now substitute r cos # D x and r sin # D y to obtain: x cos ˛ C y sin ˛ D d . Dividing by cos ˛, transferring sides and simplifying yields
x C y tan ˛ D
d
cos ˛
y tan ˛ D "x C
yD"
d
cos ˛
x
d
C
tan ˛
tan ˛ cos ˛
so
y D ." cot ˛/ x C
d
sin ˛
This equation of the line implies that L has slope m D " cot ˛ and y -intercept
d
sin ˛ .
In Exercises 41–44, find an equation in polar coordinates of the line L with the given description.
%
&
41. The point on L closest to the origin has polar coordinates 2; !9 .
SOLUTION In Example 5, it is shown that the polar equation of the line where .r; ˛/ is the point on the line closest to the origin
%
&
is r D d sec .# " ˛/. Setting .d; ˛/ D 2; !9 we obtain the following equation of the line:
!
!"
r D 2 sec # "
:
9
42. The point on L closest to the origin has rectangular coordinates ."2; 2/.
SOLUTION
!
2 <˛ <
so
We first convert the rectangular coordinates ."2; 2/ to polar coordinates .d; ˛/. This point is in the second quadrant
!. Hence,
q
p
p
."2/2 C 22 D 8 D 2 2
# $
2
!
3!
˛ D tan!1
D tan!1 ."1/ D ! " D
"2
4
4
dD
p
Substituting d D 2 2 and ˛ D
3!
4
)
in the equation r D d sec .# " ˛/ gives us
#
$
p
3!
:
r D 2 2 sec # "
4
#
$
p 3!
.d; ˛/ D 2 2;
:
4
S E C T I O N 11.3
Polar Coordinates
1409
p
43. L is tangent to the circle r D 2 10 at the point with rectangular coordinates ."2; "6/.
SOLUTION
y
x
(−2, −6)
Since L is tangent to the circle at the point ."2; "6/, this is the point on L closest to the center of the circle which is at the origin.
Therefore, we may use the polar coordinates .d; ˛/ of this point in the equation of the line:
r D d sec .# " ˛/
We thus must convert the coordinates ."2; "6/ to polar coordinates. This point is in the third quadrant so ! < ˛ <
q
p
p
d D ."2/2 C ."6/2 D 40 D 2 10
# $
"6
˛ D tan!1
D tan!1 3 % ! C 1:25 % 4:39
"2
(1)
3!
2 .
We get
Substituting in (1) yields the following equation of the line:
p
r D 2 10 sec .# " 4:39/ :
44. L has slope 3 and is tangent to the unit circle in the fourth quadrant.
SOLUTION
We denote the point of tangency by P0 D .1; ˛/ (in polar coordinates).
(1, )
Since L is the tangent line to the circle at P0 , P0 is the point on L closest to the center of the circle at the origin. Thus, the polar
equation of L is
r D sec .# " ˛/
(1)
We now must find ˛. Let ˇ be the given angle shown in the figure.
O
2 −
B
C = (1, )
By the given information, tan ˇ D 3. Also, since the point of tangency is in the fourth quadrant, ˇ must be an acute angle. Hence
tan ˇ D 3; 0 < ˇ <
Now, since
3!
2
!
) ˇ D 1:25 rad:
2
< ˛ < 2!, we have for the triangle OBC
.2! " ˛/ C
!
3!
C 1:25 D ! ) ˛ D
C 1:25 D 5:96 rad:
2
2
1410
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
Substituting into (1) we obtain the following polar equation of the tangent line:
r D sec .# " 5:96/ :
45. Show that every line that does not pass through the origin has a polar equation of the form
b
sin # " a cos #
rD
where b ¤ 0.
SOLUTION Write the equation of the line in rectangular coordinates as y D ax C b. Since the line does not pass through the
origin, we have b ¤ 0. Substitute for y and x to convert to polar coordinates, and simplify:
y D ax C b
r sin # D ar cos # C b
r.sin # " a cos #/ D b
b
sin # " a cos #
46. By the Law of Cosines, the distance d between two points (Figure 7) with polar coordinates .r; #/ and .r0 ; #0 / is
rD
d 2 D r 2 C r02 " 2rr0 cos.# " #0 /
Use this distance formula to show that
!
!"
r 2 " 10r cos # "
D 56
4
%
&
is the equation of the circle of radius 9 whose center has polar coordinates 5; !4 .
y
(r, )
d
r
r0
0
(r0,
0)
x
FIGURE 7
%
&
The distance d between a point .r; #/ on the circle and the center .r0 ; #0 / D 5; !4 is the radius 9. Setting d D 9,
r0 D 5 and #0 D !4 in the distance formula we get
SOLUTION
d 2 D r 2 C r02 " 2rr0 cos .# " #0 /
!
!"
92 D r 2 C 52 " 2 ! r ! 5 cos # "
4
Transferring sides we get
!
!"
r 2 " 10r cos # "
D 56:
4
47. For a > 0, a lemniscate curve is the set of points P such that the product of the distances from P to .a; 0/ and ."a; 0/ is a2 .
Show that the equation of the lemniscate is
.x 2 C y 2 /2 D 2a2 .x 2 " y 2 /
Then find the equation in polar coordinates. To obtain the simplest form of the equation, use the identity cos 2# D cos2 # " sin2 #.
Plot the lemniscate for a D 2 if you have a computer algebra system.
SOLUTION
We compute the distances d1 and d2 of P .x; y/ from the points .a; 0/ and ."a; 0/ respectively. We obtain:
q
q
.x " a/2 C .y " 0/2 D .x " a/2 C y 2
q
q
d2 D .x C a/2 C .y " 0/2 D .x C a/2 C y 2
d1 D
For the points P .x; y/ on the lemniscate we have d1 d2 D a2 . That is,
q
q
q)
*)
*
a2 D .x " a/2 C y 2 .x C a/2 C y 2 D
.x " a/2 C y 2 .x C a/2 C y 2
S E C T I O N 11.3
Polar Coordinates
1411
q
.x " a/2 .x C a/2 C y 2 .x " a/2 C y 2 .x C a/2 C y 4
q
)
*
D .x 2 " a2 /2 C y 2 .x " a/2 C .x C a/2 C y 4
q
%
&
D x 4 " 2a2 x 2 C a4 C y 2 x 2 " 2xa C a2 C x 2 C 2xa C a2 C y 4
q
D x 4 " 2a2 x 2 C a4 C 2y 2 x 2 C 2y 2 a2 C y 4
q
D x 4 C 2x 2 y 2 C y 4 C 2a2 .y 2 " x 2 / C a4
q
2
D .x 2 C y 2 / C 2a2 .y 2 " x 2 / C a4 :
D
Squaring both sides and simplifying yields
a4 D .x 2 C y 2 /2 C 2a2 .y 2 " x 2 / C a4
0 D .x 2 C y 2 /2 C 2a2 .y 2 " x 2 /
so
.x 2 C y 2 /2 D 2a2 .x 2 " y 2 /
We now find the equation in polar coordinates. We substitute x D r cos #, y D r sin # and x 2 C y 2 D r 2 into the equation of the
lemniscate. This gives
.r 2 /2 D 2a2 .r 2 cos2 # " r 2 sin2 #/ D 2a2 r 2 .cos2 # " sin2 #/ D 2a2 r 2 cos 2#
r 4 D 2a2 r 2 cos 2#
r D 0 is a solution, hence the origin is on the curve. For r ¤ 0 we divide the equation by r 2 to obtain r 2 D 2a2 cos 2#. This curve
also includes the origin (r D 0 is obtained for # D !4 for example), hence this is the polar equation of the lemniscate. Setting a D 2
we get r 2 D 8 cos 2# .
2
0
r2 = 8 cos 2
3
2
48.
(a)
(b)
(c)
(d)
Let c be a fixed constant. Explain the relationship between the graphs of:
y D f .x C c/ and y D f .x/ (rectangular)
r D f .# C c/ and r D f .#/ (polar)
y D f .x/ C c and y D f .x/ (rectangular)
r D f .#/ C c and r D f .#/ (polar)
SOLUTION
(a) For c > 0, y D f .x C c/ shifts the graph of y D f .x/ by c units to the left. If c < 0, the result is a shift to the right. It is a
horizontal translation.
y
f (x + c)
f(x)
c
x
(b) As in part (a), the graph of r D f .# C c/ is a shift of the graph of r D f .#/ by c units in #. Thus, the graph in polar
coordinates is rotated by angle c as shown in the following figure:
1412
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
2
f( )
0
c
f ( + c)
3
2
(c) y D f .x/ C c shifts the graph vertically upward by c units if c > 0, and downward by ."c/ units if c < 0. It is a vertical
translation.
(d) The graph of r D f .#/ C c is a shift of the graph of r D f .#/ by c units in r. In the corresponding graph, in polar coordinates,
each point with f .#/ > 0 moves on the ray connecting it to the origin c units away from the origin if c > 0 and ."c/ units toward
the origin if c < 0, and vice-versa for f .#/ < 0.
2
y
c
c
1
c
0
c
x
c
3
2
c>0
49. The Derivative in Polar Coordinates
Show that a polar curve r D f .#/ has parametric equations
x D f .#/ cos #;
y D f .#/ sin #
Then apply Theorem 2 of Section 11.1 to prove
dy
f .#/ cos # C f 0 .#/ sin #
D
dx
"f .#/ sin # C f 0 .#/ cos #
2
where f 0 .#/ D df =d# .
SOLUTION Multiplying both sides of the given equation by cos # yields r cos # D f .#/ cos #; multiplying both sides by sin #
yields r sin # D f .#/ sin #. The left-hand sides of these two equations are the x and y coordinates in rectangular coordinates, so
for any # we have x D f .#/ cos # and y D f .#/ sin #, showing that the parametric equations are as claimed. Now, by the formula
for the derivative we have
dy
y 0 .#/
D 0
dx
x .#/
We differentiate the functions x D f .#/ cos # and y D f .#/ sin # using the Product Rule for differentiation. This gives
y 0 .#/ D f 0 .#/ sin # C f .#/ cos #
x 0 .#/ D f 0 .#/ cos # " f .#/ sin #
Substituting in (1) gives
dy
f 0 .#/ sin # C f .#/ cos #
f .#/ cos # C f 0 .#/ sin #
D 0
D
:
dx
f .#/ cos # " f .#/ sin #
"f .#/ sin # C f 0 .#/ cos #
(1)
S E C T I O N 11.3
Polar Coordinates
1413
Further Insights and Challenges
50.
Let f .x/ be a periodic function of period 2!—that is, f .x/ D f .x C 2!/. Explain how this periodicity is reflected in
the graph of:
(a) y D f .x/ in rectangular coordinates
(b) r D f .#/ in polar coordinates
SOLUTION
(a) The graph of y D f .x/ on an interval of length 2! repeats itself on successive intervals of length 2!. For instance:
y
8
6
4
2
2
−12−10 −8 −6 −4 −2
−2
4
6
t
8 10 12
−4
−6
−8
(b) Shown below is the graph of the function above, this time drawn in polar coordinates. The graphs of the various branches repeat
themselves and are drawn one on the top of the other.
y
π
2
π
2
4
6
8 x
0π
3π
2
51.
Use a graphing utility to convince yourself that the polar equations r D f1 .#/ D 2 cos # " 1 and r D f2 .#/ D
2 cos # C 1 have the same graph. Then explain why. Hint: Show that the points .f1 .# C !/; # C !/ and .f2 .#/; #/ coincide.
SOLUTION
CAS.
The graphs of r D 2 cos # " 1 and r D 2 cos # C 1 in the xy -plane coincide as shown in the graph obtained using a
y
π
2
y
2
π
2
−2
1
2
3 x
0π
x
−2
3π
2
Recall that .r; #/ and ."r; # C !/ represent the same point. Replacing # by # C ! and r by ."r/ in r D 2 cos # " 1 we obtain
"r D 2 cos .# C !/ " 1
"r D "2 cos # " 1
r D 2 cos # C 1
Thus, the two equations define the same graph. (One could also convert both equations to rectangular coordinates and note that they
come out identical.)
1414
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
11.4 Area, Arc Length, and Slope in Polar Coordinates
Preliminary Questions
1. Polar coordinates are suited to finding the area (choose one):
(a) Under a curve between x D a and x D b.
(b) Bounded by a curve and two rays through the origin.
SOLUTION Polar coordinates are best suited to finding the area bounded by a curve and two rays through the origin. The formula
for the area in polar coordinates gives the area of this region.
2. Is the formula for area in polar coordinates valid if f .#/ takes negative values?
SOLUTION
The formula for the area
1
2
Z
ˇ
f .#/2 d#
˛
always gives the actual (positive) area, even if f .#/ takes on negative values.
3. The horizontal line y D 1 has polar equation r D csc #. Which area is represented by the integral
(a) !ABCD
(b) 4ABC
1
2
(c) 4ACD
Z
!=2
csc2 # d# (Figure 1)?
!=6
y
D
1
C y=1
A
B
x
!3
FIGURE 1
SOLUTION
part (c).
This integral represents an area taken from # D !=6 to # D !=2, which can only be the triangle 4ACD, as seen in
Exercises
1. Sketch the area bounded by the circle r D 5 and the rays # D
coordinates.
!
2
and # D !, and compute its area as an integral in polar
!
SOLUTION The region bounded by the circle r D 5 and the rays # D 2 and # D ! is the shaded region in the figure. The area
of the region is given by the following integral:
Z
Z
1 ! 2
1 ! 2
25 !
!"
25!
r d# D
5 d# D
!"
D
2 !=2
2 !=2
2
2
4
π
=
2
y
=π
x
2. Sketch the region bounded by the line r D sec # and the rays # D 0 and # D
in polar coordinates and using geometry.
SOLUTION
!
3.
The region bounded by the line r D sec # and the rays # D 0 and # D
Compute its area in two ways: as an integral
!
3
is shown here:
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1415
2
=
3
r = sec
=0
x=1
Using the area in polar coordinates, the area of the region is given by the following integral:
1
AD
2
Z
0
!=3
1
r d# D
2
2
Z
!=3
0
p
ˇ!=3
"
ˇ
1
1!
!
3
ˇ
sec # d# D tan # ˇ
D
tan " tan 0 D
2
2
3
2
0
2
We now compute the area using the formula for the area of a triangle. The equations of the lines # D !3 , # D 0, and r D sec # in
p
rectangular coordinates are y D 3x, y D 0 and x D 1 respectively (see Example 5 in Section 11.3
the equation of the line
! for
p "
r D sec #). Denoting the vertices of the triangle by O, A, B (see figure) we have O D .0; 0/, A D 1; 3 and B D .1; 0/. The
area of the triangle is thus
p
p
OB ! AB
1! 3
3
AD
D
D
:
2
2
2
y
y = 3x
x=1
A
y=0
O
x
B
3. Calculate the area of the circle r D 4 sin # as an integral in polar coordinates (see Figure 4). Be careful to choose the correct
limits of integration.
SOLUTION
The equation r D 4 sin # defines a circle of radius 2 tangent to the x-axis at the origin as shown in the figure:
π
=
2
y
2π
3
5π
6
=π
π
3
π
6
x
=π
The circle is traced as # varies from 0 to !. We use the area in polar coordinates and the identity
sin2 # D
1
.1 " cos 2#/
2
to obtain the following area:
'
(
Z
Z
Z !
Z !
1 ! 2
1 !
sin 2# !
AD
r d# D
.4 sin #/2 d# D 8
sin2 # d# D 4
.1 " cos 2#/ d# D 4 # "
2 0
2 0
2
0
0
0
##
$
$
sin 2!
D4
!"
" 0 D 4!:
2
4. Find the area of the shaded triangle in Figure 2 as an integral in polar coordinates. Then find the rectangular coordinates of P
and Q and compute the area via geometry.
1416
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
y
P
(
r = 4 sec θ −
π
4
)
x
Q
FIGURE 2
SOLUTION
The boundary of the region is traced as # varies from 0 to
!
2,
so the area is
ˇ
!
!
!"
! " ˇˇ!=2
16 sec2 # "
d# D 8 tan # "
D 8.1 C 1/ D 16
4
4 ˇ0
0
0
%
&
%
&
We now compute the area using the formula for area of a triangle. The line 4 sec # " !4 has a point closest to the origin at 4; !4 ,
! p p "
which is 2 2; 2 2 in rectangular coordinates. Since the polar line # D !4 (x D y in rectangular coordinates) is perpendicular to
!
p
p "
p
the line, the line must have a slope of "1. So the rectangular equation for the line is y " 2 2 D "1 x " 2 2 ; or y D "x C 4 2.
! p "
! p "
p p
2
Therefore, P is 0; 4 2 ; Q is 4 2; 0 ; and the area of the triangle is A D OP2"OQ D 4 2"4
D 16.
2
1
2
Z
!=2
r 2 d# D
1
2
Z
!=2
5. Find the area of the shaded region in Figure 3. Note that # varies from 0 to
!
2.
y
8
r = θ 2 + 4θ
x
1
2
FIGURE 3
SOLUTION
Since # varies from 0 to
1
2
Z
!=2
0
!
2,
the area is
Z
1 !=2 4
# C 8# 3 C 16# 2 d#
2 0
0
#
$ ˇ!=2
ˇ
1 1 5
16
!5
!4
!2
D
# C 2# 4 C # 3 ˇˇ
D
C
C
2 5
3
320
16
3
0
r 2 d# D
1
2
Z
!=2
.# 2 C 4#/2 d# D
6. Which interval of #-values corresponds to the the shaded region in Figure 4? Find the area of the region.
y
2
r=3−θ
x
3
FIGURE 4
SOLUTION
from 0 to 3.
We first find the interval of #. At the origin r D 0, so # D 3. At the endpoint on the x-axis, # D 0. Thus, # varies
S E C T I O N 11.4
Area, Arc Length, and Slope in Polar Coordinates
1417
y
2
r=3−
=3
=3
=0
0
x
3
Using the area in polar coordinates we obtain
ˇ
Z
Z
1 3 2
1 3
.3 " #/3 ˇˇ3
AD
r d# D
.3 " #/2 d# D "
ˇ D 4:5:
2 0
2 0
6
0
7. Find the total area enclosed by the cardioid in Figure 5.
y
x
−2
−1
FIGURE 5 The cardioid r D 1 " cos #.
SOLUTION
We graph r D 1 " cos # in r and # (cartesian, not polar, this time):
r
2
1
π
2
3π
2
π
2π
We see that as # varies from 0 to !, the radius r increases from 0 to 2, so we get the upper half of the cardioid (the lower half is
obtained as # varies from ! to 2! and consequently r decreases from 2 to 0). Since the cardioid is symmetric with respect to the
x-axis we may compute the upper area and double the result. Using
cos2 # D
cos 2# C 1
2
we get
Z
Z !
Z !!
"
1 ! 2
r d# D
.1 " cos #/2 d# D
1 " 2 cos # C cos2 # d#
2 0
0
0
$
$
Z !#
Z !#
cos 2# C 1
3
1
D
1 " 2 cos # C
d# D
" 2 cos # C cos 2# d#
2
2
2
0
0
AD2!
ˇ!
ˇ
3
1
3!
D # " 2 sin # C sin 2# ˇˇ D
2
4
2
0
The total area enclosed by the cardioid is A D 3!
2 .
8. Find the area of the shaded region in Figure 5.
SOLUTION The shaded region is traced as # varies from 0 to
AD
1
2
Z
!=2
0
r 2 d# D
1
2
Z
!=2
0
!
2.
Using the formula for the area in polar coordinates we get:
.1 " cos #/2 d# D
1
2
Z
0
!=2 !
"
1 " 2 cos # C cos2 # d#
#
$
#
$
Z
1 !=2
cos 2# C 1
1 !=2 3
1
D
1 " 2 cos # C
d# D
" 2 cos # C cos 2# d#
2 0
2
2 0
2
2
Z
D
1
2
D
1
2
#
#
$ ˇ!=2
$
$
##
ˇ
1
1
3 !
!
3#
1
D
" 2 sin # C sin 2# ˇˇ
! " 2 sin C sin ! " 0
2
4
2
2 2
2
4
0
$
3!
3!
"2 D
" 1 % 0:18
4
8
1418
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
9. Find the area of one leaf of the “four-petaled rose” r D sin 2# (Figure 6). Then prove that the total area of the rose is equal to
one-half the area of the circumscribed circle.
y
r = sin 2θ
x
FIGURE 6 Four-petaled rose r D sin 2#.
SOLUTION
We consider the graph of r D sin 2# in cartesian and in polar coordinates:
y
r
r = 1, θ =
1
A
π
4
A
x
π
4
π
2
3π
4
π
−1
We see that as # varies from 0 to !4 the radius r is increasing from 0 to 1, and when # varies from !4 to !2 , r is decreasing back to
zero. Hence, the leaf in the first quadrant is traced as # varies from 0 to !2 . The area of the leaf (the four leaves have equal areas) is
thus
Z
Z
1 !=2 2
1 !=2 2
AD
r d# D
sin 2# d#:
2 0
2 0
Using the identity
sin2 2# D
1 " cos 4#
2
we get
AD
1
2
Z
!=2 # 1
0
2
"
cos 4#
2
$
d# D
1
2
#
#
sin 4#
"
2
8
$ ˇ!=2
##
$
$
ˇ
1
!
sin 2!
!
ˇ
D
"
"
0
D
ˇ
2
4
8
8
0
The area of one leaf is A D !8 % 0:39. It follows that the area of the entire rose is !2 . Since the “radius” of the rose (the point where
# D !4 ) is 1, and the circumscribed circle is tangent there, the circumscribed circle has radius 1 and thus area !. Hence the area of
the rose is half that of the circumscribed circle.
10. Find the area enclosed by one loop of the lemniscate with equation r 2 D cos 2# (Figure 7). Choose your limits of integration
carefully.
y
−1
1
x
FIGURE 7 The lemniscate r 2 D cos 2#.
SOLUTION
%
&
We sketch the graph of r 2 D cos 2# in the r 2 ; # plane; for " !4 # # #
r2
1
−π
4
π 1
4
!
4:
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1419
We see that as # varies from " !4 to 0, r 2 increases from 0 to 1, hence r also increases from 0 to 1. Then, as # varies from 0 to
r 2 , so r decreases from 1 to 0. This gives the right-hand loop of the lemniscate.
y
r = 0, r =
π
4
π
4
θ=0
r=1
r = 0, r =
!
4,
x
π
4
Therefore, the area enclosed by the right-hand loop is:
ˇ
Z
Z
! ! ""
1 !=4 2
1 !=4
1 sin 2# ˇˇ!=4
1! !
1
r d# D
cos 2# d# D
D
sin " sin "
D
ˇ
2 !!=4
2 !!=4
2 2 !!=4
4
2
2
2
11. Sketch the spiral r D # for 0 # # # 2! and find the area bounded by the curve and the first quadrant.
SOLUTION
The spiral r D # for 0 # # # 2! is shown in the following figure in the xy-plane:
y
= /2,
r = /2
= ,
r=
= 0,
r=0
=2 ,
r=2
x
The spiral r D #
We must compute the area of the shaded region. This region is traced as # varies from 0 to !2 . Using the formula for the area in
polar coordinates we get
ˇ
Z
Z
1 !=2 2
1 !=2 2
1 # 3 ˇˇ!=2
1 ! ! "3
!3
AD
r d# D
# d# D
D
D
ˇ
2 0
2 0
2 3 0
6 2
48
12. Find the area of the intersection of the circles r D sin # and r D cos #.
SOLUTION
The region of intersection between the two circles is shown in the following figure:
y
r = sin
1
2
x
1
2
r = cos
We first find the value of # at the point of intersection (besides the origin) of the two circles, by solving the following equation for
0 # a # !2 :
sin # D cos #
tan # D 1 ) # D
!
4
We now compute the area as the sum of the two areas A1 and A2 , shown in the figure:
2
r = sin
1
2
=
A1
A2
1
2
r = cos
4
0
1420
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
Using the formula for the area in polar coordinates we get
$
Z
1
1 !=2
cos 2# d# D
.1 C cos 2#/ d#
2
2
4 !=2
!=4
!=2
#
$ˇ
##
$ #
$$
#
$
sin !2
1
sin 2# ˇˇ!=2
1
!
sin !
!
1 !
!
1
!
1
D
#C
D
C
"
C
D
"
"
D
"
ˇ
4
2
4
2
2
4
2
4 2
4
2
16 8
!=2
#
$
Z
Z
Z
1 !=4 2
1 !=4 1 1
1 !=4
A2 D
sin # d# D
" cos 2# d# D
.1 " cos 2#/ d#
2 0
2 0
2 2
4 0
#
$ˇ
##
$
$
sin !2
1
sin 2# ˇˇ!=4
1
!
!
1
D
#"
D
"
"
0
D
"
ˇ
4
2
4
4
2
16
8
0
A1 D
1
2
Z
!=2
cos2 # d# D
1
2
Z
!=2 # 1
C
Notice that A2 D A1 as shown in the figure due to symmetry. The total area enclosed by the two circles is the sum
#
$ #
$
!
1
!
1
!
1
A D A1 C A2 D
"
C
"
D " % 0:14:
16 8
16 8
8
4
13. Find the area of region A in Figure 8.
y
r = 4 cos
r=1
A
1
−1
2
4
x
FIGURE 8
SOLUTION
" !2 # x #
We first find the values of # at the points of intersection of the two circles, by solving the following equation for
!
2:
1
4 cos # D 1 ) cos # D ) #1 D cos!1
4
# $
1
4
y
= 1.32
r = 4 cos
r=1
x
= −1.32
We now compute the area using the formula for the area between two curves:
AD
Using the identity cos2 # D
1
2
cos 2" C1
2
Z
"1
!"1
!
Z
"
"
1 "1 !
.4 cos #/2 " 12 d# D
16 cos2 # " 1 d#
2 !"1
we get
ˇ"1
$
Z
ˇ
16 .cos 2# C 1/
1 "1
1
" 1 d# D
.8 cos 2# C 7/ d# D .4 sin 2# C 7#/ ˇˇ
2
2 !"1
2
!"1
!"1
p
D 4 sin 2#1 C 7#1 D 8 sin #1 cos #1 C 7#1 D 8 1 " cos2 #1 cos #1 C 7#1
AD
1
2
Z
"1
Using the fact that cos #1 D
#
1
4
we get
AD
p
15
C 7cos!1
2
# $
1
% 11:163
4
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1421
14. Find the area of the shaded region in Figure 9, enclosed by the circle r D 12 and a petal of the curve r D cos 3#. Hint: Compute
the area of both the petal and the region inside the petal and outside the circle.
y
r = cos 3
x
r=
1
2
FIGURE 9
SOLUTION
We compute the area A of the given region as the difference between the area A1 of the leaf, shown here:
2
r=
A1
0.5
0
3
2
The area, A2 , of the region inside the leaf and outside the circle, shown here:
2
r=
A2
0.5
0
3
2
Computing A1 : To determine the limits of integration we use the following graph of r D cos 3#:
r
1
−
2
−
3
−
1
2
6
−
6
9
3
2
9
−1
r D cos 3#
As # varies from " !6 to 0, r increases from 0 to 1. Then, as # varies from 0 to !6 , r decreases from 1 back to zero. Hence the leaf
is traced as # varies from " !6 to !6 . We use the formula for the area in polar coordinates to obtain
#
$
Z
Z
1 !=6 1
1
1 !=6
cos 3# d# D
C cos 6# d# D
.1 C cos 6#/ d#
2 !!=6 2
2
4 !!=6
!!=6
#
$ˇ
##
$ #
$$
1
sin 6# ˇˇ!=6
1
!
sin !
!
sin ."!/
1 2!
!
D
#C
D
C
" " C
D !
D
ˇ
4
6
4
6
6
6
6
4
6
12
!!=6
1
A1 D
2
Z
!=6
2
Computing A2 : The two curves intersect at the points where cos 3# D 12 , that is, # D ˙ !9 (see the graph of r D cos 3# in the
r# -plane). Using the formula for the area between two curves we get
# $2 !
#
$
Z
Z
1 !=9
1
1 !=9 1
1
1
2
A2 D
cos 3# "
d# D
C cos 6# "
d#
2 !!=9
2
2 !!=9 2
2
4
1422
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
D
D
1
8
Z
!=9
!!=9
#
$ˇ
1
sin 6# ˇˇ!=9
#C
ˇ
8
3
!!=9
!
"
11
! 0
p !
p
sin " 6!
!
9
1
!
3
!
3
AA D
" @" C
C
D
C
9
3
4 9
6
36
24
.1 C 2 cos 6#/ d# D
0
sin 6!
1@ !
9
C
8
9
3
The required area is the difference between A1 and A2 , that is,
!
A D A1 " A2 D
"
12
p !
p
!
3
!
3
C
D
"
% 0:102:
36
24
18
24
15. Find the area of the inner loop of the limaçon with polar equation r D 2 cos # " 1 (Figure 10).
y
1
1
x
2
−1
FIGURE 10 The limaçon r D 2 cos # " 1.
SOLUTION
We consider the graph of r D 2 cos # " 1 in cartesian and in polar, for " !2 # x #
!
2:
y
π
3
r
1
x
π
−
2
π
−
3
π
3
π
2
−1
−
π
3
r D 2 cos # " 1
As # varies from " !3 to 0, r increases from 0 to 1. As # varies from 0 to !3 , r decreases from 1 back to 0. Hence, the inner loop of
the limaçon is traced as # varies from " !3 to !3 . The area of the inner loop is thus
AD
D
1
2
1
2
Z
!=3
!!=3
Z
r 2 d# D
1
2
Z
!=3
!!=3
.2 cos # " 1/2 d# D
!=3
!!=3
.2 .cos 2# C 1/ " 4 cos # C 1/ d# D
1
2
1
2
Z
Z
!=3
!!=3
!=3
!!=3
!
"
4 cos2 # " 4 cos # C 1 d#
.2 cos 2# " 4 cos # C 3/ d#
ˇ!=3
##
$ # #
$
$$
! !"
ˇ
1
1
2!
!
2!
D .sin 2# " 4 sin # C 3#/ ˇˇ
D
sin
" 4 sin C ! " sin "
" 4 sin "
"!
2
2
3
3
3
3
!!=3
p
p
p
3 4 3
3 3
D
"
C! D!"
% 0:54
2
2
2
16. Find the area of the shaded region in Figure 10 between the inner and outer loop of the limaçon r D 2 cos # " 1.
SOLUTION
The region is shown in the figure below.
y
1
1
−1
We use the following graph.
2
3
x
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1423
r
2
−
1
3
3
−
−1
−2
−3
Graph of r D 2 cos # " 1
As # varies from !3 to !, r is negative and jrj increases from 0 to 3. This gives the outer loop of the limaçon which is in the lower
half plane. Similarly, the outer loop which is in the upper half plane is traced for "! # # # " !3 .
y − ≤
≤ − , −3 ≤ r ≤ 0
3
1
=− ,r=0
= − , r = −3
x
3 = , r = −3
3
=
3
1
,r=0
2
−1
3
≤ , −3 ≤ r ≤ 0
≤
Using symmetry with respect to the x-axis, we obtain the following for the area of the outer loop:
Z
Z !
Z ! !
"
1 ! 2
AD2!
r d# D
.2 cos # " 1/2 d# D
4 cos2 # " 4 cos # C 1 d#
2 !=3
!=3
!=3
ˇ!
Z !
Z !
ˇ
D
.2 .1 C cos 2#/ " 4 cos # C 1/ d# D
.2 cos 2# " 4 cos # C 3/ d# D sin 2# " 4 sin # C 3# ˇˇ
!=3
!=3
#
D .sin 2! " 4 sin ! C 3!/ " sin
2!
!
" 4 sin C !
3
3
!=3
$
!
p
p
p
3
3 3
" 2 3 C ! D 2! C
2
2
D 3! "
Finally, to find the area of the region between the inner and outer loop of the limaçon, we subtract the area of the inner loop,
obtained in the previous exercise, from the area of the outer loop:
p !
p !
p
3 3
3 3
2! C
" !"
D! C3 3
2
2
17. Find the area of the part of the circle r D sin # C cos # in the fourth quadrant (see Exercise 26 in Section 11.3).
SOLUTION
The value of # corresponding to the point B is the solution of r D sin # C cos # D 0 for "! # # # !.
y
B
A
C
x
That is,
sin # C cos # D 0 ) sin # D " cos # ) tan # D "1 ) # D "
!
4
At the point C , we have # D 0. The part of the circle in the fourth quadrant is traced if # varies between " !4 and 0. This leads to
the following area:
Z
Z
Z
"
1 0
1 0 ! 2
1 0
r 2 d# D
.sin # C cos #/2 d# D
sin # C 2 sin # cos # C cos2 # d#
AD
2 !!=4
2 !!=4
2 !!=4
Using the identities sin2 # C cos2 # D 1 and 2 sin # cos # D sin 2# we get:
#
$ˇ
Z
1 0
1
cos 2# ˇˇ0
AD
.1 C sin 2#/ d# D
#"
ˇ
2 !!=4
2
2
!!=4
1424
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
% & !!
#
$
cos !!
!
1 !
1
!
1
2
" " "
D
"
D " % 0:14:
4
2
2 4
2
8
4
%
&
%
&
!
18. Find the area of the region inside the circle r D 2 sin # C 4 and above the line r D sec # " !4 .
%
&
%
&
!
!
SOLUTION The line r D sec # " 4 intersects the circle r D 2 sin # C 4 when # D 0 and # D 2!.
1
D
2
#
1
0"
2
$
y
2 θ=π
2
1.5
1
0.5
θ=0
−0.5
0
0.5
1
1.5
x
2
−0.5
Thus the area of the region inside the circle and above the line is
1
2
Z
!=2 #!
0
$
Z
!
!
!
! ""2 ! !
! ""2
1 !=2
!"
!"
2 sin # C
" sec # "
d# D
4 sin2 # C
" sec2 # "
d#
4
4
2 0
4
4
ˇ
!
!
!
1!
!"
!"
! "" ˇˇ!=2
D
2t " 2 sin t C
cos t C
" tan t "
2
4
4
4 ˇ0
#
# $
# $
!
"
!
!! "
!! "
! ! ""$
1
3!
3!
!
D
! " 2 sin
cos
" tan
" "2 sin
cos
" tan "
2
4
4
4
4
4
4
1
!
.! C 1 " 1 C 1 " 1/ D
2
2
19. Find the area between the two curves in Figure 11(A).
D
y
y
r = 2 + sin 2
r = 2 + cos 2
r = sin 2
x
x
r = sin 2
(A)
(B)
FIGURE 11
SOLUTION We compute the area A between the two curves as the difference between the area A1 of the region enclosed in the
outer curve r D 2 C cos 2# and the area A2 of the region enclosed in the inner curve r D sin 2#. That is,
A D A1 " A2 :
y
r = 2 + 2cos2
r = sin2
A
A2
In Exercise 9 we showed that A2 D
!
2,
hence,
A D A1 "
We compute the area A1 .
x
!
2
(1)
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1425
y
A1
x
Using symmetry, the area is four times the area enclosed in the first quadrant. That is,
1
2
Z
Using the identity cos2 2# D
1
2
A1 D 4 !
!=2
0
r 2 d# D 2
cos 4# C
1
2
Z
!=2
0
Z
.2 C cos 2#/2 d# D 2
!=2 !
0
"
4 C 4 cos 2# C cos2 2# d#
we get
$
$
Z !=2 #
!=2 #
1
1
9
1
A1 D 2
4 C 4 cos 2# C cos 4# C
d# D 2
C cos 4# C 4 cos 2# d#
2
2
2
2
0
0
#
$ ˇ!=2
##
$
$
ˇ
9#
sin 4#
9!
sin 2!
9!
D2
C
C 2 sin 2# ˇˇ
D2
C
C 2 sin ! " 0 D
2
8
4
8
2
0
Z
(2)
Combining (1) and (2) we obtain
AD
9!
!
" D 4!:
2
2
20. Find the area between the two curves in Figure 11(B).
SOLUTION
Since
!
!
!!
"
!"
!"
2 C cos 2 # "
D 2 C cos 2# "
D 2 C cos
" 2# D 2 C sin 2#
4
2
2
it follows that the curve r D 2 C sin 2# is obtained by rotating the curve r D 2 C cos 2# by !4 about the origin. Therefore the
area between the curves r D 2 C sin 2# and r D sin 2# is the same as the area between the curves r D 2 C cos # and r D sin 2#
computed in Exercise 19. That is, A D 4!. (Notice that if the inner curve remains inside the rotated curve, the area between the
curves is not changed).
21. Find the area inside both curves in Figure 12.
y
2 + sin 2
x
2 + cos 2
FIGURE 12
SOLUTION
The area we need to find is the area of the shaded region in the figure.
y
r = 2 + sin 2
D
A
x
C
B
r = 2 + cos 2
We first find the values of # at the points of intersection A, B, C , and D of the two curves, by solving the following equation for
"! # # # !:
2 C cos 2# D 2 C sin 2#
cos 2# D sin 2#
tan 2# D 1 ) 2# D
!
!
!k
C !k ) # D C
4
8
2
1426
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
The solutions for "! # # # ! are
!
:
8
3!
BW # D " :
8
7!
CW # D " :
8
5!
DW # D
:
8
AW
#D
Using symmetry, we compute the shaded area in the figure below and multiply it by 4:
π
2
5π
8
A1
π
8
0π
π
r = 2 + cos 2
π
2
−
A D 4 ! A1 D 4 !
Z
5!=8 #
1
!
2
Z
5!=8
!=8
.2 C cos 2#/2 d# D 2
$
Z
5!=8 !
!=8
Z
"
4 C 4 cos 2# C cos2 2# d#
5!=8
1 C cos 4#
d# D
.9 C 8 cos 2# C cos 4#/ d#
2
!=8
!=8
ˇ
#
$
#
$
#
$
p
sin 4# ˇˇ5!=8
5!
!
5!
!
1
5!
!
9!
D 9# C 4 sin 2# C
D
9
"
C
4
sin
"
sin
C
sin
"
sin
D
"4 2
ˇ
4 !=8
8
8
4
4
4
2
2
2
D2
4 C 4 cos 2# C
22. Find the area of the region that lies inside one but not both of the curves in Figure 12.
SOLUTION
The area we need to find is the area of the shaded region in the following figure:
y
r = 2 + sin 2θ
A1
x
r = 2 + cos 2θ
We denote by A1 the area inside both curves. In Exercise 20 we showed that the curve r D 2 C sin 2# is obtained by rotating the
curve r D 2 C cos 2# by !4 around the origin. Hence, the areas enclosed in these curves are equal. We denote it by A2 . It follows
that the area A that we need to find is
A D 2A2 " 2A1 D 2 .A2 " A1 /
In Exercise 20 we found that A2 D
9!
2 ,
(1)
p
" 4 2. Substituting in (1) we obtain
and in Exercise 21 we showed that A1 D 9!
2
#
#
$$
p
p
9!
9!
AD2
"
"4 2
D 8 2 % 11:3:
2
2
23. Calculate the total length of the circle r D 4 sin # as an integral in polar coordinates.
SOLUTION
We use the formula for the arc length:
SD
Z
ˇ
˛
q
f .#/2 C f 0 .#/2 d#
(1)
In this case, f .#/ D 4 sin # and f 0 .#/ D 4 cos # , hence
q
q
p
f .#/2 C f 0 .#/2 D .4 sin #/2 C .4 cos #/2 D 16 D 4
The circle is traced as # is varied from 0 to !. Substituting ˛ D 0, ˇ D ! in (1) yields S D
R!
0
4 d# D 4!.
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1427
y
2
x
The circle r D 4 sin #
24. Sketch the segment r D sec # for 0 # # # A. Then compute its length in two ways: as an integral in polar coordinates and
using trigonometry.
SOLUTION
The line r D sec # has the rectangular equation x D 1. The segment AB for 0 # # # A is shown in the figure.
y
C
sec A
A
D
x
1
Using trigonometry, the length of the segment AB is
L D AB D 0B tan A D 1 ! tan A D tan A
Alternatively, we use the integral in polar coordinates with f .#/ D sec.#/ and f 0 .#/ D tan # sec # . This gives
LD
Z
Aq
.sec #/2 C .tan # sec #/2 d# D
0
Z
Ap
0
1 C tan2 # sec # d# D
Z
A
0
The two answers agree, as expected.
ˇA
ˇ
sec2 # d# D tan # ˇˇ D tan A:
0
In Exercises 25–30, compute the length of the polar curve.
25. The length of r D # 2 for 0 # # # !
SOLUTION
We use the formula for the arc length. In this case f .#/ D # 2 , f 0 .#/ D 2#, so we obtain
Z !q
Z !p
Z ! p
% &2
SD
# 2 C .2#/2 d# D
# 4 C 4# 2 d# D
# # 2 C 4 d#
0
0
0
We compute the integral using the substitution u D # 2 C 4, du D 2# d#. This gives
1
SD
2
Z
! 2 C4 p
4
ˇ 2
#
$
#
$
"3=2
"3=2
1 2 3=2 ˇˇ! C4
1 ! 2
1 ! 2
3=2
u du D ! u ˇ
D
! C4
"4
D
! C4
" 8 % 14:55
2 3
3
3
4
26. The spiral r D # for 0 # # # A
SOLUTION
We use the formula for the arc length. In this case f .#/ D #, f 0 .#/ D 1. Using integration formulas we get:
SD
Z
Ap
0
# 2 C 12 d# D
Z
Ap
0
# 2 C 1 d# D
p
Ap 2
1
D
A C 1 C ln jA C A2 C 1j
2
2
ˇA
p
ˇ
#p 2
1
# C 1 C ln j# C # 2 C 1jˇˇ
2
2
0
y
x
The spiral r D #
1428
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
27. The equiangular spiral r D e " for 0 # # # 2!
SOLUTION
Since f .#/ D e " , by the formula for the arc length we have:
LD
Z
2!
0
Z
q
f 0 .#/2 C f .#/ d# C
Z
q
% &2 % &2
e " C e " d# D
2!
0
p Z
D 2
2!
0
p
2e 2" d#
ˇ
" p !
"
p " ˇ2!
p !
"
e d# D 2e ˇˇ D 2 e 2! " e 0 D 2 e 2! " 1 % 755:9
2!
0
0
28. The inner loop of r D 2 cos # " 1 in Figure 10
SOLUTION
Also,
In Exercise 15 it is shown that the inner loop of the limaçon r D 2 cos # " 1 is traced as # varies from " !3 to
!
3.
f 0 .#/ D "2 sin #:
f .#/ D 2 cos # " 1 and
Using the integral for the arc length we obtain
LD
D
Z
!=3
!!=3
Z
!=3
!!=3
Z
q
f .#/2 C f 0 .#/2 d# D
q
.2 cos # " 1/2 C ."2 sin #/2 d#
!=3
!!=3
Z
p
4 cos2 # " 4 cos # C 1 C 4 sin2 # d# D
!!=3
29. The cardioid r D 1 " cos # in Figure 5
SOLUTION
!=3
p
5 " 4 cos # d#
In the equation of the cardioid, f .#/ D 1 " cos # . Using the formula for arc length in polar coordinates we have:
Z
LD
ˇ
˛
q
f .#/2 C f 0 .#/2 d#
(1)
We compute the integrand:
q
q
p
p
f .#/2 C f 0 .#/2 D .1 " cos #/2 C .sin #/2 D 1 " 2 cos # C cos2 # C sin2 # D 2 .1 " cos #/
We identify the interval of #. Since "1 # cos # # 1, every 0 # # # 2! corresponds to a nonnegative value of r. Hence, # varies
from 0 to 2!. By (1) we obtain
Z 2! p
LD
2.1 " cos #/ d#
0
q
p
Now, 1 " cos # D 2 sin2 .#=2/, and on the interval 0 # # # !, sin.#=2/ is nonnegative, so that 2.1 " cos #/ D 4 sin2 .#=2/ D
2 sin.#=2/ there. The graph is symmetric, so it suffices to compute the integral for 0 # # # !, and we have
ˇ
Z !p
Z !
# ˇˇ!
LD2
2.1 " cos #/ d# D 2
2 sin.#=2/ d# D 8 sin ˇ D 8
2 0
0
0
30. r D cos2 #
Since cos # D cos ."#/ and cos2 .! " #/ D cos2 # the curve is symmetric with respect to the x and y-axis. Therefore, we may compute the length as four times the length of the part of the curve in the first quadrant. We use the formula for the
arc length in polar coordinates. In this case, f .#/ D cos2 #; f 0 .#/ D 2 cos # ." sin #/, so we obtain
q
p
p
f .#/2 C f 0 .#/2 D cos4 # C 4 cos2 # sin2 # D cos # cos2 #C4 sin2 #
p
p
D cos # cos2 # C sin2 # C 3 sin2 # D cos # 1 C 3 sin2 #
SOLUTION
Thus,
LD
Z
!=2 q
f .#/2 C f 0 .#/2 d# D
0
We compute the integral using the substitution u D
p
Z
!=2
0
cos #
p
1 C 3 sin2 # d#:
3 sin # we get
# p
$ ˇ p3
p
ˇ
u
1
2
2
1 C u C ln ju C 1 C u j ˇˇ
2
2
0
0
!
p
!
!
"
p
p "
p
1
3p
1
1
D p
1 C 3 C ln
3 C 1 C 3 " 0 D 1 C p ln 2 C 3
2
2
3
2 3
1
LD p
3
Z
p
3p
1 C u2 du
1
D p
3
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
y
x
Thus the total length equals 4L D 4 C
p2
3
Graph of r D cos2 #
!
p "
ln 2 C 3 % 5:52.
In Exercises 31 and 32, express the length of the curve as an integral but do not evaluate it.
31. r D .2 " cos #/!1 ,
SOLUTION
0 # # # 2!
We have f .#/ D .2 " cos #/!1 , f 0 .#/ D ".2 " cos #/!2 sin #, hence,
r
q
q
!
"
f 2 .#/ C f 0 .#/2 D .2 " cos #/!2 C .2 " cos #/!4 sin2 # D .2 " cos #/!4 .2 " cos #/2 C sin2 #
D .2 " cos #/!2
Using the integral for the arc length we get
p
LD
32. r D sin3 t, 0 # # # 2!
SOLUTION
Z
4 " 4 cos # C cos2 # C sin2 # D .2 " cos #/!2
2!
p
5 " 4 cos #
p
5 " 4 cos #.2 " cos #/!2 d#:
0
We have f .t/ D sin3 t, f 0 .t/ D 3 sin2 t cos t, so that
q
p
p
f .t/2 C f 0 .t/2 D sin6 t C 9 sin4 t cos2 t D sin2 t sin2 t C 9 cos2 t
p
p
D sin2 t sin2 t C cos2 t C 8 cos2 t D sin2 t 1 C 8 cos2 t
Using the formula for arc length integral we get
LD
Z
2!
sin2 t
0
p
1 C 8 cos2 t dt
In Exercises 33–36, use a computer algebra system to calculate the total length to two decimal places.
33.
SOLUTION
The three-petal rose r D cos 3# in Figure 9
We have f .#/ D cos 3#, f 0 .#/ D "3 sin 3#, so that
q
p
p
p
f .#/2 C f 0 .#/2 D cos2 3# C 9 sin2 3# D cos2 3# C sin2 3# C 8 sin2 3# D 1 C 8 sin2 3#
Note that the curve is traversed completely for 0 # # # !. Using the arc length formula and evaluating with Maple gives
Z !q
Z !p
LD
f .#/2 C f 0 .#/2 d# D
1 C 8 sin2 3# d# % 6:682446608
0
34.
SOLUTION
0
The curve r D 2 C sin 2# in Figure 12
We have f .#/ D 2 C sin 2#, f 0 .#/ D 2 cos 2#, so that
q
q
p
f .#/2 C f 0 .#/2 D .2 C sin 2#/2 C 4 cos2 2# D 4 C 4 sin 2# C sin2 2# C 4 cos2 2#
p
D 4 C 4 sin 2# C sin2 2# C cos2 2# C 3 cos2 2#
p
D 5 C 4 sin 2# C 3 cos2 2#
The curve is traversed completely for 0 # # # 2!. Using the arc length formula and evaluating with Maple gives
Z 2! q
Z 2! q
LD
f .#/2 C f 0 .#/2 d# D
f .#/2 C f 0 .#/2 d# % 15:40375907
0
0
1429
1430
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
35.
The curve r D # sin # in Figure 13 for 0 # # # 4!
y
10
5
x
5
5
FIGURE 13 r D # sin # for 0 # # # 4!.
SOLUTION
We have f .#/ D # sin #, f 0 .#/ D sin # C # cos #, so that
q
q
p
f .#/2 C f 0 .#/2 D # 2 sin2 # C .sin # C # cos #/2 D # 2 sin2 # C sin2 # C 2# sin # cos # C # 2 cos2 #
p
D # 2 C sin2 # C # sin 2#
using the identities sin2 # C cos2 # D 1 and 2 sin # cos # D sin 2#. Thus by the arc length formula and evaluating with Maple, we
have
Z 4! q
Z 4! p
LD
f .#/2 C f 0 .#/2 d# D
# 2 C sin2 # C # sin 2# d# % 79:56423976
36.
rD
0
p
0
#, 0 # # # 4!
p
SOLUTION We have f .#/ D #, f 0 .#/ D
1 !1=2
,
2#
so that
r
q
1
2
0
2
f .#/ C f .#/ D # C
4#
so that by the arc length formula, evaluating with Maple, we have
LD
Z
4!
0
q
f .#/2
Cf
0 .#/2 d#
D
Z
4!
0
37. Use Eq. (8) to find the slope of the tangent line to r D # at # D
!
2
r
#C
1
d# % 30:50125041
4#
and # D !.
SOLUTION In the given curve we have r D f .#/ D # . Using Eq. (8) we obtain the following derivative, which is the slope of
the tangent line at .r; #/.
dy
f .#/ cos # C f 0 .#/ sin #
# cos # C 1 ! sin #
D
D
dx
"f .#/ sin # C f 0 .#/ cos #
"# sin # C 1 ! cos #
The slope, m, of the tangent line at # D
!
2
mD
and # D ! is obtained by substituting these values in (1). We get .# D
!
!
!
2 cos 2 C sin 2
!
!
" 2 sin 2 C cos !2
D
!
2 !0C1
" !2 ! 1 C 0
D
1
2
D" :
" !2
!
.# D !/:
mD
! cos ! C sin !
"!
D
D !:
"! sin ! C cos !
"1
38. Use Eq. (8) to find the slope of the tangent line to r D sin # at # D
SOLUTION
!
3.
We have f .#/ D sin #, f 0 .#/ D cos # and, by Eq. (8), the slope of the tangent line is
Evaluating at # D
!
3
dy
f .#/ cos # C f 0 .#/ sin #
sin # cos # C cos # sin #
sin 2#
D
D
D
dx
"f .#/ sin # C f 0 .#/ cos #
cos 2#
" sin2 # C cos2 #
gives
p
p
sin 2!
dy
32
3
D" 3
D
D
2!
dx
"1=2
cos 3
p
Thus the slope of the tangent line to r D sin # at # D !3 is " 3.
(1)
!
2 /:
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1431
39. Find the polar coordinates of the points on the lemniscate r 2 D cos 2t in Figure 14 where the tangent line is horizontal.
y
r 2 = cos (2t)
x
−1
1
FIGURE 14
This curve is defined for " !2 # 2t # !2 (where cos 2t $ 0), so for " !4 # t # !4 . For each # in that range,
p
there are two values of r satisfying the equation (˙ cos 2t ). By symmetry, we need only calculate thepcoordinates of the points
correspondingpto the positive square root (i.e. to the right of the y axis). Then the equation becomes r D cos 2t . Now, by Eq. (8),
with f .t/ D cos.2t/ and f 0 .t/ D " sin.2t/.cos.2t//!1=2 , we have
p
cos t cos.2t/ " sin.2t/ sin t.cos.2t//!1=2
dy
f .t/ cos t C f 0 .t/ sin t
D
D
p
dx
"f .t/ sin t C f 0 .t/ cos t
" sin t cos.2t/ " sin.2t/ cos t.cos.2t//!1=2
SOLUTION
The tangent line is horizontal when this derivative is zero,
p which occurs when the numerator of the fraction is zero and the denominator is not. Multiply top and bottom of the fraction by cos.2t/, and use the identities cos 2t D cos2 t " sin2 t, sin 2t D 2 sin t cos t
to get
"
cos t cos 2t " sin t sin 2t
cos t.cos2 t " 3 sin2 t/
D"
sin t cos 2t C cos t sin 2t
sin t cos 2t C cos t sin 2t
The numerator is zero when cos t D 0, so when t D
˙ !6
!
2
" !4
or t D
3!
2 , or when
tan t D ˙ p1 , so when t D ˙ !6 or t D ˙ 5!
6 . Of these
3
possibilities, only t D
lie in the range
#t #
Note that the denominator is nonzero for t D ˙ !6 , so these are the two
values of t for which the tangent line is horizontal. The corresponding values of r are solutions to
Finally, the four points are .r; t/ D
#
$
1 !
p ;
;
2 6
!
4.
! !"
!! "
1
r 2 D cos 2 !
D cos
D
6
3
2
! "! "
! !"
1
r 2 D cos 2 !
D cos "
D
6
3
2
#
$
1 !
"p ;
;
2 6
#
1
pi
p ;"
6
2
$
;
#
$
1
!
"p ; "
6
2
If desired, we can change the second and fourth points by adding ! to the angle and making r positive, to get
#
$
#
$
#
$
#
$
1 !
1 7!
1
pi
1 5!
p ;
;
p ;
;
p ;"
;
p ;
6
2 6
2 6
2
2 6
40. Find the equation in rectangular coordinates of the tangent line to r D 4 cos 3# at # D
SOLUTION
We have f .#/ D 4 cos 3# . By Eq. (8),
mD
Setting # D
!
6.
!
6
4 cos 3# cos # " 12 sin 3# sin #
:
"4 cos 3# sin # " 12 sin 3# cos #
yields
mD
% &
% & % &
cos ! " 12 sin !2 sin !6
"12 sin !6
!
1
%2! & %6! &
%! &
%! & D
D tan D p :
"12 cos !6
6
"4 cos 2 sin 6 " 12 sin 2 cos 6
3
4 cos
%! &
!
We identify the point of tangency. For # D !6 we have r D 4 cos 3!
6 D 4 cos 2 D 0. The point of tangency is the origin. The
1
x
tangent line is the line through the origin with slope p . This is the line y D p .
3
41. Use Eq. (8) to show that for r D sin # C cos #,
dy
cos 2# C sin 2#
D
dx
cos 2# " sin 2#
Then calculate the slopes of the tangent lines at points A; B; C in Figure 4.
3
1432
C H A P T E R 11
SOLUTION
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
In Exercise 49 in Section 11.3 we proved that for a polar curve r D f .#/ the following formula holds:
dy
f .#/ cos # C f 0 .#/ sin #
D
dx
"f .#/ sin # C f 0 .#/ cos #
(1)
For the given circle we have r D f .#/ D sin # C cos #, hence f 0 .#/ D cos # " sin #. Substituting in (1) we have
dy
.sin # C cos #/ cos # C .cos # " sin #/ sin #
sin # cos # C cos2 # C cos # sin # " sin2 #
D
D
dx
" .sin # C cos #/ sin # C .cos # " sin #/ cos #
" sin2 # " cos # sin # C cos2 # " sin # cos #
D
cos2 # " sin2 # C 2 sin # cos #
cos2 # " sin2 # " 2 sin # cos #
We use the identities cos2 # " sin2 # D cos 2# and 2 sin # cos # D sin 2# to obtain
dy
cos 2# C sin 2#
D
dx
cos 2# " sin 2#
(2)
dy
The derivative dx
is the slope of the tangent line at .r; #/. The slopes of the tangent lines at the points with polar coordinates
!
"
% !&
A D 1; 2 B D 0; 3!
C D .1; 0/ are computed by substituting the values of # in (2). This gives
4
%
&
%
&
ˇ
cos 2 ! !2 C sin 2 ! !2
dy ˇˇ
cos ! C sin !
"1 C 0
%
&
%
&
D
D
D
D1
dx ˇA
cos ! " sin !
"1 " 0
cos 2 ! !2 " sin 2 ! !2
!
"
!
"
ˇ
3!
cos 2 ! 3!
C sin 2 ! 3!
cos 3!
dy ˇˇ
4
4
0"1
2 C sin 2
!
"
!
"
D
D
D
D "1
3!
3!
3!
3!
dx ˇB
0
C1
cos 2 " sin 2
cos 2 ! 4 " sin 2 ! 4
ˇ
dy ˇˇ
cos .2 ! 0/ C sin .2 ! 0/
cos 0 C sin 0
1C0
D
D
D
D1
dx ˇC
cos .2 ! 0/ " sin .2 ! 0/
cos 0 " sin 0
1"0
42. Find the polar coordinates of the points on the cardioid r D 1 C cos # where the tangent line is horizontal (see Figure 15).
SOLUTION
Use Eq. (8) with f .#/ D 1 C cos # and f 0 .#/ D " sin #. Then
dy
f .#/ cos # C f 0 .#/ sin #
cos # C cos2 # " sin2 #
cos # C cos 2#
D
D
D"
dx
"f .#/ sin # C f 0 .#/ cos #
" sin # " cos # sin # " sin # cos #
sin # C sin 2#
The tangent line is horizontal when the numerator is zero but the denominator is not. The numerator is zero when cos # C cos 2# D
0. But
#
$
1
cos # C cos 2# D cos # C 2 cos2 # " 1 D cos # "
.cos # C 1/
2
So for 0 # # < 2!, the numerator is zero when # D ! and when # D ˙ !3 . For the latter two points, the denominator is nonzero,
so the tangent is horizontal at the points
#
$
#
$ #
$
3 !
3 !
3 5!
.r; #/ D
;
;
;"
D
;
2 3
2
3
2 3
When # D !, both numerator and denominator vanish. However, using L’Hôpital’s Rule, we have
" lim
" !!
cos # C cos 2#
" sin # " 2 sin 2#
D " lim
D0
sin # C sin 2#
" !! cos # C 2 cos 2#
so that the tangent is defined at # D !, and it is horizontal. Thus the tangent is also horizontal at the point
.r; #/ D .0; !/
Further Insights and Challenges
43. Suppose that the polar coordinates of a moving particle at time t are .r.t/; #.t//. Prove that the particle’s speed is equal to
q
.dr=dt /2 C r 2 .d#=dt /2 .
SOLUTION
The speed of the particle in rectangular coordinates is:
ds
D
dt
q
x 0 .t/2 C y 0 .t/2
(1)
Area, Arc Length, and Slope in Polar Coordinates
S E C T I O N 11.4
1433
We need to express the speed in polar coordinates. The x and y coordinates of the moving particles as functions of t are
x.t/ D r.t/ cos #.t/;
y.t/ D r.t/ sin #.t/
We differentiate x.t/ and y.t/, using the Product Rule for differentiation. We obtain (omitting the independent variable t)
x 0 D r 0 cos # " r .sin #/ # 0
y 0 D r 0 sin # " r .cos #/ # 0
Hence,
%
&2 %
&2
2
2
x 0 C y 0 D r 0 cos # " r# 0 sin # C r 0 sin # C r# 0 cos #
2
2
2
2
D r 0 cos2 # " 2r 0 r# 0 cos # sin # C r 2 # 0 sin2 # C r 0 sin2 # C 2r 0 r# 0 sin2 # cos # C r 2 # 0 cos2 #
!
"
!
"
2
2
2
2
D r 0 cos2 # C sin2 # C r 2 # 0 sin2 # C cos2 # D r 0 C r 2 # 0
(2)
Substituting (2) into (1) we get
ds
D
dt
q
r 02 C r 2# 02 D
s
#
dr
dt
$2
#
C r2
$2
d#
dt
44.
Compute the speed at time t D 1 of a particle whose polar coordinates at time t are r D t, # D t (use Exercise 43).
What would the speed be if the particle’s rectangular coordinates were x D t, y D t? Why is the speed increasing in one case and
constant in the other?
SOLUTION
By Exercise 43 the speed of the particle is
ds
D
dt
In this case r D t and # D t so
dr
dt
D 1 and
d"
dt
s
#
dr
dt
$2
C r2
#
d#
dt
$2
(1)
D 1. Substituting into (1) gives the following function of the speed:
ds
D
dt
q
1 C r.t/2
The speed expressed in rectangular coordinates is
ds
D
dt
q
x 0 .t/2 C y 0 .t/2
If x D t and y D t, then x 0 .t/ D 1 and y 0 .t/ D 1. So the speed of the particle at time t is
p
p
ds
D 12 C 12 D 2
dt
p
On the curve x D t, y D t the particle travels the same distance t 2 for all time intervals $t , hence, it has a constant speed.
However, on the spiral r D t, # D t the particle travels greater distances for time intervals .t; t C t/ as t increases, hence the speed
is an increasing function of t.
y
y
Δt
t + Δt
Δt
2Δt
x
Δt
t
Δt
t
x D t, y D t
45.
(a)
(b)
(c)
(d)
Δt
t + Δt
x
r D t, # D t
We investigate how the shape of the limaçon curve r D b C cos # depends on the constant b (see Figure 15).
Show that the constants b and "b yield the same curve.
Plot the limaçon for b D 0; 0:2; 0:5; 0:8; 1 and describe how the curve changes.
Plot the limaçon for b D 1:2; 1:5; 1:8; 2; 2:4 and describe how the curve changes.
Use Eq. (8) to show that
#
$
b cos # C cos 2#
dy
csc #
D"
dx
b C 2 cos #
1434
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
(e) Find the points where the tangent line is vertical. Note that there are three cases: 0 # b < 2, b D 1, and b > 2. Do the plots
constructed in (b) and (c) reflect your results?
y
y
y
1
1
1
2
3
x
1
1
r = 1 + cos
2
3
x
1
r = 1.5 + cos
2
3
x
r = 2.3 + cos
FIGURE 15
SOLUTION
(a) If .r; #/ is on the curve r D "b C cos #, then so is ."r; # C !/ since they represent the same point. Thus
"r D "b C cos.# C !/
"r D "b " cos #
r D b C cos #
Thus the same set of points lie on the graph of both equations, so they define the same curve.
(b)
y
y
y
y
y
1
1
0.5
0.5
0.5
0.5
0.5
0
1
x
−1
−1
−0.5
−0.5
b = 0.2
b=0
2
−0.5
−0.5
−0.5
1
1
1
1
x
x
x
x
0
b = 0.5
b=1
b = 0.8
For 0 < b < 1, there is a “loop” inside the curve. For b D 0, the curve is a circle, although actually for 0 # # # 2! the circle is
traversed twice, so in fact the loop is as large as the circle and overlays it. When b D 1, the loop is pinched to a point.
(c)
y
y
y
1.5
1
1
0.5
0.5
1
2
−0.5
1
b = 1.2
b = 1.5
1
x
−0.5
1
−2
2
x
−0.5
1
2
b = 1.8
−2
x
3
1
2
3
−1
−1
−1.5
−1.5
−1.5
2
1.5
1
0.5
−1
−1
−1
2
y
2
x
x
−0.5
y
2
1.5
1
0.5
−2
b=2
b = 2.4
For b between 1 and 2, the pinch at b D 1 smooths out into a concavity in the curve, which decreases in size. By b D 2 it appears
to be gone; further increases in b push the left-hand section of the curve out, making it more convex.
(d) By Eq. (8), with f .#/ D b C cos # and f 0 .#/ D " sin #, we have (using the double-angle identities for sin and cos)
dy
f .#/ cos # C f 0 .#/ sin #
.b C cos #/ cos # " sin2 #
b cos # C cos 2#
D
D
D
0
dx
"f .#/ sin # C f .#/ cos #
".b C cos #/ sin # " sin # cos #
"b sin # " 2 sin # cos #
#
$
b cos # C cos 2#
b cos # C cos 2#
D"
D"
csc #
sin #.b C 2 cos #/
b C 2 cos #
(e) From part (d), the tangent line is vertical when either csc # is undefined or when b C 2 cos # D 0 (as long as the numerator
b cos # C cos 2# ¤ 0). Consider first the case when csc # is undefined, so that # D 0 or # D !. If # D 0, the numerator of the
fraction is b C 1 ¤ 0 and the denominator is b C 2 ¤ 0, so that the tangent is vertical here.
For any b; the limaçon has a vertical tangent at .b C cos 0; 0/ D .b C 1; 0/
If # D !, the numerator of the fraction is 1 " b and the denominator is b C 2 ¤ 0. As long as b ¤ 1, the numerator does not vanish
and we have found a point of vertical tangency. If b D 1, then by L’Hôpital’s Rule,
$
#
$
#
b cos # C cos 2#
sin t C sin 2t
b cos # C cos 2#
csc
#
D
"
lim
D lim
D0
" lim
b C 2 cos #
" !! .b C 2 cos #/ sin #
" !! 2 cos2 t " 2 sin2 t C cos t
" !!
S E C T I O N 11.5
Vectors in the Plane
1435
so that the tangent is not vertical here. Thus
If b ¤ 1; the limaçon has a vertical tangent at .b C cos !; !/ D .b " 1; !/
Next consider the possibility that b C 2 cos # D 0; this happens when cos # D " b2 . First assume that 0 # b < 2. This equation
! "
! "
holds for two values of #: cos!1 " b2 and " cos!1 " b2 . Neither of these angles is 0 or !, so that csc # is defined. Additionally,
the numerator is
b cos # C cos 2# D b cos # C 2 cos2 # " 1 D "
b2
b2
C2!
" 1 D "1
2
4
so that the numerator does not vanish. Thus
For 0 # b < 2; the limaçon has a vertical tangent at
#
#
$$
#
#
$$
b
b
b
b
; cos!1 "
and
; " cos!1 "
2
2
2
2
Next assume that b D 2; then cos # D "1 holds for # D !; we have considered that case above. Finally assume that b > 2; then
cos # D " b2 has no solutions. Thus, in summary, vertical tangents of the limaçon occur as follows:
#
#
$$ #
#
$$
b
b
b
b
0 # b < 2; b ¤ 1 W
; cos!1 "
;
; " cos!1 "
; .b " 1; !/; .b C 1; 0/
2
2
2
2
#
#
$$ #
#
$$
b
b
b
b
bD1W
; cos!1 "
;
; " cos!1 "
; .b C 1; 0/
2
2
2
2
b$2W
.b C 1; 0/; .b " 1; !/
These do correspond to the figures in parts (b) and (c).
11.5 Vectors in the Plane
Preliminary Questions
1.
(a)
(b)
(c)
(d)
Answer true or false. Every nonzero vector is:
Equivalent to a vector based at the origin.
Equivalent to a unit vector based at the origin.
Parallel to a vector based at the origin.
Parallel to a unit vector based at the origin.
SOLUTION
(a) This statement is true. Translating the vector so that it is based on the origin, we get an equivalent vector based at the origin.
(b) Equivalent vectors have equal lengths, hence vectors that are not unit vectors, are not equivalent to a unit vector.
(c) This statement is true. A vector based at the origin such that the line through this vector is parallel to the line through the given
vector, is parallel to the given vector.
(d) Since parallel vectors do not necessarily have equal lengths, the statement is true by the same reasoning as in (c).
2. What is the length of "3a if kak D 5?
SOLUTION
Using properties of the length we get
k"3ak D j"3jkak D 3kak D 3 ! 5 D 15
3. Suppose that v has components h3; 1i. How, if at all, do the components change if you translate v horizontally two units to the
left?
SOLUTION
Translating v D h3; 1i yields an equivalent vector, hence the components are not changed.
4. What are the components of the zero vector based at P D .3; 5/?
SOLUTION
The components of the zero vector are always h0; 0i, no matter where it is based.
5. True or false?
(a) The vectors v and "2v are parallel.
(b) The vectors v and "2v point in the same direction.
SOLUTION
(a) The lines through v and "2v are parallel, therefore these vectors are parallel.
(b) The vector "2v is a scalar multiple of v, where the scalar is negative. Therefore "2v points in the opposite direction as v.
1436
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
6. Explain the commutativity of vector addition in terms of the Parallelogram Law.
To determine the vector v C w, we translate w to the equivalent vector w0 whose tail coincides with the head of v.
The vector v C w is the vector pointing from the tail of v to the head of w0 .
SOLUTION
w'
v
v
w
+ v
+
w
v'
w
To determine the vector w C v, we translate v to the equivalent vector v0 whose tail coincides with the head of w. Then w C v is the
vector pointing from the tail of w to the head of v0 . In either case, the resulting vector is the vector with the tail at the basepoint of
v and w, and head at the opposite vertex of the parallelogram. Therefore v C w D w C v.
Exercises
1. Sketch the vectors v1 ; v2 ; v3 ; v4 with tail P and head Q, and compute their lengths. Are any two of these vectors equivalent?
SOLUTION
v1
v2
v3
v4
P
.2; 4/
."1; 3/
."1; 3/
.4; 1/
Q
.4; 4/
.1; 3/
.2; 4/
.6; 3/
Using the definitions we obtain the following answers:
""!
v1 D PQ D h4 " 2; 4 " 4i D h2; 0i
p
kv1 k D 22 C 02 D 2
y
P
v1
v2 D h1 " ."1/; 3 " 3i D h2; 0i
p
kv2 k D 22 C 02 D 2
y
Q
P
v2
Q
x
x
v3 D h2 " ."1/; 4 " 3i D h3; 1i
p
p
kv3 k D 32 C 12 D 10
y
v4 D h6 " 4; 3 " 1i D h2; 2i
p
p
p
kv4 k D 22 C 22 D 8 D 2 2
y
Q
P
Q
v3
P
x
v4
x
v1 and v2 are parallel and have the same length, hence they are equivalent.
2. Sketch the vector b D h3; 4i based at P D ."2; "1/.
SOLUTION The vector b D h3; 4i based at P has terminal point Q, located 3 units to the right and 4 units up from P . Therefore
Q D ."2 C 3; "1 C 4/ D .1; 3/. The vector equivalent to b is PQ shown in the figure.
S E C T I O N 11.5
Vectors in the Plane
1437
y
Q = (1, 3)
x
P = (−2, −1)
3. What is the terminal point of the vector a D h1; 3i based at P D .2; 2/? Sketch a and the vector a0 based at the origin and
equivalent to a.
SOLUTION The terminal point Q of the vector a is located 1 unit to the right and 3 units up from P D .2; 2/. Therefore,
Q D .2 C 1; 2 C 3/ D .3; 5/. The vector a0 equivalent to a based at the origin is shown in the figure, along with the vector a.
y
Q
a
a0 P
0
x
""!
4. Let v D PQ, where P D .1; 1/ and Q D .2; 2/. What is the head of the vector v0 equivalent to v based at .2; 4/? What is the
head of the vector v0 equivalent to v based at the origin? Sketch v, v0 , and v0 .
SOLUTION
We first find the components of v:
""!
v D PQ D h1; 1i
Since v0 is equivalent to v, the two vectors have the same components, hence the head of v0 is located one unit to the right and one
unit up from .2; 4/. This is the point .3; 5/. The head of v0 is located one unit to the right and one unit up from the origin; that is,
the head is at the point .1; 1/.
y
(3, 5)
(2, 4)
v1
Q
P
v
v0
(0, 0)
x
""!
In Exercises 5–8, find the components of PQ.
5. P D .3; 2/,
SOLUTION
""!
Using the definition of the components of a vector we have PQ D h2 " 3; 7 " 2i D h"1; 5i.
6. P D .1; "4/,
SOLUTION
Q D .1; "4/
""!
By the definition of the components of a vector, we obtain PQ D h1 " 3; "4 " 5i D h"2; "9i.
8. P D .0; 2/,
SOLUTION
Q D .3; 5/
""!
""!
The components of PQ are PQ D h3 " 1; 5 " ."4/i D h2; 9i.
7. P D .3; 5/,
SOLUTION
Q D .2; 7/
Q D .5; 0/
""!
""!
The components of the vector PQ are PQ D h5 " 0; 0 " 2i D h5; "2i.
1438
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
In Exercises 9–14, calculate.
9. h2; 1i C h3; 4i
Using vector algebra we have h2; 1i C h3; 4i D h2 C 3; 1 C 4i D h5; 5i.
SOLUTION
10. h"4; 6i " h3; "2i
h"4; 6i " h3; "2i D h"4 " 3; 6 " ."2/i D h"7; 8i
SOLUTION
11. 5 h6; 2i
5h6; 2i D h5 ! 6; 5 ! 2i D h30; 10i
SOLUTION
12. 4.h1; 1i C h3; 2i/
Using vector algebra we obtain
SOLUTION
13.
D
" 12 ; 53
E
D
C 3;
10
3
E
4 .h1; 1i C h3; 2i/ D 4h1 C 3; 1 C 2i D 4h4; 3i D h4 ! 4; 4 ! 3i D h16; 12i
+
, +
, +
, +
,
1 5
10
1
5
10
5
The vector sum is " ;
C 3;
D " C 3; C
D ;5 .
2 3
3
2
3
3
2
14. hln 2; ei C hln 3; !i
SOLUTION
SOLUTION
The vector sum is hln 2; ei C hln 3; !i D hln 2 C ln 3; e C !i D hln 6; e C !i.
15. Which of the vectors (A)–(C) in Figure 1 is equivalent to v " w?
v
w
(A)
(B)
(C)
FIGURE 1
SOLUTION The vector "w has the same length as w but points in the opposite direction. The sum v C ."w/, which is the
difference v " w, is obtained by the parallelogram law. This vector is the vector shown in (b).
−w
v−w
v
w
−w
16. Sketch v C w and v " w for the vectors in Figure 2.
v
w
FIGURE 2
SOLUTION
The vector v C w is obtained by the parallelogram law:
w
v+w
v
w
Since v " w D v C ."w/, we first sketch the vector "w, which has the same length as w but points to the opposite direction. Then
we add "w to v using the parallelogram law. This gives:
−w
v
w
v−w
−w
17. Sketch 2v, "w, v C w, and 2v " w for the vectors in Figure 3.
y
5
4
3
v = 〈2, 3〉
2
1
w = 〈1, 4〉
1 2
3 4
5 6
FIGURE 3
x
S E C T I O N 11.5
SOLUTION
Vectors in the Plane
The scalar multiple 2v points in the same direction as v and its length is twice the length of v. It is the vector
2v D h4; 6i.
y
y
5
5
4
3
4
3
2v
2
1
2
1
1
2
3
4
5
6
x
v
1
2 3
4 5
6
x
"w has the same length as w but points to the opposite direction. It is the vector "w D h"4; "1i.
y
w
x
−w
The vector sum v C w is the vector:
v C w D h2; 3i C h4; 1i D h6; 4i:
This vector is shown in the following figure:
y
v+w
v
w
x
The vector 2v " w is
2v " w D 2h2; 3i " h4; 1i D h4; 6i " h4; 1i D h0; 5i
It is shown next:
y
2v − w
x
18. Sketch v D h1; 3i, w D h2; "2i, v C w, v " w.
SOLUTION
1439
We compute the sum v C w and the difference v " w and then sketch the vectors. This gives:
v C w D h1; 3i C h2; "2i D h1 C 2; 3 " 2i D h3; 1i
v " w D h1; 3i " h2; "2i D h1 " 2; 3 C 2i D h"1; 5i
1440
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
v−w
v
v+w
x
w
19. Sketch v D h0; 2i, w D h"2; 4i, 3v C w, 2v " 2w.
SOLUTION
We compute the vectors and then sketch them:
3v C w D 3h0; 2i C h"2; 4i D h0; 6i C h"2; 4i D h"2; 10i
2v " 2w D 2h0; 2i " 2h"2; 4i D h0; 4i " h"4; 8i D h4; "4i
y
3v + w
w
v
x
2v − 2w
20. Sketch v D h"2; 1i, w D h2; 2i, v C 2w, v " 2w.
SOLUTION
We compute the linear combinations v C 2w and v " 2w and then sketch the vectors:
v C 2w D h"2; 1i C 2h2; 2i D h"2; 1i C h4; 4i D h2; 5i
v " 2w D h"2; 1i " 2h2; 2i D h"2; 1i " h4; 4i D h"6; "3i
y
v + 2w
w
v
x
v − 2w
21. Sketch the vector v such that v C v1 C v2 D 0 for v1 and v2 in Figure 4(A).
y
y
3
v2
v3
v1
v4
1
1
−3
v2
x
x
v1
(A)
(B)
FIGURE 4
SOLUTION Since v C v1 C v2 D 0, we have that v D "v1 " v2 , and since v1 D h1; 3i and v2 D h"3; 1i, then v D "v1 " v2 D
h2; "4i, as seen in this picture.
S E C T I O N 11.5
Vectors in the Plane
1441
y
3
v2
v1
1
x
1 2
−3
v
−4
22. Sketch the vector sum v D v1 C v2 C v3 C v4 in Figure 4(B).
SOLUTION If we place the vectors v1 ; v2 ; v3 ; v4 tip to tail, as shown in the following figure, it is easy to sketch in the sum
v1 C v2 C v3 C v4 , as shown.
y
v4
v1 + v2 + v3 + v4
v3
x
v1
v2
""!
23. Let v D PQ, where P D ."2; 5/, Q D .1; "2/. Which of the following vectors with the given tails and heads are equivalent
to v?
(a) ."3; 3/, .0; 4/
(b) .0; 0/, .3; "7/
(c) ."1; 2/, .2; "5/
(d) .4; "5/, .1; 4/
SOLUTION Two vectors are equivalent if they have the same components. We thus compute the vectors and check whether this
condition is satisfied.
""!
v D PQ D h1 " ."2/; "2 " 5i D h3; "7i
(a) h0 " ."3/; 4 " 3i D h3; 1i
(c) h2 " ."1/; "5 " 2i D h3; "7i
We see that the vectors in (b) and (c) are equivalent to v.
(b) h3 " 0; "7 " 0i D h3; "7i
(d) h1 " 4; 4 " ."5/i D h"3; 9i
24. Which of the following vectors are parallel to v D h6; 9i and which point in the same direction?
(a) h12; 18i
(b) h3; 2i
(c) h2; 3i
(d) h"6; "9i
(e) h"24; "27i
(f) h"24; "36i
SOLUTION Two vectors are parallel if they are scalar multiples of each other. The vectors point in the same direction if the
multiplying scalar is positive. We use this to obtain the following conclusions:
(a) h12; 18i D 2h6; 9i D 2v ) both vectors point in the same direction.
(b) h3; 2i is not a scalar multiple of v, hence the vectors are not parallel.
(c) h2; 3i D 13 h6; 9i D 13 v ) both vectors point in the same direction.
(d) h"6; "9i D "h6; 9i D "v ) parallel to v and points in the opposite direction.
(e) h"24; "27i is not a scalar multiple of v, hence the vectors are not parallel.
(f) h"24; "36i D "4h6; 9i D "4v ) parallel to v and points in the opposite direction.
""!
""!
In Exercises 25–28, sketch the vectors AB and PQ, and determine whether they are equivalent.
25. A D .1; 1/, B D .3; 7/,
SOLUTION
P D .4; "1/,
Q D .6; 5/
We compute the vectors and check whether they have the same components:
""!
AB D h3 " 1; 7 " 1i D h2; 6i
""!
PQ D h6 " 4; 5 " ."1/i D h2; 6i
)
The vectors are equivalent.
26. A D .1; 4/, B D ."6; 3/, P D .1; 4/, Q D .6; 3/
""!
""!
SOLUTION We compute AB and PQ and see if they have the same components:
""!
AB D h"6 " 1; 3 " 4i D h"7; "1i
""!
PQ D h6 " 1; 3 " 4i D h5; "1i
)
The vectors are not equivalent.
1442
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
(1, 4)
(−6, 3)
(6, 3)
A, P
B
Q
x
27. A D ."3; 2/, B D .0; 0/,
P D .0; 0/, Q D .3; "2/
""!
""!
SOLUTION We compute the vectors AB and PQ :
""!
AB D h0 " ."3/; 0 " 2i D h3; "2i
""!
PQ D h3 " 0; "2 " 0i D h3; "2i
)
The vectors are equivalent.
28. A D .5; 8/, B D .1; 8/, P D .1; 8/, Q D ."3; 8/
""!
""!
SOLUTION Computing AB and PQ gives:
""!
AB D h1 " 5; 8 " 8i D h"4; 0i
""!
PQ D h"3 " 1; 8 " 8i D h"4; 0i
(−3, 8)
Q
y
(1, 8)
B, P
)
The vectors are equivalent.
(5, 8)
A
x
""!
""!
In Exercises 29–32, are AB and PQ parallel? And if so, do they point in the same direction?
29. A D .1; 1/, B D .3; 4/,
P D .1; 1/, Q D .7; 10/
""!
""!
SOLUTION We compute the vectors AB and PQ:
""!
AB D h3 " 1; 4 " 1i D h2; 3i
""!
PQ D h7 " 1; 10 " 1i D h6; 9i
""!
Since AB D 13 h6; 9i, the vectors are parallel and point in the same direction.
30. A D ."3; 2/, B D .0; 0/,
SOLUTION
P D .0; 0/,
We compute the two vectors:
Q D .3; 2/
""!
AB D h0 " ."3/; 0 " 2i D h3; "2i
""!
PQ D h3 " 0; 2 " 0i D h3; 2i
The vectors are not scalar multiples of each other, hence they are not parallel.
31. A D .2; 2/, B D ."6; 3/,
P D .9; 5/, Q D .17; 4/
""!
""!
SOLUTION We compute the vectors AB and PQ:
""!
AB D h"6 " 2; 3 " 2i D h"8; 1i
""!
PQ D h17 " 9; 4 " 5i D h8; "1i
""!
""!
Since AB D "PQ, the vectors are parallel and point in opposite directions.
32. A D .5; 8/, B D .2; 2/, P D .2; 2/, Q D ."3; 8/
""!
""!
SOLUTION Computing AB and PQ gives:
""!
AB D h2 " 5; 2 " 8i D h"3 " 6i
""!
PQ D h"3 " 2; 8 " 2i D h"5; 6i
The vectors are not scalar multiples of each other, hence they are not parallel.
S E C T I O N 11.5
Vectors in the Plane
1443
In Exercises 33–36, let R D ."2; 7/. Calculate the following.
""!
33. The length of OR
q
p
""!
""!
SOLUTION Since OR D h"2; 7i, the length of the vector is kORk D
."2/2 C 72 D 53.
"!
34. The components of u D PR, where P D .1; 2/
SOLUTION
We compute the components of the vector to obtain:
"!
u D PR D h"2 " 1; 7 " 2i D h"3; 5i
"!
35. The point P such that PR has components h"2; 7i
SOLUTION
Denoting P D .x0 ; y0 / we have:
"!
PR D h"2 " x0 ; 7 " y0 i D h"2; 7i
Equating corresponding components yields:
" 2 " x0 D "2
7 " y0 D 7
)
x0 D 0; y0 D 0
)
P D .0; 0/
""!
36. The point Q such that RQ has components h8; "3i
SOLUTION
We denote Q D .x0 ; y0 / and have:
""!
RQ D hx0 " ."2/; y0 " 7i D hx0 C 2; y0 " 7i D h8; "3i
Equating the corresponding components of the two vectors yields:
x0 C 2 D 8
y0 " 7 D "3
)
x0 D 6; y0 D 4
)
Q D .6; 4/
In Exercises 37–42, find the given vector.
37. Unit vector ev where v D h3; 4i
SOLUTION
The unit vector ev is the following vector:
ev D
We find the length of v D h3; 4i:
kvk D
Thus
38. Unit vector ew where w D h24; 7i
SOLUTION
p
p
32 C 42 D 25 D 5
+
,
1
3 4
ev D h3; 4i D ;
:
5
5 5
The unit vector ew is the following vector:
ew D
We find the length of w D h24; 7i:
kwk D
Thus
1
v
kvk
1
w
kwk
p
p
242 C 72 D 625 D 25
ew D
+
,
1
24 7
h24; 7i D
;
:
25
25 25
39. Vector of length 4 in the direction of u D h"1; "1i
D
E
p
p
1
1
SOLUTION Since kuk D ."1/2 C ."1/2 D 2, the unit vector in the direction of u is eu D " p ; " p . We multiply eu by
2
2
4 to obtain the desired vector:
+
, D p
p E
1
1
4eu D 4 " p ; " p D "2 2; "2 2
2
2
1444
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
40. Unit vector in the direction opposite to v D h"2; 4i
SOLUTION We first compute the unit vector ev in the direction of v and then multiply by "1 to obtain a unit vector in the opposite
direction. This gives:
+
, +
,
1
1
1
2
4
1
2
ev D
vD p
h"2; 4i D p h"2; 4i D " p ; p D " p ; p
kvk
20
2 5 2 5
5
5
."2/2 C 42
The desired vector is thus
41. Unit vector e making an angle of
SOLUTION
4!
7
+
, +
,
1
2
1
2
"ev D " " p ; p D p ; " p :
5
5
5
5
with the x-axis
The unit vector e is the following vector:
+
,
4!
4!
e D cos
; sin
D h"0:22; 0:97i:
7
7
42. Vector v of length 2 making an angle of 30ı with the x-axis
SOLUTION
The desired vector is
ı
ı
v D 2hcos 30 ; sin 30 i D 2
*p
+
Dp E
3 1
;
D
3; 1 :
2 2
43. Find all scalars % such that % h2; 3i has length 1.
SOLUTION
We have:
The scalar % must satisfy
p
p
k%h2; 3ik D j%jkh2; 3ik D j%j 22 C 32 D j%j 13
p
j%j 13 D 1
)
1
j%j D p
13
44. Find a vector v satisfying 3v C h5; 20i D h11; 17i.
1
1
%1 D p ; %2 D " p
13
13
SOLUTION Write v D hx; yi to get the equation 3 hx; yi C h5; 20i D h11; 17i, which gives us 3x C 5 D 11 (and thus x D 2)
and also 3y C 20 D 17 (and so y D "1). Thus, v D h2; "1i.
45. What are the coordinates of the point P in the parallelogram in Figure 5(A)?
y
y
(7, 8)
P
(−1, b)
(2, 3)
(5, 4)
(−3, 2)
(2, 2)
(a, 1)
x
(A)
(B)
FIGURE 5
SOLUTION
We denote by A, B, C the points in the figure.
y
C (7, 8)
P (x0, y0)
B (5, 4)
A (2, 2)
x
x
S E C T I O N 11.5
Vectors in the Plane
1445
Let P D .x0 ; y0 /. We compute the following vectors:
""!
P C D h7 " x0 ; 8 " y0 i
""!
AB D h5 " 2; 4 " 2i D h3; 2i
""!
""!
The vectors P C and AB are equivalent, hence they have the same components. That is:
7 " x0 D 3
8 " y0 D 2
)
x0 D 4; y0 D 6
)
P D .4; 6/
46. What are the coordinates a and b in the parallelogram in Figure 5(B)?
SOLUTION
We denote the points in the figure by A, B, C and D.
y
B (−1, b)
C (2, 3)
A (−3, 2)
D (a, 1)
x
We compute the following vectors:
""!
AB D h"1 " ."3/; b " 2i D h2; b " 2i
""!
DC D h2 " a; 3 " 1i D h2 " a; 2i
""! ""!
Since AB D DC , the two vectors have the same components. That is,
2D2"a
b"2D2
aD0
)
bD4
""!
""!
47. Let v D AB and w D AC , where A; B; C are three distinct points in the plane. Match (a)–(d) with (i)–(iv). (Hint: Draw a
picture.)
(a) "w
(b) "v
(c) w " v
(d) v " w
""!
"!
""!
"!
(i) CB
(ii) CA
(iii) BC
(iv) BA
SOLUTION
"!
(a) "w has the same length as w and points in the opposite direction. Hence: "w D CA.
C
−w
A
"!
(b) "v has the same length as v and points in the opposite direction. Hence: "v D BA.
B
−v
A
(c) By the parallelogram law we have:
""! "! ""!
BC D BA C AC D "v C w D w " v
That is,
""!
w " v D BC
B
→
−v + w = BC
−v
A
w
C
1446
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
(d) By the parallelogram law we have:
""! "! ""!
CB D CA C AB D "w C v D v " w
That is,
""!
v " w D CB:
B
→
−w + v = CB
v
A
C
−w
48. Find the components and length of the following vectors:
(a) 4i C 3j
(b) 2i " 3j
(c) i C j
(d) i " 3j
SOLUTION
(a) Since i D h1; 0i and j D h0; 1i, using vector algebra we have:
4i C 3j D 4h1; 0i C 3h0; 1i D h4; 0i C h0; 3i D h4 C 0; 0 C 3i D h4; 3i
The length of the vector is:
k4i C 3jk D
p
42 C 32 D 5
(b) We use vector algebra and the definition of the standard basis vector to compute the components of the vector 2i " 3j:
2i " 3j D 2h1; 0i " 3h0; 1i D h2; 0i " h0; 3i D h2 " 0; 0 " 3i D h2; "3i
The length of this vector is:
k2i " 3jk D
(c) We find the components of the vector i C j:
q
p
22 C ."3/2 D 13
i C j D h1; 0i C h0; 1i D h1 C 0; 0 C 1i D h1; 1i
The length of this vector is:
ki C jk D
p
p
12 C 12 D 2
(d) We find the components of the vector i " 3j, using vector algebra:
i " 3j D h1; 0i " 3h0; 1i D h1; 0i " h0; 3i D h1 " 0; 0 " 3i D h1; "3i
The length of this vector is
ki " 3jk D
q
p
12 C ."3/2 D 10
In Exercises 49–52, calculate the linear combination.
49. 3j C .9i C 4j/
SOLUTION
We have:
" 23 i C
%
&
5 12 j " 12 i
50.
SOLUTION
We have:
51. .3i C j/ " 6j C 2.j " 4i/
SOLUTION
3j C .9i C 4j/ D 3 h0; 1i C 9 h1; 0i C 4 h0; 1i D h9; 7i
+
,
%1
&
%1
3
1
5
3
1 &
" i C 5 j " i D " h1; 0i C 5 h0; 1i " h1; 0i D "4;
2
2
2
2
2
2
2
We have:
.3i C j/ " 6j C 2.j " 4i/ D .h3; 0i C h0; 1i/ " h0; 6i C 2.h0; 1i " h4; 0i/ D h"5; "3i
S E C T I O N 11.5
Vectors in the Plane
1447
52. 3.3i " 4j/ C 5.i C 4j/
SOLUTION
We have:
3.3i " 4j/ C 5.i C 4j/ D 3.h3; 0i " h0; 4i/ C 5.h1; 0i C h0; 4i/ D h14; 8i
53. For each of the position vectors u with endpoints A, B, and C in Figure 6, indicate with a diagram the multiples rv and sw
""!
such that u D rv C sw. A sample is shown for u D OQ.
y
A
w
B
C
v
rv
x
Q
sw
FIGURE 6
SOLUTION
See the following three figures:
y
y
y
A
sw
w
w
w
B
v
rv
C
v
sw
x
rv
x
sw
v
x
rv
54. Sketch the parallelogram spanned by v D h1; 4i and w D h5; 2i. Add the vector u D h2; 3i to the sketch and express u as a
linear combination of v and w.
SOLUTION
We have
u D h2; 3i D rv C sw D rh1; 4i C sh5; 2i
which becomes the two equations
2 D r C 5s
3 D 4r C 2s
Solving the first equation for r gives
r D 2 " 5s
and substituting that into the first equation gives
3 D 4.2 " 5s/ C 2s D 8 " 18s
So 18s D 5, so s D 5=18, and thus r D 11=18. In other words,
u D h2; 3i D
11
5
h1; 4i C h5; 2i
18
18
as seen in this picture:
y
v
u
w
x
1448
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
In Exercises 55 and 56, express u as a linear combination u D rv C sw. Then sketch u; v; w, and the parallelogram formed by rv
and sw.
55. u D h3; "1i; v D h2; 1i, w D h1; 3i
SOLUTION
We have
u D h3; "1i D rv C sw D rh2; 1i C sh1; 3i
which becomes the two equations
3 D 2r C s
"1 D r C 3s
Solving the second equation for r gives r D "1 " 3s, and substituting that into the first equation gives 3 D 2."1 " 3s/ C s D
"2 " 6s C s, so 5 D "5s, so s D "1, and thus r D 2. In other words,
u D h3; "1i D 2h2; 1i " 1h1; 3i
as seen in this sketch:
y
w
v
x
u
56. u D h6; "2i; v D h1; 1i, w D h1; "1i
SOLUTION
We have
u D h6; "2i D rv C sw D rh1; 1i C sh1; "1i
which becomes the two equations
6DrCs
"2 D r " s
Adding gives 4 D 2r, so r D 2 and thus s D 4. In other words,
u D h6; "2i D 2h1; 1i C 4h1; "1i
as seen in this sketch:
y
v
x
w
u
57. Calculate the magnitude of the force on cables 1 and 2 in Figure 7.
65
25
Cable 1
Cable 2
50 kg
FIGURE 7
S E C T I O N 11.5
SOLUTION
#
#
Vectors in the Plane
1449
The three forces acting on the point P are:
The force F of magnitude 490 newtons that acts vertically downward.
The forces F1 and F2 that act through cables 1 and 2 respectively.
y
F1
F2
115°
25°
x
P
F
Since the point P is not in motion we have
F1 C F2 C F D 0
(1)
We compute the forces. Letting kF1 k D f1 and kF2 k D f2 we have:
F1 D f1 hcos 115ı ; sin 115ı i D f1 h"0:423; 0:906i
F2 D f2 hcos 25ı ; sin 25ı i D f2 h0:906; 0:423i
Substituting the forces in (1) gives
F D h0; "490i
f1 h"0:423; 0:906i C f2 h0:906; 0:423i C h0; "490i D h0; 0i
h"0:423f1 C 0:906f2 ; 0:906f1 C 0:423f2 " 490i D h0; 0i
We equate corresponding components and get
"0:423f1 C 0:906f2 D 0
0:906f1 C 0:423f2 " 490 D 0
By the first equation, f2 D 0:467f1 . Substituting in the second equation and solving for f1 yields
0:906f1 C 0:423 ! 0:467f1 " 490 D 0
1:104f1 D 490
)
f1 D 443:84; f2 D 0:467f1 D 207:27
We conclude that the magnitude of the force on cable 1 is f1 D 443:84 newtons and the magnitude of the force on cable 2 is
f2 D 207:27 newtons.
58. Determine the magnitude of the forces F1 and F2 in Figure 8, assuming that there is no net force on the object.
F2
20 kg
45°
30°
F1
FIGURE 8
SOLUTION We denote kF1 k D f1 and kF2 k D f2 . Note that the third force is of magnitude 20 kg D 196 newtons. It is
convenient (but not necessary) to redraw the vectors as being centered at the object, giving us the following figure.
F1
210
x
F3
196
−45
F2
Since there is no net force on the object, we have
F1 C F2 C F3 D 0
We find the forces:
F1 D f1 h0; 1i D h0; f1 i
(1)
1450
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
ı
ı
F2 D f2 hcos."45 /; sin."45 /i D f2
*p
p +
2
2
;"
D h0:707f2 ; "0:707f2 i
2
2
F3 D 196hcos 210ı ; sin 210ı i D h"169:74; "98i
We substitute the forces in (1):
h0; f1 i C h0:707f2 ; "0:707f2 i C h"169:74; "98i D h0; 0i
h0:707f2 " 169:74; f1 " 0:707f2 " 98i D h0; 0i
Equating corresponding components we obtain
0:707f2 " 169:74 D 0
f1 " 0:707f2 " 98 D 0
The first equation gives f2 D 240:08. Substituting in the second equation and solving for f1 gives
f1 " 0:707 ! 240:08 " 98 D 0 ) f1 D 267:74
The magnitude of the forces F1 and F2 are f1 D 267:74 newtons and f2 D 240:08 newtons respectively.
59. A plane flying due east at 200 km/h encounters a 40-km/h wind blowing in the north-east direction. The resultant velocity of
the plane is the vector sum v D v1 C v2 , where v1 is the velocity vector of the plane and v2 is the velocity vector of the wind
(Figure 9). The angle between v1 and v2 is !4 . Determine the resultant speed of the plane (the length of the vector v).
40 km/h
v2
v
v1
200 km/h
FIGURE 9
SOLUTION The resultant speed of the plane is the length of the sum vector v D v1 C v2 . We place the xy-coordinate system as
shown in the figure, and compute the components of the vectors v1 and v2 . This gives
v1 D hv1 ; 0i
*
p
p +
!
!E
2
2
v2 D v2 cos ; v2 sin
D v2 !
; v2 !
4
4
2
2
D
y
v2
v2
π
4
v1
We now compute the sum v D v1 C v2 :
v D hv1 ; 0i C
*p
2v2
;
2
p
2v2
2
x
v1
+
D
*p
+
p
2
2
v2 C v1 ;
v2
2
2
The resultant speed is the length of v, that is,
v
!2
!2 s
u p
p
p
q
u
p
v22
v2
2v
2v
2
2
2
t
2
v D kvk D
C v1 C
D
C v1 C 2 !
v2 v1 C 2 D v12 C v22 C 2v1 v2
2
2
2
2
2
Finally, we substitute the given information v1 D 200 and v2 D 40 in the equation above, to obtain
q
p
v D 2002 C 402 C 2 ! 200 ! 40 % 230 km=hr
S E C T I O N 11.5
Vectors in the Plane
1451
Further Insights and Challenges
In Exercises 60–62, refer to Figure 10, which shows a robotic arm consisting of two segments of lengths L1 and L2 .
y
L2
L1
2
1
P
r
1
x
FIGURE 10
""!
60. Find the components of the vector r D OP in terms of #1 and #2 .
SOLUTION
We denote by A the point in the figure.
y
A
90° −
2
1
90° −
2
P
1
O
2
− 90°
x
By the parallelogram law we have
"! ""!
r D OA C AP
"!
""!
We find the vectors OA and AP :
"!
# The vector OA has length L1 and it makes an angle of 90ı " #1 with the x-axis.
""!
# The vector AP has length L2 and it makes an angle of ".90ı " #2 / D #2 " 90ı with the x-axis.
Hence,
˝
˛
"!
OA D L1 cos.90ı " #1 /; sin.90ı " #1 / D L1 hsin #1 ; cos #1 i D hL1 sin #1 ; L1 cos #1 i
˝
˛
""!
AP D L2 cos.#2 " 90ı /; sin.#2 " 90ı / D L2 hsin #2 ; " cos #2 i D hL2 sin #2 ; "L2 cos #2 i
Substituting into (1) we obtain
r D hL1 sin #1 ; L1 cos #1 i C hL2 sin #2 " L2 cos #2 i
r D hL1 sin #1 C L2 sin #2 ; L1 cos #1 " L2 cos # 2 i
Thus, the x component of r is L1 sin #1 C L2 sin #2 and the y component is L1 cos #1 " L2 cos #2 .
61. Let L1 D 5 and L2 D 3. Find r for #1 D
SOLUTION
!
3 , #2
D
!
4.
In Exercise 60 we showed that
r D hL1 sin #1 C L2 sin #2 ; L1 cos #1 " L2 cos # 2 i
Substituting the given information we obtain
* p
p
p +
!
!
!
!E
5 3
3 2 5 3 2
r D 5 sin C 3 sin ; 5 cos " 3 cos
D
C
; "
% h6:45; 0:38i
3
4
3
4
2
2 2
2
D
62. Let L1 D 5 and L2 D 3. Show that the set of points reachable by the robotic arm with #1 D #2 is an ellipse.
SOLUTION
Substituting L1 D 5, L2 D 3, and #1 D #2 D # in the formula for r obtained in Exercise 60 we get
r D hL1 sin #1 C L2 sin #2 ; L1 cos #1 " L2 cos #2 i
D h5 sin # C 3 sin #; 5 cos # " 3 cos #i D h8 sin #; 2 cos #i
Thus, the x and y components of r are
x D 8 sin #; y D 2 cos #
(1)
1452
so
x
8
C H A P T E R 11
D sin #,
y
2
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
D cos #. Using the identity sin2 # C cos2 # D 1 we get
! x "2
8
which is the formula of an ellipse.
C
! y "2
D 1;
2
63. Use vectors to prove that the diagonals AC and BD of a parallelogram bisect each other (Figure 11). Hint: Observe that the
midpoint of BD is the terminal point of w C 12 .v " w/.
1
(v − w)
2
C
D
1
(v + w)
2
w
B
v
A
FIGURE 11
SOLUTION
We denote by O the midpoint of BD. Hence,
1 ""!
""!
DO D DB
2
C
v
D
O
w
w
B
v
A
Using the Parallelogram Law we have
""! ""! ""! ""! 1 ""!
AO D AD C DO D AD C DB
2
""!
""!
Since AD D w and DB D v " w we get
1
wCv
""!
AO D w C .v " w/ D
2
2
""! ""! ""!
On the other hand, AC D AD C DC D w C v, hence the midpoint O 0 of the diagonal AC is the terminal point of
(1)
wCv
2 .
That is,
""!0
wCv
AO D
2
C
v
D
(2)
O'
w
B
v
A
We combine (1) and (2) to conclude that O and O 0 are the same point. That is, the diagonal AC and BD bisect each other.
64. Use vectors to prove that the segments joining the midpoints of opposite sides of a quadrilateral bisect each other (Figure 12).
Hint: Show that the midpoints of these segments are the terminal points of
1
.2u C v C z/
4
1
.2v C w C u/
4
and
z
w
u
v
FIGURE 12
S E C T I O N 11.5
Vectors in the Plane
1453
SOLUTION We denote by A, B, C , D the corresponding points in the figure and by E, F , G, H the midpoints of the sides AB,
BC , CD and AD, respectively. Also, O is the midpoint of FH and O 0 is the midpoint of EG.
C
z
w
F
G
B
O
u
E
H
D
v
A
We must show that O and O 0 are the same point. Using the Parallelogram Law we have
1
1 ""!
""! ""! ""!
AO D AH C HO D v C HF
2
2
1
1
""! ""! ""! ""!
HF D HA C AB C BF D " v C u C z
2
2
Hence,
#
$
1
1
1
1
1
1
1
1
""!
AO D v C
" v C u C z D v C u C z D .2u C v C z/
2
2
2
2
4
2
4
4
(1)
C
z
w
F
G
B
u
Oʹ′
E
H
D
v
A
Similarly,
""!0 ""! ""! ""!0
1
AO D AD C DG C GO D v C w C
2
1
""! ""! ""! ""!
GE D GD C DA C AE D " w " v C
2
1 ""!
GE
2
1
u
2
Hence,
#
$
""!0
1
1
1
1
1
1
1
1
AO D v C w C
" w " v C u D v C w C u D .2v C w C u/
2
2
2
2
2
4
4
4
(2)
""! ""!
To show that AO D AO 0 we must express z in terms of u, v and w. We have
vCw"z"uD0 )z D vCw"u
Substituting into (1) we get
1
1
""!
AO D .2u C v C .v C w " u// D .2v C w C u/
(3)
4
4
""!
""!
By (2) and (3) we conclude that AO D AO 0 . It means that the points O and O 0 are the same point, in other words, the segment
FH and EG bisect each other.
65. Prove that two vectors v D ha; bi and w D hc; d i are perpendicular if and only if
ac C bd D 0
Suppose that the vectors v and w make angles #1 and #2 , which are not
x-axis. Then their components satisfy
SOLUTION
a D kvk cos #1
b D kvk sin #1
c D kwk cos #2
d D kwk sin #2
)
b
sin #1
D
D tan #1
a
cos #1
)
d
sin #2
D
D tan #2
c
cos #2
!
2
or
3!
2 ,
respectively, with the positive
1454
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
v
w
1
2
That is, the vectors v and w are on the lines with slopes
satisfy
b d
! D "1
a c
b
a
and
)
x
d
c , respectively.
bd D "ac
)
The lines are perpendicular if and only if their slopes
ac C bd D 0
We now consider the case where one of the vectors, say v, is perpendicular to the x-axis. In this case a D 0, and the vectors are
perpendicular if and only if w is parallel to the x-axis, that is, d D 0. So ac C bd D 0 ! c C b ! 0 D 0.
11.6 Dot Product and the Angle between Two Vectors
Preliminary Questions
1. Is the dot product of two vectors a scalar or a vector?
SOLUTION
The dot product of two vectors is the sum of products of scalars, hence it is a scalar.
2. What can you say about the angle between a and b if a ! b < 0?
a"b
Since the cosine of the angle between a and b satisfies cos # D kakkbk
, also cos # < 0. By definition 0 # # # !, but
since cos # < 0 then # is in .!=2; !". In other words, the angle between a and b is obtuse.
SOLUTION
3. Which property of dot products allows us to conclude that if v is orthogonal to both u and w, then v is orthogonal to u C w?
SOLUTION One property is that two vectors are orthogonal if and only if the dot product of the two vectors is zero. The second
property is the Distributive Law. Since v is orthogonal to u and w, we have v ! u D 0 and v ! w D 0. Therefore,
v ! .u C w/ D v ! u C v ! w D 0 C 0 D 0
We conclude that v is orthogonal to u C w.
4. Which is the projection of v along v: (a) v or (b) ev ?
SOLUTION
The projection of v along itself is v, since
vjj D
!v ! v"
v!v
vDv
Also, the projection of v along ev is the same answer, v, because
#
$
v ! ev
vjj D
ev D kvkev D v
ev ! ev
5. Let ujj be the projection of u along v. Which of the following is the projection u along the vector 2v and which is the projection
of 2u along v?
(a) 12 ujj
(b) ujj
(c) 2ujj
SOLUTION
Since ujj is the projection of u along v, we have,
ujj D
!u ! v"
v!v
v
The projection of u along the vector 2v is
#
$
#
$
#
$
!u ! v"
u ! 2v
2u ! v
4u ! v
2v D
2v D
vD
v D ujj
2v ! 2v
4v ! v
4v ! v
v!v
That is, ujj is the projection of u along 2v, so our answer is (b) for the first part. Notice that the projection of u along v is the
projection of u along the unit vector ev , hence it depends on the direction of v rather than on the length of v. Therefore, the
projection of u along v and along 2v is the same vector.
On the other hand, the projection of 2u along v is as follows:
#
$
!u ! v"
2u ! v
vD2
v D 2ujj
v!v
v!v
giving us answer (c) for the second part.
S E C T I O N 11.6
Dot Product and the Angle between Two Vectors
6. Which of the following is equal to cos #, where # is the angle between u and v?
(a) u ! v
(b) u ! ev
SOLUTION
1455
(c) eu ! ev
By the Theorems on the Dot Product and the Angle Between Vectors, we have
cos # D
u!v
u
v
D
!
D eu ! ev
kukkvk
kuk kvk
The correct answer is (c).
Exercises
In Exercises 1–4, compute the dot product.
1. h3; 1i ! h4; "7i
SOLUTION
2.
˝1
1
6; 2
The dot product of the two vectors is the following scalar:
h3; 1i ! h4; "7i D 3 ! 4 C 1 ! ."7/ D 5
˛ ˝ 1˛
! 3; 2
SOLUTION
The dot product is
+
3. i ! j
SOLUTION
, +
,
1 1
1
1
1 1
1
1
3
;
! 3;
D !3C ! D C D
6 2
2
6
2 2
2
4
4
By the orthogonality of i and j, we have i ! j D 0
4. j ! j
SOLUTION
Since j has length 1, we have j ! j D 1
In Exercises 5–6, determine whether the two vectors are orthogonal and, if not, whether the angle between them is acute or obtuse.
˝
˛ ˝1 7˛
4
5. 12
5 ; "5 ,
2;"4
SOLUTION
We find the dot product of the two vectors:
+
, +
,
# $ # $
12 4
1 7
12 1
4
7
12
28
13
;" ! ;" D
! C "
! "
D
C
D
5
5
2 4
5 2
5
4
10
20
5
The dot product is positive, hence the angle between the vectors is acute.
6. h12; 6i,
SOLUTION
h2; "4i
Since h12; 6i ! h2; "4i D 12 ! 2 C 6 ! ."4/ D 0, the vectors are orthogonal.
In Exercises 7–8, find the angle between the vectors. Use a calculator if necessary.
p
p ˛
˝ p ˛ ˝
7. 2; 2 , 1 C 2; 1 " 2
D p E
D
p
p E
SOLUTION We write v D 2; 2 and w D 1 C 2; 1 " 2 . To use the formula for the cosine of the angle # between two
vectors we need to compute the following values:
p
p
kvk D 4 C 2 D 6
q
p
p
p
kwk D .1 C 2/2 C .1 " 2/2 D 6
p
p
p
v!wD2C2 2C 2"2D3 2
Hence,
p
p
v!w
3 2
2
cos # D
D p p D
kvkkwk
2
6 6
and so,
# D cos!1
p
2
D !=4
2
1456
C H A P T E R 11
˝ p ˛
8. 5; 3 ,
˝p
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
˛
3; 2
D p E
Dp E
We denote v D 5; 3 and w D
3; 2 . To use the formula for the cosine of the angle # between two vectors we
need to compute the following values:
p
p
kvk D 25 C 3 D 28
p
p
kwk D 3 C 4 D 7
p
p
p
v!wD 5! 3C 3!2 D 7 3
SOLUTION
Hence,
cos # D
p
p
v!w
7 3
3
D p p D
kvkkwk
2
28 7
and so,
!1
# D cos
p
3
D !=6
2
In Exercises 9–12, simplify the expression.
9. .v " w/ ! v C v ! w
SOLUTION
By properties of the dot product we obtain
.v " w/ ! v C v ! w D v ! v " w ! v C v ! w D kvk2 " v ! w C v ! w D kvk2
10. .v C w/ ! .v C w/ " 2v ! w
SOLUTION
Using properties of the dot product we obtain
.v C w/ ! .v C w/ " 2v ! w D v ! .v C w/ C w ! .v C w/ " 2v ! w D v ! v C v ! w C w ! v C w ! w " 2v ! w
D kvk2 C v ! w C v ! w C kwk2 " 2v ! w D kvk2 C kwk2
11. .v C w/ ! v " .v C w/ ! w
SOLUTION
We use properties of the dot product to write
.v C w/ ! v " .v C w/ ! w D v ! v C w ! v " v ! w " w ! w
D kvk2 C w ! v " w ! v " kwk2 D kvk2 " kwk2
12. .v C w/ ! v " .v " w/ ! w
SOLUTION
By properties of the dot product we get
.v C w/ ! v " .v " w/ ! w D .v C w/ ! v " w ! .v " w/
Dv!vCw!v"w!vCw!w
D v ! v C w ! w D kvk2 C kwk2
In Exercises 13–16, use the properties of the dot product to evaluate the expression, assuming that u ! v D 2, kuk D 1, and kvk D 3.
13. u ! .4v/
SOLUTION
Using properties of the dot product we get
u ! .4v/ D 4.u ! v/ D 4 ! 2 D 8:
14. .u C v/ ! v
SOLUTION
Using the distributive law and the dot product relation with length we obtain
.u C v/ ! v D u ! v C v ! v D u ! v C kvk2 D 2 C 32 D 11:
15. 2u ! .3u " v/
SOLUTION
By properties of the dot product we obtain
2u ! .3u " v/ D .2u/ ! .3u/ " .2u/ ! v D 6.u ! u/ " 2.u ! v/
D 6kuk2 " 2.u ! v/ D 6 ! 12 " 2 ! 2 D 2
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
16. .u C v/ ! .u " v/
SOLUTION
We use the distributive law, commutativity and the relation with length to write
.u C v/ ! .u " v/ D u ! .u " v/ C v ! .u " v/ D kuk2 " u ! v C u ! v " kvk2
D kuk2 " kvk2 D 12 " 32 D "8
17. Find the angle between v and w if v ! w D "kvk kwk.
SOLUTION
Using the formula for dot product, and the given equation v ! w D "kvk kwk, we get:
kvk kwk cos # D "kvk kwk;
which implies cos # D "1, and so the angle between the two vectors is # D !.
18. Find the angle between v and w if v ! w D 12 kvk kwk.
SOLUTION
Using the formula for dot product, and the given equation v ! w D 12 kvk kwk, we get:
kvk kwk cos # D
which implies cos # D
1
2,
1
kvk kwk;
2
and so the angle between the two vectors is # D !=3.
!
3.
32 C 52
19. Assume that kvk D 3, kwk D 5 and that the angle between v and w is # D
2
(a) Use the relation kv C wk D .v C w/ ! .v C w/ to show that kv C
(b) Find kv C wk.
SOLUTION
wk2
D
C 2v ! w.
For part (a), we use the distributive property to get:
kv C wk2 D .v C w/ ! .v C w/
D v!vCv!wCw!vCw!w
D kvk2 C 2v ! w C kwk2
D 32 C 52 C 2v ! w
For part (b), we use the definition of dot product on the previous equation to get:
kv C wk2 D 32 C 52 C 2v ! w
D 34 C 2 ! 3 ! 5 ! cos !=3
Thus, kv C wk D
p
D 34 C 15 D 49
49 D 7:
20. Assume that kvk D 2, kwk D 3, and the angle between v and w is 120ı . Determine:
(a) v ! w
(b) k2v C wk
(c) k2v " 3wk
SOLUTION
(a) We use the relation between the dot product and the angle between two vectors to write
# $
1
v ! w D kvkkwk cos # D 2 ! 3 cos 120ı D 6 ! "
D "3
2
(b) By the relation of the dot product with length and by properties of the dot product we have
k2v C wk2 D .2v C w/ ! .2v C w/ D 4v ! v C 2v ! w C 2w ! v C w ! w
D 4kvk2 C 4v ! w C kwk2
We now substitute v ! w D "3 from part (a) and the given information, obtaining
k2v C wk2 D 4 ! 22 C 4."3/ C 32 D 13
)
k2v C wk D
p
13 % 3:61
(c) We express the length in terms of a dot product and use properties of the dot product. This gives
k2v " 3wk2 D .2v " 3w/ ! .2v " 3w/ D 4v ! v " 6v ! w " 6w ! v C 9w ! w
D 4kvk2 " 12v ! w C 9kwk2
Substituting v ! w D "3 from part (a) and the given values yields
k2v " 3wk2 D 4 ! 22 " 12."3/ C 9 ! 32 D 133
)
k2v " 3wk D
p
133 % 11:53
1457
1458
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
21. Show that if e and f are unit vectors such that ke C fk D 32 , then ke " fk D
SOLUTION
p
7
2 .
Hint: Show that e ! f D 18 .
We use the relation of the dot product with length and properties of the dot product to write
9=4 D ke C fk2 D .e C f/ ! .e C f/ D e ! e C e ! f C f ! e C f ! f
D kek2 C 2e ! f C kfk2 D 12 C 2e ! f C 12 D 2 C 2e ! f
We now find e ! f:
9=4 D 2 C 2e ! f
e ! f D 1=8
)
Hence, using the same method as above, we have:
ke " fk2 D .e " f/ ! .e " f/ D e ! e " e ! f " f ! e C f ! f
D kek2 " 2e ! f C kfk2 D 12 " 2e ! f C 12 D 2 " 2e ! f D 2 " 2=8 D 7=4:
Taking square roots, we get:
ke " fk D
p
7
2
22. Find k2e " 3fk assuming that e and f are unit vectors such that ke C fk D
SOLUTION
p
3=2.
We use the relation of the dot product with length and properties of the dot product to write
3=2 D ke C fk2 D .e C f/ ! .e C f/ D e ! e C e ! f C f ! e C f ! f
D kek2 C 2e ! f C kfk2 D 12 C 2e ! f C 12 D 2 C 2e ! f
We now find e ! f:
3=2 D 2 C 2e ! f
e ! f D "1=4
)
Hence, using the same method as above, we have:
k2e " 3fk2 D .2e " 3f/ ! .2e " 3f/
D k2ek2 " 2 ! 2e ! 3f C k3fk2 D 22 " 12e ! f C 32 D 13 C 3 D 16:
Taking square roots, we get:
k2e " 3fk D 4
23. Find the angle # in the triangle in Figure 1.
y
(0, 10)
(10, 8)
(3, 2)
x
FIGURE 1
SOLUTION
We denote by u and v the vectors in the figure.
y
(0, 10)
v
(10, 8)
u
(3, 2)
x
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1459
Hence,
cos # D
v!u
kvkkuk
(1)
We find the vectors v and u, and then compute their length and the dot product v ! u. This gives
v D h0 " 10; 10 " 8i D h"10; 2i
u D h3 " 10; 2 " 8i D h"7; "6i
q
p
kvk D ."10/2 C22 D 104
q
p
kuk D ."7/2 C ."6/2 D 85
v ! u D h"10; 2i ! h"7; "6i D ."10/ ! ."7/ C 2 ! ."6/ D 58
Substituting these values in (1) yields
cos # D p
58
p % 0:617
104 85
Hence the angle of the triangle is 51:91ı .
24. Find all three angles in the triangle in Figure 2.
y
(2, 7)
(6, 3)
x
(0, 0)
FIGURE 2
SOLUTION
We denote by u, v and w the vectors and by #1 , #2 , and #3 the angles shown in the figure. We compute the vectors:
u D h2; 7i
v D h6; 3i
w D h6 " 2; 3 " 7i D h4; "4i
y
(2, 7)
3
w
u
2
1
(6, 3)
v
x
(0, 0)
Since the angles are acute the cosines are positive, so we have
cos #1 D
ju ! vj
;
kukkvk
cos #2 D
jv ! wj
;
kvkkwk
cos #3 D 180 " .#1 C #2 /
We compute the lengths and the dot products in (1):
u ! v D h2; 7i ! h6; 3i D 2 ! 6 C 7 ! 3 D 33
v ! w D h6; 3i ! h4; "4i D 6 ! 4 C 3 ! ."4/ D 12
(1)
1460
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
p
p
22 C 72 D 53
p
p
kvk D 62 C 32 D 45
q
p
kwk D 42 C ."4/2 D 32
kuk D
Substituting in (1) and solving for acute angles yields
33
cos #1 D p p % 0:676
53 45
)
#1 % 47:47ı
12
cos #2 D p p % 0:316
45 32
)
#2 % 71:58ı
The sum of the angles in a triangle is 180ı , hence
#3 D 180ı " .47:47 C 71:58/ % 60:95ı :
In Exercises 25–26, find the projection of u along v.
25. u D h2; 5i,
SOLUTION
v D h1; 1i
We first compute the following dot products:
u ! v D h2; 5i ! h1; 1i D 7
v ! v D kvk2 D 12 C 12 D 2
The projection of u along v is the following vector:
ujj D
26. u D h2; "3i,
SOLUTION
v D h1; 2i
!u ! v"
v!v
vD
+
,
7
7 7
vD ;
2
2 2
We first compute the following dot products:
u ! v D h2; "3i ! h1; 2i D "4
v ! v D kvk2 D 12 C 22 D 5
The projection of u along v is the following vector:
ujj D
27. Find the length of OP in Figure 3.
!u ! v"
v!v
vD
+
,
"4
"4 "8
vD
;
5
5 5
y
u = !3, 5"
u⊥
v = !8, 2"
P
O
x
FIGURE 3
SOLUTION
This is just the component of u D h3; 5i along v D h8; 2i. We first compute the following dot products:
u ! v D h3; 5i ! h8; 2i D 34
v ! v D kvk2 D 82 C 22 D 68
The component of u along v is the length of the projection of u along v
-! u ! v " 34 p
34
v- D
kvk D
68
v!v
68
68
28. Find ku? k in Figure 3.
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1461
SOLUTION From the previous problem (see solution above) we know that the component of u along v is 1=2, and thus the
projection is uk D h4; 1i. Using the standard formula for u? , we obtain
u? D u " uk D h3; 5i " h4; 1i D h"1; 4i
In Exercises 29–30, find the decomposition a D ajj C a? with respect to b.
29. a D h1; 0i,
b D h1; 1i
SOLUTION
Step 1. We compute a ! b and b ! b
a ! b D h1; 0i ! h1; 1i D 1 ! 1 C 0 ! 1 D 1
b ! b D kbk2 D 12 C 12 D 2
Step 2. We find the projection of a along b:
ajj D
#
$
+
,
a!b
1
1 1
b D h1; 1i D ;
b!b
2
2 2
Step 3. We find the orthogonal part as the difference:
a? D a " ajj D h1; 0i "
+
, +
,
1 1
1 1
;
D ;"
2 2
2 2
Hence,
a D ajj C a? D
30. a D h2; "3i,
SOLUTION
+
, +
,
1 1
1 1
;
C ;" :
2 2
2 2
b D h5; 0i
We first compute a ! b and b ! b to find the projection of a along b:
a ! b D h2; "3i ! h5; 0i D 10
b ! b D kbk2 D 52 C 02 D 25
Hence,
ajj D
#
$
a!b
10
bD
h5; 0i D h2; 0i
b!b
25
We now find the vector a? orthogonal to b by computing the difference:
a " ajj D h2; "3i " h2; 0i D h0; "3i
Thus, we have
31. Let e" D hcos #; sin #i. Show that e" ! e
SOLUTION
a D ajj C a? D h2; 0i C h0; "3i
D cos.# "
/ for any two angles # and
.
First, e" is a unit vector since by a trigonometric identity we have
ke" k D
p
p
cos2 # C sin2 # D 1 D 1
The cosine of the angle ˛ between e" and the vector i in the direction of the positive x-axis is
cos ˛ D
e" ! i
D e" ! i D ..cos #/i C .sin #/j/ ! i D cos #
ke" k ! kik
The solution of cos ˛ D cos # for angles between 0 and ! is ˛ D #. That is, the vector e" makes an angle # with the x-axis. We
now use the trigonometric identity
cos # cos
C sin # sin
D cos.# "
/
to obtain the following equality:
e" ! e
D hcos #; sin #i ! hcos ; sin i D cos # cos
C sin # sin
D cos.# "
/
1462
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
32.
Let v and w be vectors in the plane.
(a) Use Theorem 2 to explain why the dot product v ! w does not change if both v and w are rotated by the same angle #.
Dp p E
(b) Sketch the vectors e1 D h1; 0i and e2 D 22 ; 22 , and determine the vectors e01 ; e02 obtained by rotating e1 ; e2 through an
angle
!
4.
Verify that e1 ! e2 D e01 ! e02 .
SOLUTION
(a) By Theorem 2,
v ! w D kvkkwk cos ˛
where ˛ is the angle between v and w. Since rotation by an angle # does not change the angle between the vectors, nor the norms
of the vectors, the dot product v ! w remains unchanged.
y
e2 =
〈
2
2
,
2
2
〉
x
e1 = 〈1, 0〉
(b) Notice from the picture that if weD p
rotate
e E by !=4, we get e2 , and when we rotate e2p by the same
amount
we get a unit
p 1
p
p
2
2
2
2
2
0
0
vector along the y axis. Thus, e1 D 2 ; 2 and e2 D h0; 1i. Note that e1 ! e2 D 1 ! 2 C 0 ! 2 D 2 and e01 ! e02 D
0!
p
2
2
C1!
p
2
2
D
p
2
2 .
Thus, e1 ! e2 D e01 ! e02 .
33.
Let v and w be nonzero vectors and set u D ev C ew . Use the dot product to show that the angle between u and v is
equal to the angle between u and w. Explain this result geometrically with a diagram.
SOLUTION We denote by ˛ the angle between u and v and by ˇ the angle between u and w. Since ev and ew are vectors in the
directions of v and w respectively, ˛ is the angle between u and ev and ˇ is the angle between u and ew . The cosines of these angles
are thus
u ! ev
u ! ev
u ! ew
u ! ew
cos ˛ D
D
I cos ˇ D
D
kukkev k
kuk
kukkew k
kuk
To show that cos ˛ D cos ˇ (which implies that ˛ D ˇ) we must show that
u ! ev D u ! ew :
We compute the two dot products:
u ! ev D .ev C ew / ! ev D ev ! ev C ew ! ev D 1 C ew ! ev
u ! ew D .ev C ew / ! ew D ev ! ew C ew ! ew D ev ! ew C 1
We see that u ! ev D u ! ew . We conclude that cos ˛ D cos ˇ, hence ˛ D ˇ. Geometrically, u is a diagonal in the rhombus OABC
(see figure), hence it bisects the angle ^AOC of the rhombus.
v
A
B
ev
O
34.
ample.
SOLUTION
u
ew
C
w
Let v, w, and a be nonzero vectors such that v ! a D w ! a. Is it true that v D w? Either prove this or give a counterexThe equality v ! a D w ! a is equivalent to the following equality:
v!aD w!a
v!a"w!aD 0
.v " w/ ! a D 0
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1463
That is, v " w is orthogonal to a rather than v D w. Consider the following counterexample:
a D h1; 1iI
v D h1; 0iI
w D h0; 1i
Obviously, v ¤ w, but v ! a D w ! a since
v ! a D h1; 0i ! h1; 1i D 1 ! 1 C 1 ! 0 D 1
w ! a D h0; 1i ! h1; 1i D 0 ! 1 C 1 ! 1 D 1
35. Calculate the force (in newtons) required to push a 40-kg wagon up a 10ı incline (Figure 4).
40 kg
10°
FIGURE 4
SOLUTION Gravity exerts a force Fg of magnitude 40g newtons where g D 9:8. The magnitude of the force required to push
the wagon equals the component of the force Fg along the ramp. Resolving Fg into a sum Fg D Fjj C F? , where Fjj is the force
along the ramp and F? is the force orthogonal to the ramp, we need to find the magnitude of Fjj . The angle between Fg and the
ramp is 90ı " 10ı D 80ı . Hence,
Fjj D kFg k cos 80ı D 40 ! 9:8 ! cos 80ı % 68:07 N:
F||
80°
10°
Fg
F
Therefore the minimum force required to push the wagon is 68.07 N. (Actually, this is the force required to keep the wagon from
sliding down the hill; any slight amount greater than this force will serve to push it up the hill.)
36. A force F is applied to each of two ropes (of negligible weight) attached to opposite ends of a 40-kg wagon and making an
angle of 35ı with the horizontal (Figure 5). What is the maximum magnitude of F (in newtons) that can be applied without lifting
the wagon off the ground?
F
F
35°
35°
40 kg
FIGURE 5
SOLUTION With two ropes at either end, both at the same angle with the horizontal and both with the same force, pulling on the
40-kg wagon, each rope will need to lift 20 kg. Let’s look at the situation on the right-hand side of the wagon. We resolve the force
F on the right-hand rope into a sum F D Fjj C F? where Fjj is the horizontal force and F? is the force orthogonal to the ground.
The wagon will not be lifted off the ground if the magnitude of F? , that is the component of F along the direction orthogonal to the
ground, is equal to (but not more than) the magnitude of the force due to gravity from 20 kg (remember, each rope needs to only
lift half of the wagon, and remember also that the acceleration due to gravity is 9.8 meters per second squared). That is,
(1)
20.9:8/ D kF? k
The angle between F and a vector orthogonal to the ground is
90ı
" 35ı
D
55ı ,
hence, 20.9:8/ D 196 D
F
F⊥
55°
35°
F||
Fg
This gives us
196 D kFk cos 55ı
)
kFk D
196
% 341 Newtons
cos 55ı
The maximum force that can be applied is of magnitude 341 newtons on each rope.
kFk cos 55ı
1464
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
37. A light beam travels along the ray determined by a unit vector L, strikes a flat surface at point P , and is reflected along the ray
determined by a unit vector R, where #1 D #2 (Figure 6). Show that if N is the unit vector orthogonal to the surface, then
R D 2.L ! N/N " L
Incoming light
Reflected light
N
L
R
1
2
P
FIGURE 6
SOLUTION
We denote by W a unit vector orthogonal to N in the direction shown in the figure, and let #1 D #2 D #.
Incoming light
Reflected light
N
L
R
W
We resolve the unit vectors R and L into a sum of forces along N and W. This gives
R D cos.90 " #/W C cos #N D sin #W C cos #N
L D " cos.90 " #/W C cos #N D " sin #W C cos #N
(1)
N
W
Now, since
L ! N D kLkkNk cos # D 1 ! 1 cos # D cos #
N
N
L
R
90 +
90 −
W
W
we have by (1):
2.L ! N/N " L D .2 cos #/N " L D .2 cos #/N " .." sin #/W C .cos #/N/
D .2 cos #/N C .sin #/W " .cos #/N D .sin #/W C .cos #/N D R
38. Verify the Distributive Law:
u ! .v C w/ D u ! v C u ! w
SOLUTION
We denote the components of the vectors u, v, and w by
u D ha1 ; a2 iI
v D hb1 ; b2 iI
w D hc1 ; c2 i
We compute the left-hand side:
u ! .v C w/ D ha1 ; a2 i .hb1 ; b2 i C hc1 ; c2 i/
D ha1 ; a2 i ! hb1 C c1 ; b2 C c2 i
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1465
D ha1 .b1 C c1 /; a2 .b2 C c2 /i
Using the distributive law for scalars and the definitions of vector sum and the dot product we get
u ! .v C w/ D ha1 b1 C a1 c1 ; a2 b2 C a2 c2 i
D ha1 b1 ; a2 b2 i C ha1 c1 ; a2 c2 i
D ha1 ; a2 i ! hb1 ; b2 i C ha1 ; a2 i ! hc1 ; c2 i
Du!vCu!w
39. Verify that .%v/ ! w D %.v ! w/ for any scalar %.
SOLUTION
We denote the components of the vectors v and w by
v D ha1 ; a2 i w D hb1 ; b2 i
Thus,
.%v/ ! w D .%ha1 ; a2 i/ ! hb1 ; b2 i D h%a1 ; %a2 i ! hb1 ; b2 i
D %a1 b1 C %a2 b2
Recalling that %, ai , and bi are scalars and using the definitions of scalar multiples of vectors and the dot product, we get
.%v/ ! w D %.a1 b1 C a2 b2 / D % .ha1 ; a2 i ! hb1 ; b2 i/ D %.v ! w/
Further Insights and Challenges
40. Prove the Law of Cosines, c 2 D a2 C b 2 " 2ab cos #, by referring to Figure 7. Hint: Consider the right triangle 4PQR.
R
a
c
a sin
P
Q
b
b − a cos
FIGURE 7
SOLUTION
""! "! "!
We denote the vertices of the triangle by S, Q, and R. Since RQ D RS C SQ, we have
""! 2 ""! ""! !"! "!" !"! "!"
c 2 D kRQk D RQ ! RQ D RS C SQ ! RS C SQ
"! "! "! "! "! "! "! "!
D RS ! RS C RS ! SQ C SQ ! RS C SQ ! SQ
"! 2
"! "!
"! 2
D kRSk C 2RS ! SQ C kSQk
"! "!
c 2 D a2 C 2RS ! SQ C b 2
(1)
R
a
c
P
S
Q
b
"! "!
"!
"!
We find the dot product RS ! SQ. The angle between the vectors RS and SQ is #, hence,
"! "!
"!
"!
SR ! SQ D kSRk ! kSQk cos # D ab cos #:
Therefore,
"! "!
"! "!
RS ! SQ D "SR ! SQ D "ab cos #
Substituting (2) in (1) yields
c 2 D a2 " 2ab cos # C b 2 D a2 C b 2 " 2ab cos #:
(Note that we did not need to use the point P .)
(2)
1466
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
41. In this exercise, we prove the Cauchy–Schwarz inequality: If v and w are any two vectors, then
jv ! wj # kvk kwk
6
(a) Let f .x/ D kxv C wk2 for x a scalar. Show that f .x/ D ax 2 C bx C c, where a D kvk2 , b D 2v ! w, and c D kwk2 .
(b) Conclude that b 2 " 4ac # 0. Hint: Observe that f .x/ $ 0 for all x.
SOLUTION
(a) We express the norm as a dot product and compute it:
f .x/ D kxv C wk2 D .xv C w/ ! .xv C w/
D x 2 v ! v C xv!w C xw ! v C w ! w D kvk2 x 2 C 2.v ! w/x C kwk2
Hence, f .x/ D ax 2 C bx C c, where a D kvk2 , b D 2v ! w, and c D kwk2 .
(b) If f has distinct real roots x1 and x2 , then f .x/ is negative for x between x1 and x2 , but this is impossible since f is the
square of a length.
y
x1
x2
x
f (x) = ax2 + bx + c, a > 0
Using properties of quadratic functions, it follows that f has a nonpositive discriminant. That is, b 2 " 4ac # 0. Substituting the
values for a, b, and c, we get
4.v ! w/2 " 4kvk2 kwk2 # 0
.v ! w/2 # kvk2 kwk2
Taking the square root of both sides we obtain
jv ! wj # kvkkwk
42. Use (6) to prove the Triangle Inequality
kv C wk # kvk C kwk
Hint: First use the Triangle Inequality for numbers to prove
j.v C w/ ! .v C w/j # j.v C w/ ! vj C j.v C w/ ! wj
SOLUTION
Using the relation between the length and dot product we have
kv C wk2 D .v C w/ ! .v C w/ D v ! v C v ! w C w ! v C w ! w
D kvk2 C 2v ! w C kwk2
(1)
Obviously, v ! w # jv ! wj. Also, by the Cauchy–Schwarz inequality jv ! wj # kvkkwk. Therefore, v ! w # kvkkwk, and combining
this with (1) we get
kvCwk2 D kvk2 C 2v ! w C kwk2 # kvk2 C 2kvkkwk C kwk2 D .kvk C kwk/2
That is,
kvCwk2 # .kvk C kwk/2
Taking the square roots of both sides and recalling that the length is nonnegative, we get
kvCwk # kvk C kwk
Dot Product and the Angle between Two Vectors
S E C T I O N 11.6
1467
43. This exercise gives another proof of the relation between the dot product and the angle # between two vectors v D ha1 ; b1 i
and w D ha2 ; b2 i in the plane. Observe that v D kvk hcos #1 ; sin #1 i and w D kwk hcos #2 ; sin #2 i, with #1 and #2 as in Figure 8.
Then use the addition formula for the cosine to show that
v ! w D kvk kwk cos #
y
y
a2
y
w
w
a1
b2
2
v
1
x
v
b1
x
x
=
2
−
1
FIGURE 8
SOLUTION
Using the trigonometric function for angles in right triangles, we have
a2 D kvk sin #1 ;
a1 D kvk cos #1
b2 D kwk sin #2 ;
b1 D kwk cos #2
Hence, using the given identity we obtain
v ! w D ha1 ; a2 i ! hb1 ; b2 i D a1 b1 C a2 b2 D kvk cos #1 kwk cos #2 C kvk sin #1 kwk sin #2
D kvkkwk.cos #1 cos #2 C sin #1 sin #2 / D kvkkwk cos.#1 " #2 /
That is,
v ! w D kvkkwk cos.#/
44. Let v D hx; yi and
v" D hx cos # C y sin #; "x sin # C y cos #i
Prove that the angle between v and v" is #.
SOLUTION The dot product of the vectors v and v" is
v ! v" D hx; yi ! hx cos # C y sin #; "x sin # C y cos #i
D x.x cos # C y sin #/ C y."x sin # C y cos #/
D x 2 cos # C xy sin # " xy sin # C y 2 cos #
D .x 2 C y 2 / cos #
That is,
v ! v" D .x 2 C y 2 / cos #
(1)
On the other hand, if ˛ denotes the angle between v and v" , we have
v ! v" D kvkkv" k cos ˛
(2)
We compute the lengths. Using the identities cos2 # C sin2 # D 1 and 2 sin # cos # D sin 2#, we obtain
q
p
kvk D hx; yi D x 2 C y 2
q
kv" k D .x cos # C y sin #/2 C ."x sin # C y cos #/2
q
D x 2 cos2 # C xy sin 2# C y 2 sin2 # C x 2 sin2 # " xy sin 2# C y 2 cos2 #
q
q
q
D x 2 .cos2 # C sin2 #/ C y 2 .sin2 # C cos2 #/ D x 2 ! 1 C y 2 ! 1 D x 2 C y 2
Substituting the lengths in (2) yields
v ! v" D
We now equate (1) and (3) to obtain
q
q
x 2 C y 2 ! x 2 C y 2 cos ˛ D .x 2 C y 2 / cos ˛
.x 2 C y 2 / cos # D .x 2 C y 2 / cos ˛
cos # D cos ˛
The solution for angles between 0ı and 180ı is ˛ D 0. That is, the angle between v and v" is #.
(3)
1468
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
11.7 Calculus of Vector-Valued Functions
Preliminary Questions
1. State two forms of the Product Rule for vector-valued functions.
SOLUTION
The Product Rule for scalar multiple f .t/ of a vector-valued function r.t/ states that:
d
f .t/r.t/ D f .t/r0 .t/ C f 0 .t/r.t/
dt
The Product Rule for dot products states that:
d
r1 .t/ ! r2 .t/ D r1 .t/ ! r02 .t/ C r01 .t/ ! r2 .t/
dt
In Questions 2–5, indicate whether the statement is true or false, and if it is false, provide a correct statement.
2. The derivative of a vector-valued function is defined as the limit of the difference quotient, just as in the scalar-valued case.
SOLUTION
The statement is true. The derivative of a vector-valued function r.t/ is defined a limit of the difference quotient:
r .t C h/ " r.t/
t !0
h
r0 .t/ D lim
in the same way as in the scalar-valued case.
3. There are two Chain Rules for vector-valued functions: one for the composite of two vector-valued functions and one for the
composite of a vector-valued and a scalar-valued function.
SOLUTION This statement is false. A vector-valued function r.t/ is a function whose domain is a set of real numbers and whose
range consists of position vectors. Therefore, if r1 .t/ and r2 .t/ are vector-valued functions, the composition “.r1 ! r2 /.t/ D
r1 .r2 .t//” has no meaning since r2 .t/ is a vector and not a real number. However, for a scalar-valued function f .t/, the composition
r.f .t// has a meaning, and there is a Chain Rule for differentiability of this vector-valued function.
4. The terms “velocity vector” and “tangent vector” for a path r.t/ mean one and the same thing.
SOLUTION
This statement is true.
5. The derivative of a vector-valued function is the slope of the tangent line, just as in the scalar case.
SOLUTION The statement is false. The derivative of a vector-valued function is again a vector-valued function, hence it cannot
be the slope of the tangent line (which is a scalar). However, the derivative, r0 .t0 / is the direction vector of the tangent line to the
curve traced by r.t/, at r.t0 /.
6. State whether the following derivatives of vector-valued functions r1 .t/ and r2 .t/ are scalars or vectors:
&
d
d %
(a)
r1 .t/
(b)
r1 .t/ ! r2 .t/
dt
dt
SOLUTION
(a) vector, (b) scalar
Exercises
In Exercises 1–6, evaluate the limit.
D
E
1. lim t 2 ; 4t
t !3
SOLUTION
By the theorem on vector-valued limits we have:
,
D
E +
lim t 2 ; 4t D lim t 2 ; lim 4t D h9; 12i :
t !3
t !3
t !3
2. lim sin 2ti C cos tj
t !!
SOLUTION
We compute the limit of each component. That is:
#
$
#
$
lim .sin 2ti C cos tj/ D lim sin 2t i C lim cos t j
t !!
t !!
t !!
D .sin 2!/ i C .cos !/ j D "j:
S E C T I O N 11.7
Calculus of Vector-Valued Functions
1469
3. lim e 2t i C ln.t C 1/j
t !0
SOLUTION
4. lim
t !0
+
Computing the limit of each component, we obtain:
$
#
$
!
" #
lim e 2t i C ln .t C 1/ j D lim e 2t i C lim ln.t C 1/ j D e 0 i C .ln 1/j D i
1
;
t C1
SOLUTION
et
t !0
"1
t
t !0
,
t !0
We use the theorem on vector-valued limits and L’Hôpital’s rule to write:
, +
, +
,
1
et " 1
1
et " 1
et
;
D lim
; lim
D 1; lim
D h1; 1i :
t !0 t C 1
t !0 t C 1 t !0
t !0 1
t
t
˝
˛
r.t C h/ " r.t/
5. Evaluate lim
for r.t/ D t !1 ; sin t .
h
h!0
lim
SOLUTION
+
This limit is the derivative
dr
.
dt
Using componentwise differentiation yields:
+ !
, +
,
" d
r .t C h/ " r.t/
dr
d
1
D
D
t !1 ;
.sin t/ D " 2 ; cos t :
h
dt
dt
dt
t
h!0
lim
6. Evaluate lim
t !0
SOLUTION
r.t/
for r.t/ D hsin t; 1 " cos ti.
t
r.t /
t !0 t
Since r.0/ D hsin 0; 1 " cos 0i D h0; 0i we may think of the limit lim
componentwise differentiation. That is,
as a derivative and compute it using
ˇ
+
,ˇ
ˇ
r.t/
r.t/ " r.0/
r .0 C h/ " r.0/
d r ˇˇ
d
d
lim
D lim
D lim
D
D
.sin t/ ;
.1 " cos t/ ˇˇ
t !0 t
t !0
t
h
dt ˇt D0
dt
dt
h!0
t D0
ˇ
ˇ
D hcos t; sin ti ˇˇ
D hcos 0; sin 0i D h1; 0i
t D0
In Exercises 7–12, compute the derivative.
˝
˛
7. r.t/ D t; t 2
Using componentwise differentiation we get:
+
,
dr
d
d
D
.t/; .t 2 / D h1; 2ti
dt
dt
dt
˝
p ˛
8. r.t/ D 7 " t; 4 t
SOLUTION
Using componentwise differentiation we get:
+
, D
,
E +
dr
d
d p
2
D
.7 " t/; .4 t/ D "1; 2t !1=2 D "1; p
dt
dt
dt
t
˝ 3s !s ˛
9. r.s/ D e ; e
SOLUTION
Using componentwise differentiation we get:
+
, D
E
dr
d 3s d !s
D
.e /; .e / D 3e 3s ; "e !s
ds
ds
ds
D
E
3t !4 6!t
10. b.t/ D e
;e
SOLUTION
SOLUTION
Using componentwise differentiation we get:
+
, D
E
db
d 3t !4 d 6!t
D
.e
/; .e
/ D 3e 3t !4 ; "e 6!t
dt
dt
dt
11. c.t/ D t !1 i
SOLUTION
Using componentwise differentiation we get:
%
&0
c0 .t/ D t !1 i D "t !2 i
1470
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
12. a.#/ D .cos 3#/i C .sin2 #/j
SOLUTION
Using componentwise differentiation we get:
13. Calculate
r0 .t/
SOLUTION
We perform the differentiation componentwise to obtain:
and
r00 .t/
˝
˛
for r.t/ D t; t 2 .
a0 .#/ D "3 sin 3#i C 2 sin # cos #j
˝
˝
˛
00 ˛
r0 .t/ D .t/0 ; .t 2 / D 1; 2t
We now differentiate the derivative vector to find the second derivative:
˛
d ˝
1; 2t D h0; 2i :
dt
˝
˛
14. Sketch the curve r.t/ D 1 " t 2 ; t for "1 # t # 1. Compute the tangent vector at t D 1 and add it to the sketch.
r00 .t/ D
SOLUTION
We find that
˛
d ˝
1 " t 2 ; t D h"2t; 1i
dt
r0 .t/ D
and so at t D 1, we have
r0 .1/ D h"2; 1i
To graph r.t/, we note that it satisfies x D 1 " y 2 . The sketch is shown here, along with the tangent vector at t D 1.
y
r'(1) = 〈−2, 1〉
t=1
t=0
x
t = −1
˝
˛
˝
˛
15. Sketch the curve r1 .t/ D t; t 2 together with its tangent vector at t D 1. Then do the same for r2 .t/ D t 3 ; t 6 .
SOLUTION
and so r2 0 .1/
˝
˛
Note that r1 0 .t/ D h1; 2ti and so r1 0 .1/ D h1; 2i. The graph of r1 .t/ satisfies y D x 2 . Likewise, r2 0 .t/ D 3t 2 ; 6t 5
D h3; 6i. The graph of r2 .t/ also satisfies y D x 2 . Both graphs and tangent vectors are given here.
1
2
r 1(t)
r 2(t)
˝
˛
16. Sketch the cycloid r.t/ D t " sin t; 1 " cos t together with its tangent vectors at t D
SOLUTION
!
3
and
3!
4 .
The tangent vector r0 .t/ is the following vector:
r0 .t/ D
d
ht " sin t; 1 " cos ti D h1 " cos t; sin ti
dt
Substituting the given values gives the following vectors:
* p +
!
!E
1
3
r
D 1 " cos ; sin
D
;
3
3
3
2 2
p p +
# $ +
, *
3!
3!
2
2
0 3!
r
D 1 " cos
; sin
D 1C
;
4
4
4
2
2
0
!! "
D
The cycloid r.t/ D ht " sin t; 1 " cos ti and the two tangent vectors are shown in the following figure:
y
〈 12, 23 〉
y=2
t=
t=0
3
〈1 +
2
2
2 , 2
〉
t = 34
t=2
x
S E C T I O N 11.7
In Exercises 17–18, evaluate
˝
˛
17. r.t/ D t 2 ; 1 " t ,
SOLUTION
Calculus of Vector-Valued Functions
1471
d
r.g.t// using the Chain Rule.
dt
g.t/ D e t
We first differentiate the two functions:
r0 .t/ D
E
d D2
t ; 1 " t D h2t; "1i
dt
g 0 .t/ D
d t
.e / D e t
dt
Using the Chain Rule we get:
˝
˛
18. r.t/ D t 2 ; t 3 , g.t/ D sin t
SOLUTION
E
˝
˛ D
d
r.g.t// D g 0 .t/r0 .g.t// D e t 2e t ; "1 D 2e 2t ; "e t
dt
We first differentiate the two functions:
r0 .t/ D
E
d D 2 3E D
t ; t D 2t; 3t 2
dt
g 0 .t/ D cos t
Using the Chain Rule we get:
D
E D
E
d
r.g.t// D g 0 .t/r0 .g.t// D cos t 2 sin t; 3 sin2 t D 2 sin t cos t; 3 sin2 t cos t
dt
In Exercises 19–20, find a parametrization of the tangent line at the point indicated.
˝
˛
19. r.t/ D t 2 ; t 4 , t D "2
SOLUTION
The tangent line has the following parametrization:
`.t/ D r."2/ C tr0 ."2/
We compute the vectors r."2/ and r0 ."2/:
˝
˛
r."2/ D ."2/2 ; ."2/4 D h4; 16i
r0 .t/ D
Substituting in (1) gives:
˛
d ˝ 2 4˛ ˝
t ; t D 2t; 4t 3
dt
)
r0 ."2/ D h"4; "32i
`.t/ D h4; 16i C t h"4; "32i D h4 " 4t; 16 " 32ti
The parametrization for the tangent line is, thus,
x D 4 " 4t;
y D 16 " 32t;
"1 < t < 1:
To find a direct relation between y and x, we express t in terms of x and substitute in y D 16 " 32t. This gives:
x D 4 " 4t ) t D
x"4
:
"4
Hence,
y D 16 " 32t D 16 " 32 !
The equation of the tangent line is y D 8x " 16.
˝
˛
20. r.t/ D cos 2t; sin 3t , t D !4
x "4
D 16 C 8.x " 4/ D 8x " 16:
"4
(1)
1472
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
SOLUTION
The tangent line is parametrized by:
`.t/ D r
We compute the vectors in the above parametrization:
r
!! "
4
C tr0
!! "
(1)
4
+
,
˝
!
3! ˛
1
D cos ; sin
D 0; p
4
2
4
2
˝
˛
˝
˛
d
r0 .t/ D
cos 2t; sin 3t D "2 sin 2t; 3 cos 3t
dt
, +
,
! " +
!
3!
"3
0 !
)
r
D "2 sin ; 3 cos
D "2; p
4
2
4
2
!! "
Substituting the vectors in (1) we obtain the following parametrization:
+
,
+
, +
,
1
"3
1
`.t/ D 0; p C t "2; p D "2t; p .1 " 3t/
2
2
2
In Exercises 21–28, evaluate the integrals.
Z 3D
E
21.
8t 2 " t; 6t 3 C t dt
!1
SOLUTION
Vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each com-
ponent.
ˇ
#
$ #
$
8 3 t 2 ˇˇ3
9
8 1
212
8t " t dt D t " ˇ D 72 "
" " "
D
3
2
2
3
2
3
!1
!1
ˇ
#
$
#
$
Z 3
3
t 2 ˇ3
243
9
3
1
6t 3 C t dt D t 4 C ˇˇ D
C
"
C
D 124
2
2 !1
2
2
2
2
!1
Z
3
2
Therefore,
22.
Z
0
1+
Z
,
3
!1
D
2
E
3
8t " t; 6t C t dt D
*Z
3
!1
2
8t " t dt;
Z
3
!1
3
+
6t C t dt D
+
,
212
; 124
3
1
s
;
ds
1 C s2 1 C s2
The vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each
component. For the second integral we use the substitution t D 1 C s 2 , dt D 2s ds. We get:
ˇ1
Z 1
ˇ
ds
!1
ˇ D tan!1 .1/ " tan1 .0/ D ! " 0 D !
D
tan
.s/
ˇ
2
4
4
1
C
s
0
0
ˇ2
Z 1
Z 2 # $
Z 2
ˇ
s
1 dt
1
dt
1
1
1
ds D
D
D ln t ˇˇ D .ln 2 " ln 1/ D ln 2
2
t
2
2
t
2
2
2
1
C
s
0
1
1
1
SOLUTION
Therefore,
Z
1+
0
23.
Z
2
!2
*Z
+ +
,
,
Z 1
1
1
s
ds
s ds
! 1
;
ds
D
;
D
;
ln
2
2
2
4 2
1 C s2 1 C s2
0 1Cs
0 1Cs
% 3
&
u i C u5 j du
SOLUTION
The vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each
component.
ˇ
u4 ˇˇ2
16 16
ˇ D 4 " 4 D0
4
!2
!2
ˇ2
Z 2
6
u ˇˇ
64 64
u5 du D
ˇ D 6 " 6 D0
6
!2
!2
Z
2
u3 du D
Calculus of Vector-Valued Functions
S E C T I O N 11.7
1473
Therefore,
24.
Z
0
Z
1!
2
!2
% 3
&
u i C u5 j du D
"
2
te !t i C t ln.t 2 C 1/j dt
SOLUTION
s D "t 2 , ds
Z
2
!2
3
!
u du i C
Z
2
!2
5
!
u du j D 0i C 0j
We compute the integral of each component. The integral of the first component is computed using the substitution
D "2t dt. This gives
ˇ
$
Z 1
Z !1 #
Z
&
&
ds
1 0 s
1 ˇ0
1%
1%
2
te !t dt D
es "
D
e ds D e s ˇˇ D e 0 " e !1 D 1 " e !1
2
2 !1
2 !1
2
2
0
0
For the integral of the second component we use the substitution s D t 2 C 1 , ds D 2t dt. This gives:
ˇ2
Z 1
Z 2
Z
ˇ
ds
1 2
1
1
t ln.t 2 C 1/ dt D
ln s
D
ln s ds D s.ln s " 1/ˇˇ D .2.ln 2 " 1/ " 1.ln 1 " 1//
2
2 1
2
2
0
1
1
D ln 2 " 1 C
1
1
D ln 2 "
2
2
Hence,
25.
Z
Z
1
1D
0
+ !
,
E
" 1
1
2
te !t ; t ln.t 2 C 1/ dt D
1 " e !1 ; " C ln 2 :
2
2
h2t; 4ti dt
0
SOLUTION
The vector valued integration is defined via componentwise integration. Therefore,
*Z
+ * ˇ
ˇ +
Z 1
Z 1
1
ˇ1 2 ˇ1
2ˇ
2t dt;
4t dt D t ˇ ; 2t ˇˇ D h1; 2i
h2t; 4ti dt D
0
26.
Z
+
1
1=2
0
0
0
0
,
1 1
;
du
u2 u4
SOLUTION
The vector valued integration is defined via componentwise integration. Computing the integral of each component
we get:
Z
1
1=2
Z
1
1=2
ˇ
1
"1 ˇˇ1
du D
D "1 " ."2/ D 1
u ˇ1=2
u2
ˇ
1
"1 ˇˇ1
"1 "8
7
du D 3 ˇ
D
"
D
3
3
3
u4
3u 1=2
Therefore,
Z
1
1=2
27.
Z
1
4%
+
*Z
+ +
,
,
Z 1
1 1
1 1
1
7
;
du
D
du;
du
D
1;
2
4
3
u2 u4
1=2 u
1=2 u
p &
t !1 i C 4 t j dt
SOLUTION
We perform the integration componentwise. Computing the integral of each component we get:
Z
4
t
1
Z
4
1
!1
ˇ4
ˇ
dt D ln t ˇˇ D ln 4 " ln 1 D ln 4
1
ˇ4
"
p
ˇ
2
8 ! 3=2
56
4 t dt D 4 ! t 3=2 ˇˇ D
4
"1 D
3
3
3
1
Hence,
Z
4!
1
p "
56
t !1 i C 4 tj dt D .ln 4/ i C j
3
1474
28.
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
Z
t
0
%
&
3si C 6s 2 j ds
SOLUTION
We first compute the integral of each component:
ˇ
3 2 ˇˇt
3
3s ds D s ˇ D t 2
2
2
0
0
ˇt
Z t
6 ˇ
6s 2 ds D s 3 ˇˇ D 2t 3
3 0
0
Z
Hence,
Z
t
0
!
#Z
"
3si C 6s 2 j dt D
t
t
0
$
#Z
3s ds i C
t
0
$
#
$
% &
3 2
6s 2 ds j D
t i C 2t 3 j
2
In Exercises 29–32, find both the general solution of the differential equation and the solution with the given initial condition.
29.
dr
D h1 " 2t; 4ti,
dt
SOLUTION
r.0/ D h3; 1i
We first find the general solution by integrating ddtr :
+Z
,
Z
Z
˝
˛
r.t/ D h1 " 2t; 4ti dt D
.1 " 2t/ dt; 4t dt D t " t 2 ; 2t 2 C c
Since r.0/ D h3; 1i, we have:
Substituting in (1) gives the solution:
(1)
˝
˛
r.0/ D 0 " 02 ; 2 ! 02 C c D h3; 1i ) c D h3; 1i
˝
˛
˝
˛
r.t/ D t " t 2 ; 2t 2 C h3; 1i D "t 2 C t C 3; 2t 2 C 1
% &
30. r0 .t/ D hsin 3t; sin 3ti, r !2 D h2; 4i
SOLUTION
We first integrate the vector r0 .t/ to find the general solution:
+Z
,
Z
Z
r.t/ D hsin 3t; sin 3ti dt D
sin 3t dt; sin 3t dt
+
,
1
1
D " cos 3t; " cos 3t C c
3
3
(1)
(2)
Substituting the initial condition we obtain:
+
,
1
!
1
!
r.!=2/ D " cos ; " cos
C c D h0; 0i C c D h2; 4i
3
2
3
2
Hence,
c D h2; 4i " h0; 0i D h2; 4i
Substituting in (2) we obtain the solution:
+
,
+
,
1
1
1
1
r.t/ D " cos 3t; " cos 3t C h2; 4i D " cos 3t C 2; " cos 3t C 4
3
3
3
3
31. r00 .t/ D h0; 2i, r.3/ D h1; 1i, r0 .3/ D h0; 0i
SOLUTION
To find the general solution we first find r0 .t/ by integrating r00 .t/:
Z
Z
r0 .t/ D r00 .t/ dt D h0; 2i dt D h0; 2ti C c1
We now integrate r0 .t/ to find the general solution r.t/:
Z
Z
D
E
r.t/ D r0 .t/ dt D .h0; 2ti C c1 / dt D 0; t 2 C c1 t C c2
We substitute the initial conditions in (1) and (2). This gives:
r0 .3/ D h0; 6i C c1 D h0; 0i
)
c1 D h0; "6i
(1)
(2)
Calculus of Vector-Valued Functions
S E C T I O N 11.7
1475
r.3/ D h0; 9i C c1 .3/ C c2 D h1; 1i
h0; 9i C h0; "18i C c2 D h1; 1i
c2 D h1; 10i
)
Combining with (2) we obtain the following solution:
˝
˛
32. r00 .t/ D e t ; sin t ,
SOLUTION
r.0/ D h1; 0i ;
D
E
r.t/ D 0; t 2 C t h0; "6i C h1; 10i
D
E
D 1; t 2 " 6t C 10
r0 .0/ D h0; 2i
We perform integration componentwise on r00 .t/ to obtain:
Z
˝ t
˛
˝
˛
r0 .t/ D
e ; sin t dt D e t ; " cos t C c1
We now integrate r0 .t/ to obtain the general solution:
Z
%˝ t
˛
&
˝
˛
r.t/ D
e ; " cos t C c1 dt D e t ; " sin t C c1 t C c2
(1)
(2)
Now, we substitute the initial conditions r.0/ D h1; 0i and r0 .0/ D h0; 2i into (1) and (2) and solve for the vectors c1 and c2 . We
obtain:
r0 .0/ D h1; "1i C c1 D h0; 2i
r.0/ D h1; 0i C c2 D h1; 0i
c1 D h"1; 3i
)
)
c2 D h0; 0i
Finally we combine the above to obtain the solution:
˝
˛
˝
˛
r.t/ D e t ; " sin t C h"1; 3i t C h0; 0i D e t " t; " sin t C 3t
33. Find the location at t D 3 of a particle whose path (Figure 1) satisfies
+
,
dr
1
D 2t "
;
2t
"
4
;
dt
.t C 1/2
r.0/ D h3; 8i
y
10
(3, 8)
t=0
5
t=3
x
5
10
15
20
25
FIGURE 1 Particle path.
SOLUTION
To determine the position of the particle in general, we perform integration componentwise on r0 .t/ to obtain:
Z
r.t/ D r0 .t/ dt
Z +
D
2t "
,
1
; 2t " 4 dt
.t C 1/2
+
,
1
D t2 C
; t 2 " 4t C c1
t C1
Using the initial condition, observe the following:
r.0/ D h1; 0i C c1 D h3; 8i
)
c1 D h2; 8i
Therefore,
+
r.t/ D t 2 C
,
+
,
1
1
; t 2 " 4t C h2; 8i D t 2 C
C 2; t 2 " 4t C 8
t C1
t C1
and thus, the location of the particle at t D 3 is r.3/ D h45=4; 5i D h11:25; 5i
1476
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
34. Find the location and velocity at t D 4 of a particle whose path satisfies
D
E
dr
D 2t !1=2 ; 6 ;
dt
SOLUTION
r.1/ D h4; 9i
The velocity of this particle at t D 4 is exactly:
D
E
r0 .4/ D 2.4/!1=2 ; 6 D h1; 6i
To determine the location of the particle at any general t, we will perform integration componentwise on r0 .t/:
Z
Z D
E
r.t/ D r0 .t/ dt D
2t !1=2 ; 6 dt
Using the initial condition, observe the following:
D p
E
D 4 t; 6t C c1
r.1/ D h4; 6i C c1 D h4; 9i
)
c1 D h0; 3i
Therefore,
D p
E
D p
E
r.t/ D 4 t; 6t C h0; 3i D 4 t; 6t C 3
Then the location of this particle at t D 4 is:
35. Find all solutions to
SOLUTION
gives:
r0 .t/
r.4/ D h8; 27i
D v with initial condition r.1/ D w, where v and w are constant vectors in R2 .
We denote the components of the constant vector v by v D hv1 ; v2 i and integrate to find the general solution. This
r.t/ D
Z
v dt D
Z
hv1 ; v2 i dt D
+Z
v1 dt;
Z
v2 dt
,
D hv1 t C c1 ; v2 t C c2 i D t hv1 ; v2 i C hc1 ; c2 i
We let c D hc1 ; c2 i and obtain:
r.t/ D tv C c D c C tv
Notice that the solutions are the vector parametrizations of all the lines with direction vector v.
We are also given the initial condition that r.1/ D w, using this information we can determine:
r.1/ D .1/v C c D w
Therefore c D w " v and we get:
r.t/ D .w " v/ C tv D .t " 1/v C w
36. Let u be a constant vector in R2 . Find the solution of the equation r0 .t/ D .sin t/u satisfying r0 .0/ D 0.
SOLUTION
We first integrate to find the general solution. Denoting u D hu1 ; u2 i we get:
Z
Z
r.t/ D .sin t/ u dt D hu1 sin t; u2 sin ti dt
D
+Z
u1 sin t dt;
, + Z
,
Z
u2 sin t dt D u1 sin t dt; u2 sin t dt
Z
D h"u1 cos t C c1 ; "u2 cos t C c2 i D " cos t hu1 ; u2 i C hc1 ; c2 i
Letting c D hc1 ; c2 i we obtain the following solutions:
r.t/ D ." cos t/ u C c
Since r0 .0/ D 0 we have r0 .0/ D sin 0 ! u D 0.
S E C T I O N 11.7
Calculus of Vector-Valued Functions
1477
37. Find all solutions to r0 .t/ D 2r.t/ where r.t/ is a vector-valued function.
˝
˛
We denote the components of r.t/ by r.t/ D hx.t/; y.t/i. Then, r0 .t/ D x 0 .t/; y 0 .t/ . Substituting in the differential
equation we get:
˝ 0
˛
x .t/; y 0 .t/ D 2 hx.t/; y.t/i
SOLUTION
Equating corresponding components gives:
x 0 .t/ D 2x.t/
y 0 .t/
D 2y.t/
)
x.t/ D c1 e 2t
y.t/ D c2 e 2t
We denote the constant vector by c D hc1 ; c2 i and obtain the following solutions:
˝
˛
r.t/ D c1 e 2t ; c2 e 2t D e 2t hc1 ; c2 i D e 2t c
38. Show that w.t/ D hsin.3t C 4/; sin.3t " 2/i satisfies the differential equation w00 .t/ D "9w.t/.
SOLUTION
We differentiate the vector w.t/ twice:
w0 .t/ D h3 cos .3t C 4/ ; 3 cos .3t " 2/i
w00 .t/ D
d % 0 &
w .t/ D h"9 sin .3t C 4/ ; "9 sin .3t " 2/i
dt
D "9 hsin .3t C 4/ ; sin .3t " 2/i D "9w.t/
We thus showed that w00 .t/ D "9w.t/
Further Insights and Challenges
39. Let r.t/ D hx.t/; y.t/i trace a plane curve C. Assume that x 0 .t0 / ¤ 0. Show that the slope of the tangent vector r0 .t0 / is equal
to the slope dy=dx of the curve at r.t0 /.
SOLUTION
By the Chain Rule we have
dy
dy dx
D
!
dt
dx dt
Hence, at the points where
dx
dt
¤ 0 we have:
dy
D
dx
dy
dt
dx
dt
D
y 0 .t/
x 0 .t/
The line `.t/ D ha; bi C tr0 .t0 / passes through .a; b/ at t D 0. It holds that:
`.0/ D ha; bi C 0r0 .t0 / D ha; bi
That is, .a; b/ is the terminal point of the vector `.0/, hence the line passes through
ˇ .a; b/. The line has the direction vector
˝
˛
0 .t /
dy ˇ
0
r0 .t0 / D x 0 .t0 /; y 0 .t0 / , therefore the slope of the line is yx 0 .t
which
is
equal
to
.
/
dx ˇ
0
40. Verify the Sum and Product Rules for derivatives of vector-valued functions.
SOLUTION
t Dt0
We first verify the Sum Rule stating:
.r1 .t/ C r2 .t//0 D r01 .t/ C r02 .t/
Let r1 .t/ D hx1 .t/; y1 .t/i and r2 .t/ D hx2 .t/; y2 .t/i. Then,
d
hx1 .t/ C x2 .t/; y1 .t/ C y2 .t/i
dt
˝
˛
D .x1 .t/ C x2 .t//0 ; .y1 .t/ C y2 .t//0
˝
˛
D x10 .t/ C x20 .t/; y10 .t/ C y20 .t/
˝
˛ ˝
˛
D x10 .t/; y10 .t/ C x20 .t/; y20 .t/ D r01 .t/ C r02 .t/
.r1 .t/ C r2 .t// 0 D
The Product Rule states that for any differentiable scalar-valued function f .t/ and differentiable vector-valued function r.t/, it
holds that:
d
f .t/r.t/ D f .t/r0 .t/ C f 0 .t/r.t/
dt
1478
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
To verify this rule, we denote r.t/ D hx.t/; y.t/; z.t/i. Then,
d
d
f .t/r.t/ D
hf .t/x.t/; f .t/y.t/i
df
dt
Applying the Product Rule for scalar functions for each component we get:
˝
˛
d
f .t/r.t/ D f .t/x 0 .t/ C f 0 .t/x.t/; f .t/y 0 .t/ C f 0 .t/y.t/
dt
˝
˛ ˝
˛
D f .t/x 0 .t/; f .t/y 0 .t/ C f 0 .t/x.t/; f 0 .t/y.t/
˝
˛
D f .t/ x 0 .t/; y 0 .t/ C f 0 .t/ hx.t/; y.t/i D f .t/r0 .t/ C f 0 .t/r.t/
41. Verify the Chain Rule for vector-valued functions.
SOLUTION
Let g.t/ and r.t/ D hx.t/; y.t/i be differentiable scalar and vector valued functions respectively. We must show that:
d
r .g.t// D g 0 .t/r0 .g.t// :
dt
We have
r .g.t// D hx .g.t// ; y .g.t//i
We differentiate the vector componentwise, using the Chain Rule for scalar functions. This gives:
+
,
˝
˛
d
d
d
r .g.t// D
.x .g.t/// ;
.y .g.t/// D g 0 .t/x 0 .g.t// ; g 0 .t/y 0 .g.t//
dt
dt
dt
˝
˛
D g 0 .t/ x 0 .g.t// ; y 0 .g.t// D g 0 .t/r0 .g.t//
42. Verify the linearity properties
Z
Z
cr.t/ dt D c
%
&
r1 .t/ C r2 .t/ dt D
Z
Z
r.t/ dt
r1 .t/ dt C
(c any constant)
Z
r2 .t/ dt
SOLUTION We denote the components of the vectors by r.t/ D hx.t/; y.t/i; r1 .t/ D hx1 .t/; y1 .t/i; r2 .t/ D hx2 .t/; y2 .t/i.
Using vector operations, componentwise integration and the linear properties for scalar functions, we obtain:
+Z
,
Z
Z
Z
cr.t/ dt D hcx.t/; cy.t/i dt D
cx.t/ dt; cy.t/ dt
+ Z
,
+Z
,
Z
Z
D c x.t/ dt; c y.t/ dt D c
x.t/ dt; y.t/ dt
Dc
Z
Z
hx.t/; y.t/i dt D c
r.t/ dt
Next we prove the second linear property:
Z
Z
.r1 .t/ C r2 .t// dt D hx1 .t/ C x2 .t/; y1 .t/ C y2 .t/i dt
D
D
D
D
+Z
+Z
+Z
Z
.x1 .t/ C x2 .t// dt;
x1 .t/ dt C
x1 .t/ dt;
Z
Z
Z
.y1 .t/ C y2 .t// dt
x2 .t/ dt;
Z
y1 .t/ dt C
Z
Z
hx2 .t/; y2 .t/i dt D
43. Prove the Substitution Rule (where g.t/ is a differentiable scalar function):
b
a
y2 .t/ dt
,
, +Z
,
Z
y1 .t/ dt C
x2 .t/ dt; y2 .t/ dt
hx1 .t/; y1 .t/i dt C
Z
,
r.g.t//g 0 .t/ dt D
Z
g !1 .b/
g !1 .a/
r.u/ du
Z
r1 .t/ dt C
Z
r2 .t/ dt
Chapter Review Exercises
1479
SOLUTION (Note that an early edition of the textbook had the integral limits as g.a/ and g.b/; they should actually be g !1 .a/
and g !1 .b/.) We denote the components of the vector-valued function by r.t/ dt D hx.t/; y.t/i. Using componentwise integration
we have:
Z
Write
Z
b
x.t/ dt as
a
Z
b
a
b
a
r.t/ dt D
*Z
b
x.t/ dt;
a
Z
b
y.t/ dt
a
+
x.s/ ds. Let s D g.t/, so ds D g 0 .t/ dt. The substitution gives us
procedure for the other integral gives us:
*Z
Z
b
a
r.t/ dt D
D
D
g !1 .b/
0
x .g.t// g .t/ dt;
g !1 .a/
Z
Z
Z
g !1 .b/
y .g.t// g .t/ dt
g !1 .a/
g !1 .b/ ˝
g !1 .a/
g !1 .b/
g !1 .a/
0
˛
x .g.t// g 0 .t/; y .g.t// g 0 .t/ dt
hx .g.t// ; y .g.t//i g 0 .t/ dt D
44. Prove that if kr.t/k # K for t 2 Œa; b", then
Z
g !1 .b/
g !1 .a/
Z
g !1 .b/
g !1 .a/
x.g.t//g 0 .t/ dt. A similar
+
r .g.t// g 0 .t/ dt
-Z
- b
r.t/ dt - # K.b " a/
- a
-
Z b
Think of r.t/ as a velocity vector. Then,
r.t/ dt gives the displacement vector from the location at time t D a to
a
-Z
- b
the time t D b, and so r.t/ dt - gives the length of this displacement vector. But, since speed is kr.t/k which is less than or
- a
equal to K, then in the interval a # t # b, the object
-Z can move-a total distance not more than K.b " a/. Thus, the length of the
- b
displacement vector is # K.b " a/, which gives us r.t/ dt - # K.b " a/, as desired.
- a
SOLUTION
CHAPTER REVIEW EXERCISES
1. Which of the following curves pass through the point .1; 4/?
(a) c.t/ D .t 2 ; t C 3/
(b) c.t/ D .t 2 ; t " 3/
(c) c.t/ D .t 2 ; 3 " t/
(d) c.t/ D .t " 3; t 2 /
To check whether it passes through the point .1; 4/, we solve the equations c.t/ D .1; 4/ for the given curves.
(a) Comparing the second coordinate of the curve and the point yields:
SOLUTION
t C3D 4
t D1
We substitute t D 1 in the first coordinate, to obtain
t 2 D 12 D 1
Hence the curve passes through .1; 4/.
(b) Comparing the second coordinate of the curve and the point yields:
t "3D4
t D7
We substitute t D 7 in the first coordinate to obtain
t 2 D 72 D 49 ¤ 1
Hence the curve does not pass through .1; 4/.
1480
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
(c) Comparing the second coordinate of the curve and the point yields
3"t D4
t D "1
We substitute t D "1 in the first coordinate, to obtain
t 2 D ."1/2 D 1
Hence the curve passes through .1; 4/.
(d) Comparing the first coordinate of the curve and the point yields
t "3D1
t D4
We substitute t D 4 in the second coordinate, to obtain:
t 2 D 42 D 16 ¤ 4
Hence the curve does not pass through .1; 4/.
2. Find parametric equations for the line through P D .2; 5/ perpendicular to the line y D 4x " 3.
SOLUTION
equation
The line perpendicular to y D 4x " 3 at P D .2; 5/ is the line of slope " 14 passing through P . This line has the
1
y " 5 D " .x " 2/
4
A bit of calculation shows that the parametric equations of the line are
#
$
1
c.t/ D 2 C t; 5 " t
4
or
x D 2Ct
1
y D5" t
4
3. Find parametric equations for the circle of radius 2 with center .1; 1/. Use the equations to find the points of intersection of the
circle with the x- and y-axes.
SOLUTION
Using the standard technique for parametric equations of curves, we obtain
c.t/ D .1 C 2 cos t; 1 C 2 sin t/
We compare the x coordinate of c.t/ to 0:
1 C 2 cos t D 0
1
2
2!
t D˙
3
cos t D "
Substituting in the y coordinate yields
p
$
p
2!
3
1 C 2 sin ˙
D1˙2
D1˙ 3
3
2
p
Hence, the intersection points with the y-axis are .0; 1 ˙ 3/. We compare the y coordinate of c.t/ to 0:
#
1 C 2 sin t D 0
1
2
!
t D"
6
sin t D "
or
7
!
6
Substituting in the x coordinates yields
p
! !"
p
3
1 C 2 cos "
D1C2
D1C 3
6
2
Chapter Review Exercises
#
7
1 C 2 cos
!
6
$
D 1 " 2 cos
Hence, the intersection points with the x-axis are .1 ˙
p
!! "
6
D1"2
1481
p
p
3
D1" 3
2
3; 0/.
4. Find a parametrization c.t/ of the line y D 5 " 2x such that c.0/ D .2; 1/.
SOLUTION The line is passing through P D .0; 5/ with slope "2, hence (by one of the examples in section 11.1) it has the
parametrization
c.t/ D .t; 5 " 2t/
This parametrization does not satisfy c.0/ D .2; 1/. We replace the parameter t by a parameter s, so that t D s C ˇ, to obtain
another parametrization for the line:
c $ .s/ D .s C ˇ; 5 " 2.s C ˇ// D .s C ˇ; 5 " 2ˇ " 2s/
(1)
We require that c $ .0/ D .2; 1/. That is,
c $ .0/ D .ˇ; 5 " 2ˇ/ D .2; 1/
or
ˇD2
5 " 2ˇ D 1
)
ˇD2
Substituting in (1) gives the parametrization
c $ .s/ D .s C 2; 1 " 2s/
5. Find a parametrization c.#/ of the unit circle such that c.0/ D ."1; 0/.
SOLUTION
The unit circle has the parametrization
c.t/ D .cos t; sin t/
This parametrization does not satisfy c.0/ D ."1; 0/. We replace the parameter t by a parameter # so that t D # C ˛, to obtain
another parametrization for the circle:
c $ .#/ D .cos.# C ˛/; sin.# C ˛//
(1)
We need that c $ .0/ D .1; 0/, that is,
c $ .0/ D .cos ˛; sin ˛/ D ."1; 0/
Hence
cos ˛ D "1
sin ˛ D 0
)
˛D!
Substituting in (1) we obtain the following parametrization:
c $ .#/ D .cos.# C !/; sin.# C !//
6. Find a path c.t/ that traces the parabolic arc y D x 2 from .0; 0/ to .3; 9/ for 0 # t # 1.
SOLUTION The second coordinates of the points on the parabolic arc are the square of the first coordinates. Therefore the points
on the arc have the form:
c.t/ D .˛t; ˛ 2 t 2 /
We need that c.1/ D .3; 9/. That is,
c.1/ D .˛; ˛ 2 / D .3; 9/ ) ˛ D 3
Substituting in (1) gives the following parametrization:
c.t/ D .3t; 9t 2 /
7. Find a path c.t/ that traces the line y D 2x C 1 from .1; 3/ to .3; 7/ for 0 # t # 1.
(1)
1482
C H A P T E R 11
SOLUTION
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
Solution 1: By one of the examples in section 11.1, the line through P D .1; 3/ with slope 2 has the parametrization
c.t/ D .1 C t; 3 C 2t/
But this parametrization does not satisfy c.1/ D .3; 7/. We replace the parameter t by a parameter s so that t D ˛s C ˇ. We get
c $ .s/ D .1 C ˛s C ˇ; 3 C 2.˛s C ˇ// D .˛s C ˇ C 1; 2˛s C 2ˇ C 3/
We need that c $ .0/ D .1; 3/ and c $ .1/ D .3; 7/. Hence,
c $ .0/ D .1 C ˇ; 3 C 2ˇ/ D .1; 3/
c $ .1/ D .˛ C ˇ C 1; 2˛ C 2ˇ C 3/ D .3; 7/
We obtain the equations
1Cˇ D 1
3 C 2ˇ D 3
)
˛CˇC1D 3
ˇ D 0; ˛ D 2
2˛ C 2ˇ C 3 D 7
Substituting in (1) gives
c $ .s/ D .2s C 1; 4s C 3/
Solution 2: The segment from .1; 3/ to .3; 7/ has the following vector parametrization:
.1 " t/ h1; 3i C t h3; 7i D h1 " t C 3t; 3.1 " t/ C 7ti D h1 C 2t; 3 C 4ti
The parametrization is thus
c.t/ D .1 C 2t; 3 C 4t/
8. Sketch the graph c.t/ D .1 C cos t; sin 2t/ for 0 # t # 2! and draw arrows specifying the direction of motion.
SOLUTION
From x D 1 C cos t we have x " 1 D cos t. We substitute this in the y coordinate to obtain
q
p
p
y D sin 2t D 2 sin t cos t D ˙2 sin2 t cos t D ˙2 1 " cos2 t cos t D ˙2 1 " .x " 1/2 .x " 1/
p
We can see that the graph is symmetric with respect to the x-axis, hence we plot the function y D 2 1 " .x " 1/2 .x " 1/ and
reflect it with respect to the x-axis. When t D 0 we have c.0/ D .2; 0/. when t increases near 0, cos t is decreasing and sin 2t is
increasing, hence the general direction at the point .2; 0/ is upwards and left. As t approaches !=2, the x-coordinate decreases to 1
and the y-coordinate to 0. Likewise, as t moves from !=2 to !, the x-coordinate moves to 0 while the y-coordinate falls to "1 and
then rises to 0. The resulting graph is seen here in the corresponding figure.
y
1
x
1
2
−1
Plot of Exercise 8
In Exercises 9–12, express the parametric curve in the form y D f .x/.
9. c.t/ D .4t " 3; 10 " t/
SOLUTION
We use the given equation to express t in terms of x.
x D 4t " 3
4t D x C 3
tD
xC3
4
Chapter Review Exercises
1483
Substituting in the equation of y yields
y D 10 " t D 10 "
xC3
x
37
D" C
4
4
4
That is,
yD"
x
37
C
4
4
10. c.t/ D .t 3 C 1; t 2 " 4/
The parametric equations are x D t 3 C 1 and y D t 2 " 4. We express t in terms of x:
SOLUTION
x D t3 C 1
t3 D x " 1
t D .x " 1/1=3
Substituting in the equation of y yields
y D t 2 " 4 D .x " 1/2=3 " 4
That is,
#
2
1
11. c.t/ D 3 " ; t 3 C
t
t
SOLUTION
y D .x " 1/2=3 " 4
$
We use the given equation to express t in terms of x:
x D3"
2
t
2
D3"x
t
2
tD
3"x
Substituting in the equation of y yields
yD
12. x D tan t,
SOLUTION
y D sec t
#
2
3"x
$3
C
1
8
3"x
D
C
2=.3 " x/
2
.3 " x/3
We use the trigonometric identity
1 C tan2 t D sec2 t
Substituting the parametric equations x D tan t and y D sec t we obtain
1 C x2 D y2
or
p
y D ˙ x2 C 1
In Exercises 13–16, calculate dy=dx at the point indicated.
13. c.t/ D .t 3 C t; t 2 " 1/,
SOLUTION
dy
:
dx
t D3
The parametric equations are x D t 3 C t and y D t 2 " 1. We use the theorem on the slope of the tangent line to find
dy
D
dx
We now substitute t D 3 to obtain
dy
dt
dx
dt
D
2t
3t 2 C 1
ˇ
dy ˇˇ
2!3
3
D
D
ˇ
2
dx t D3
14
3!3 C1
1484
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
14. c.#/ D .tan2 #; cos #/,
SOLUTION
#D
!
4
The parametric equations are x D tan2 #, y D cos #. We use the theorem on the slope of the tangent line to find
dy
D
dx
We now substitute # D
!
4
15. c.t/ D .e t " 1; sin t/,
SOLUTION
dy
d"
dx
d"
" sin #
cos3 #
D
"
2
2 tan # sec2 #
to obtain
ˇ
cos3 !4
dy ˇˇ
1
D"
D" p
ˇ
dx " D!=4
2
4 2
t D 20
We use the theorem for the slope of the tangent line to find
dy
D
dx
We now substitute t D 20:
16. c.t/ D .ln t; 3t 2 " t/,
SOLUTION
D
dy
:
dx
dy
dt
dx
dt
dy
:
dx
.sin t/0
cos t
D t
.e t " 1/0
e
D
ˇ
dy ˇˇ
cos 20
D 20
ˇ
dx t D0
e
P D .0; 2/
The parametric equations are x D ln t, y D 3t 2 " t. We use the theorem for the slope of the tangent line to find
dy
D
dx
dy
dt
dx
dt
6t " 1
D
1
t
D 6t 2 " t
dy
:
dx
(1)
We now must identify the value of t corresponding to the point P D .0; 2/ on the curve. We solve the following equations:
2
ln t D 0
)
3t " t D 2
Substituting t D 1 in (1) we obtain
17.
SOLUTION
t D1
ˇ
dy ˇˇ
D 6 ! 12 " 1 D 5
dx ˇP
Find the point on the cycloid c.t/ D .t " sin t; 1 " cos t/ where the tangent line has slope 21 .
Since x D t " sin t and y D 1 " cos t, the theorem on the slope of the tangent line gives
dy
D
dx
The points where the tangent line has slope
1
2
dy
dt
dx
dt
are those where
D
dy
dx
sin t
1 " cos t
D 12 . We solve for t:
dy
1
D
dx
2
sin t
1
D
1 " cos t
2
2 sin t D 1 " cos t
p
p
We let u D sin t. Then cos t D ˙ 1 " sin2 t D ˙ 1 " u2 . Hence
p
2u D 1 ˙ 1 " u2
We transfer sides and square to obtain
p
˙ 1 " u2 D 2u " 1
1 " u2 D 4u2 " 4u C 1
5u2 " 4u D u.5u " 4/ D 0
(1)
Chapter Review Exercises
u D 0; u D
1485
4
5
We find t by the relation u D sin t:
u D 0W
sin t D 0 ) t D 0; t D !
uD
sin t D
4
W
5
4
) t % 0:93; t % 2:21
5
These correspond to the points .0; 1/, .!; 2/, .0:13; 0:40/, and .1:41; 1:60/, respectively, for 0 < t < 2!.
18. Find the points on .t C sin t; t " 2 sin t/ where the tangent is vertical or horizontal.
SOLUTION
We use the theorem for the slope of the tangent line to find
dy
D
dx
dy
dt
dx
dt
D
dy
:
dx
1 " 2 cos t
1 C cos t
We find the values of t for which the denominator is zero. We ignore the numerator, since when 1 C cos t D 0, 1 " 2 cos t D 3 ¤ 0.
1 C cos t D 0
cos t D "1
t D ! C 2!k
where k 2 Z
We now find the values of t for which the numerator is 0:
1 " 2 cos t D 0
1 D 2 cos t
1
D cos t
2
!
t D ˙ C 2!k
3
where k 2 Z
Note that the denominator is not zero at these points. Thus, we have vertical tangents at t D ! C 2!k and horizontal tangents at
t D ˙!=3 C 2!k.
19. Find the equation of the Bézier curve with control points
P0 D ."1; "1/;
P1 D ."1; 1/;
P2 D .1; 1/;
P3 .1; "1/
SOLUTION We substitute the given points in the appropriate formulas in the text to find the parametric equations of the Bézier
curve. We obtain
x.t/ D ".1 " t/3 " 3t.1 " t/2 C t 2 .1 " t/ C t 3
D ".1 " 3t C 3t 2 " t 3 / " .3t " 6t 2 C 3t 3 / C .t 2 " t 3 / C t 3
D ."2t 3 C 4t 2 " 1/
y.t/ D ".1 " t/3 C 3t.1 " t/2 C t 2 .1 " t/ " t 3
D ".1 " 3t C 3t 2 " t 3 / C .3t " 6t 2 C 3t 3 / C .t 2 " t 3 / " t 3
20. Find the speed at t D
SOLUTION
of a particle whose position at time t seconds is c.t/ D .sin 4t; cos 3t/.
We use the parametric definition to find the speed. We obtain
ds
D
dt
At time t D
!
4
D .2t 3 " 8t 2 C 6t " 1/
!
4
q
q
p
2
2
..sin 4t/0 / C ..cos 3t/0 / D .4 cos 4t/2 C ."3 sin 3t/2 D 16 cos2 4t C 9 sin2 3t
the speed is
r
r
ˇ
p
ds ˇˇ
1
3!
2 ! C 9 sin2
D
16
cos
D
16 C 9 ! D 20:5 % 4:53
ˇ
dt t D!=4
4
2
21. Find the speed (as a function of t) of a particle whose position at time t seconds is c.t/ D .sin t C t; cos t C t/. What is the
particle’s maximal speed?
1486
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
SOLUTION
We use the parametric definition to find the speed. We obtain
ds
D
dt
D
q
q
..sin t C t/0 /2 C ..cos t C t/0 /2 D .cos t C 1/2 C .1 " sin t/2
p
cos2 t C 2 cos t C 1 C 1 " 2 sin t C sin2 t D
We now differentiate the speed function to find its maximum:
p
3 C 2.cos t " sin t/
!p
"0
d 2s
" sin t " cos t
D
3 C 2.cos t " sin t/ D p
2
dt
3 C 2.cos t " sin t/
We equate the derivative to zero, to obtain the maximum point:
d 2s
D0
dt 2
" sin t " cos t
p
D0
3 C 2.cos t " sin t/
" sin t " cos t D 0
" sin t D cos t
sin."t/ D cos."t/
!
"t D C !k
4
!
t D " C !k
4
Substituting t in the function of speed we obtain the value of the maximal speed:
v
u
p
p !! q
r
!
p
!
!" u
2
2
t
3 C 2 cos " " sin "
D 3C2
" "
D 3C2 2
4
4
2
2
22. Find the length of .3e t " 3; 4e t C 7/ for 0 # t # 1.
SOLUTION
We use the formula for arc length, to obtain
sD
D
Z
Z
1q
..3e t " 3/0 /2 C ..4e t C 7/0 /2 dt D
0
1p
9e 2t
0
C 16e 2t
dt D
Z
1p
25e 2t
0
Z
1q
.3e t /2 C .4e t /2 dt
0
dt D
In Exercises 23 and 24, let c.t/ D .e !t cos t; e !t sin t/.
Z
1
0
t
ˇ1
ˇ
tˇ
5e dt D 5e ˇ D 5.e " 1/
0
23. Show that c.t/ for 0 # t < 1 has finite length and calculate its value.
SOLUTION
We use the formula for arc length, to obtain:
Z 1q
sD
..e !t cos t/0 /2 C ..e !t sin t/0 /2 dt
0
D
D
D
D
Z
0
Z
0
Z
1q
."e !t cos t " e !t sin t/2 C ."e !t sin t C e !t cos t/2 dt
1q
e !2t .cos t C sin t/2 C e !2t .cos t " sin t/2 dt
1
0
Z
0
1
e !t
e !t
p
p
cos2 t C 2 sin t cos t C sin2 t C cos2 t " 2 sin t cos t C sin2 t dt
2dt D
ˇ1
$
#
p
p
ˇ
2."e !t /ˇˇ D " 2 lim e !t " e 0
p
p
D " 2.0 " 1/ D 2
0
t !1
24. Find the first positive value of t0 such that the tangent line to c.t0 / is vertical, and calculate the speed at t D t0 .
Chapter Review Exercises
SOLUTION
line:
ˇ dy ˇ
ˇ D 1. We first find
The curve has a vertical tangent where lim ˇ dx
t !t0
dy
D
dx
dy
dt
dx
dt
D"
D
dy
dx
1487
using the theorem for the slope of a tangent
.e !t sin t/0
"e !t sin t C e !t cos t
D
!t
0
.e cos t/
"e !t cos t " e !t sin t
cos t " sin t
sin t " cos t
D
cos t C sin t
sin t C cos t
ˇ dy ˇ
ˇ D 1. In our case, this happens when the denominator is 0, but the numerator is not, thus:
We now search for t0 such that lim ˇ dx
t !t0
sin t0 C cos t0 D 0
cos t0 D " sin t0
cos "t0 D sin "t0
!
"t0 D " !
4
3
t0 D !
4
We now use the formula for the speed, to find the speed at t0 .
q
ds
D ..e !t sin t/0 /2 C ..e !t cos t/0 /2
dt
q
D ."e !t cos t " e !t sin t/2 C ."e !t sin t C e !t cos t/2
q
D e !2t .cos t C sin t/2 C e !2t .cos t " sin t/2
p
p
D e !t cos2 t C 2 sin t cos t C sin2 t C cos2 t " 2 sin t cos t C sin2 t D e !t 2
Next we substitute t D 43 !, to obtain
e !t0
p
p
2 D e !3!=4 2
25.
Plot c.t/ D .sin 2t; 2 cos t/ for 0 # t # !. Express the length of the curve as a definite integral, and approximate it
using a computer algebra system.
SOLUTION
We use a CAS to plot the curve. The resulting graph is shown here.
y
2
1
x
−2
−1
1
2
−1
−2
Plot of the curve .sin 2t; 2 cos t/
To calculate the arc length we use the formula for the arc length to obtain
Z !q
Z
sD
.2 cos 2t/2 C ."2 sin t/2 dt D 2
0
0
!
p
cos2 2t C sin2 t dt
We use a CAS to obtain s D 6:0972.
26. Convert the points .x; y/ D .1; "3/, .3; "1/ from rectangular to polar coordinates.
SOLUTION
We convert the given points from cartesian coordinates to polar coordinates. For the first point we have
rD
q
q
p
x 2 C y 2 D 12 C ."3/2 D 10
# D arctan
y
D arctan "3 D 5:034
x
1488
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
For the second point we have
rD
q
x2 C y2 D
q
p
32 C ."1/2 D 10
y
"1
# D arctan D arctan
D "0:321; 5:961
x
3
% ! & % 5! &
27. Convert the points .r; #/ D 1; 6 , 3; 4 from polar to rectangular coordinates.
SOLUTION
We convert the points from polar coordinates to cartesian coordinates. For the first point we have
p
!
3
x D r cos # D 1 ! cos D
6
2
!
1
y D r sin # D 1 ! sin D
6
2
For the second point we have
p
5!
3 2
x D r cos # D 3 cos
D"
4
2
p
5!
3 2
y D r sin # D 3 sin
D"
4
2
2
28. Write .x C y/ D xy C 6 as an equation in polar coordinates.
SOLUTION
We use the formula for converting from cartesian coordinates to polar coordinates to substitute r and # for x and y:
.x C y/2 D xy C 6
x 2 C 2xy C y 2 D xy C 6
x 2 C y 2 D "xy C 6
r 2 D ".r cos #/.r sin #/ C 6
r 2 D "r 2 cos # sin # C 6
r 2 .1 C sin # cos #/ D 6
r2 D
r2 D
r2 D
6
1 C sin # cos #
6
1C
sin 2"
2
12
2 C sin 2#
2 cos #
as an equation in rectangular coordinates.
cos # " sin #
SOLUTION We use the formula for converting from polar coordinates to cartesian coordinates to substitute x and y for r and #:
29. Write r D
rD
q
x2 C y2 D
q
x2 C y2 D
2 cos #
cos # " sin #
2r cos #
r cos # " r sin #
2x
x "y
4
is the polar equation of a line.
7 cos # " sin #
SOLUTION We use the formula for converting from polar coordinates to cartesian coordinates to substitute x and y for r and #:
30. Show that r D
rD
4
7 cos # " sin #
1D
4
7r cos # " r sin #
1D
4
7x " y
Chapter Review Exercises
1489
7x " y D 4
y D 7x " 4
We obtained a linear function. Since the original equation in polar coordinates represents the same curve, it represents a straight
line as well.
31.
Convert the equation
9.x 2 C y 2 / D .x 2 C y 2 " 2y/2
to polar coordinates, and plot it with a graphing utility.
SOLUTION
We use the formula for converting from cartesian coordinates to polar coordinates to substitute r and # for x and y:
9.x 2 C y 2 / D .x 2 C y 2 " 2y/2
9r 2 D .r 2 " 2r sin #/2
3r D r 2 " 2r sin #
3 D r " 2 sin #
r D 3 C 2 sin #
The plot of r D 3 C 2 sin # is shown here:
5
4
r = 3 + 2sin
3
2
1
0
−1
−2
−4
−3
−2
0
−1
1
2
3
4
Plot of r D 3 C 2 sin #
!
3
32. Calculate the area of the circle r D 3 sin # bounded by the rays # D
SOLUTION
We use the formula for area in polar coordinates to obtain
1
AD
2
Z
9
4
#
D
2!=3
!=3
9
.3 sin #/ d# D
2
2
Z
2!=3
!=3
#
$$
!
1
4!
2!
9
"
sin
" sin
D
3
2
3
3
4
2!
3 .
!
ˇ
9
sin 2# ˇˇ2!=3
.1 " cos 2#/ d# D
#"
4
2 ˇ!=3
!=3
p
p !!
p !
!
1
3
3
9 !
3
"
"
"
D
C
3
2
2
2
4 3
2
9
sin # d# D
4
2
and # D
Z
2!=3
33. Calculate the area of one petal of r D sin 4# (see Figure 1).
y
y
y
x
n = 2 (4 petals)
x
x
n = 4 (8 petals)
n = 6 (12 petals)
FIGURE 1 Plot of r D sin.n#/.
SOLUTION
We use a CAS to generate the plot, as shown here.
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1 −0.8 −0.4
r = 4 sin
0
0.4
Plot of r D sin 4#
0.8 1
1490
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
#
. We now use the formula for area in polar coordinates to obtain
4
ˇ !
Z
Z
1 !=4 2
1 !=4
1
sin 8# ˇˇ!=4
AD
sin 4# d# D
.1 " cos 8#/ d# D
#"
2 0
4 0
4
8 ˇ0
We can see that one leaf lies between the rays # D 0 and # D
!
1
!
" .sin 2! " sin 0/ D
16 32
16
34. The equation r D sin.n#/, where n $ 2 is even, is a “rose” of 2n petals (Figure 1). Compute the total area of the flower, and
show that it does not depend on n.
D
SOLUTION We calculate the total area of the flower, that is, the area between the rays # D 0 and # D 2!, using the formula for
area in polar coordinates:
ˇ !
Z
Z
1 2! 2
1 2!
1
sin 4n# ˇˇ2!
AD
sin 2n# d# D
.1 " cos 4n#/ d# D
#"
2 0
4 0
4
4n ˇ0
D
Since the area is
!
1
!
"
.sin 8n! " sin 0/ D
2
16n
2
!
for every n 2 Z, the area is independent of n.
2
35. Calculate the total area enclosed by the curve r 2 D cos #e sin " (Figure 2).
y
1
x
1
−1
FIGURE 2 Graph of r 2 D cos #e sin " .
SOLUTION
obtain:
Note that this is defined only for # between "!=2 and !=2. We use the formula for area in polar coordinates to
AD
1
2
Z
!=2
!!=2
r 2 d# D
1
2
Z
!=2
cos #e sin " d#
!!=2
We evaluate the integral by making the substitution x D sin # dx D cos # d#:
ˇ
Z
"
1 !=2
1 x ˇˇ1
1!
sin "
AD
cos #e
d# D e ˇ D
e " e !1
2 !!=2
2
2
!1
36. Find the shaded area in Figure 3.
y
1
−2
r = 1 + cos 2
1
−1
2
x
−1
FIGURE 3
SOLUTION
We first find the points of intersection between the unit circle and the function.
1 D 1 C cos 2#
cos 2# D 0
!
2# D C ! n
2
!
!
#D C n
4
2
Chapter Review Exercises
1491
We now find the area of the shaded figure in the first quadrant. This has two parts. The first, from 0 to !=4, is just an octant of the
unit circle, and thus has area !=8. The second, from !=4 to !=2, is found as follows:
Z
Z
1 !=2
1 !=2 3
1
1 C 2 cos 2# C cos2 2# d# D
C 2 cos 2# C cos 4# d#
2
2
2
2
!=4
!=4
!=4
#
$ ˇ!=2
#
$
ˇ
1 3#
1
1 3!
D
C sin 2# C sin 4# ˇˇ
D
"1
2 2
8
2 8
!=4
AD
1
2
Z
!=2
.1 C cos 2#/2 d# D
The total area in the first quadrant is thus
5!
16
" 12 ; multiply by 2 to get the total area of
5!
8
" 1.
37. Find the area enclosed by the cardioid r D a.1 C cos #/, where a > 0.
SOLUTION The graph of r D a .1 C cos #/ in the r# -plane for 0 # # # 2! and the cardioid in the xy-plane are shown in the
following figures:
y
r
θ=
2a
π
,r=a
2
θ = π, r = 0
a
θ=
π
2
π
3π
2
θ=0
r = 2a
3π
,r=a
2
x
2π
r D a .1 C cos #/
The cardioid r D a .1 C cos #/, a > 0
As # varies from 0 to ! the radius r decreases from 2a to 0, and this gives the upper part of the cardioid.
The lower part is traced as # varies from ! to 2! and consequently r increases from 0 back to 2a. We compute the area enclosed
by the upper part of the cardioid and the x-axis, using the following integral (we use the identity cos2 # D 12 C 12 cos 2# ):
1
2
Z
!
0
1
2
r 2 d# D
Z
!
0
a2
2
a2 .1 C cos #/2 d# D
Z
!
0
!
"
1 C 2 cos # C cos2 # d#
#
$
$
Z #
1
1
a2 ! 3
1
D
1 C 2 cos # C C cos 2# d# D
C 2 cos # C cos 2# d#
2 0
2
2
2 0
2
2
'
( ˇ!
'
(
2
2
ˇ
a
3#
1
a 3!
1
3!a2
D
C 2 sin # C sin 2# ˇˇ D
C 2 sin ! C sin 2! " 0 D
2
2
4
2
2
4
4
0
a2
Z
!
Using symmetry, the total area A enclosed by the cardioid is
3!a2
3!a2
D
4
2
38. Calculate the length of the curve with polar equation r D # in Figure 4.
AD2!
y
r=
2
x
FIGURE 4
SOLUTION
We get
The interval of # values is 0 # # # !. We use the formula for the arc length in polar coordinates, with r D f .#/ D #.
SD
D
Z
!
0
#p
2
q
f .#/2 C f 0 .#/2 d# D
Z
!
0
Z
q
# 2 C .# 0 /2 d# D
!
0
p
# 2 C 1 d#
ˇ ˇˇ!
"
p
p
1 ˇˇ
!p 2
1 !
ˇ
# 2 C 1 C ln ˇ# C # 2 C 1ˇ ˇˇ
D
! C 1 C ln ! C ! 2 C 1
2
2
2
" D0
1492
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
In Exercises 39–44, let v D h"2; 5i and w D h3; "2i.
39. Calculate 5w " 3v and 5v " 3w.
SOLUTION
We use the definition of basic vector operations to compute the two linear combinations:
5w " 3v D 5h3; "2i " 3h"2; 5i D h15; "10i C h6; "15i D h21; "25i
5v " 3w D 5h"2; 5i " 3h3; "2i D h"10; 25i C h"9; 6i D h"19; 31i
40. Sketch v, w, and 2v " 3w.
SOLUTION
We have,
2v " 3w D 2h"2; 5i " 3h3; "2i D h"4; 10i C h"9; 6i D h"13; 16i
The vectors v, w and 2v " 3w are shown in the figure below:
y
2v − 3w
2v
v
−3w
x
w
41. Find the unit vector in the direction of v.
SOLUTION
The unit vector in the direction of v is
ev D
1
v
kvk
We compute the length of v:
kvk D
Hence,
ev D
42. Find the length of v C w.
SOLUTION
q
p
."2/2 C 52 D 29
+
,
v
h"2; 5i
"2
5
D p
D p ;p
:
kvk
29
29
29
We first compute the sum v C w:
v C w D h"2; 5i C h3; "2i D h"2 C 3; 5 " 2i D h1; 3i
Using the definition of the length of a vector we obtain
kv C wk D kh1; 3ik D
43. Express i as a linear combination rv C sw.
SOLUTION
p
p
12 C 32 D 10:
We use basic properties of vector algebra to write
i D rv C sw
(1)
h1; 0i D rh"2; 5i C sh3; "2i D h"2r C 3s; 5r " 2si
The vector are equivalent, hence,
1 D "2r C 3s
0 D 5r " 2s
The second equation implies that s D 52 r. We substitute in the first equation and solve for r:
5
1 D "2r C 3 ! r
2
Chapter Review Exercises
11
r
2
2
rD
)
11
1493
1D
sD
5 2
5
!
D
2 11
11
Substituting in (1) we obtain
iD
2
5
v C w:
11
11
44. Find a scalar ˛ such that kv C ˛wk D 6.
SOLUTION
We compute the vector v C ˛w:
v C ˛w D h"2; 5i C ˛h3; "2i D h"2 C 3˛; 5 " 2˛i
The length of v C ˛w is
q
p
."2 C 3˛/2 C .5 " 2˛/2 D 4 " 12˛ C 9˛ 2 C 25 " 20˛ C 4˛ 2
p
D 13˛ 2 " 32˛ C 29
kv C ˛wk D
We obtain the following equation:
p
13˛ 2 " 32˛ C 29 D 6
Solving for ˛ yields
13˛ 2 " 32˛ C 29 D 36
˛1;2
13˛ 2 " 32˛ " 7 D 0
q
p
32 ˙ 322 " 4 ! 13 ! ."7/
16 ˙ 347
D
D
26
13
The two solutions are thus
p
16 ˙ 347
:
13
""!
""!
45. If P D .1; 4/ and Q D ."3; 5/, what are the components of PQ? What is the length of PQ?
˛D
SOLUTION
By the Definition of Components of a Vector we have
""!
PQ D h"3 " 1; 5 " 4i D h"4; 1i
""!
The length of PQ is
p
-""!- q
-PQ- D ."4/2 C 12 D 17:
""!
""!
""!
""!
46. Let A D .2; "1/, B D .1; 4/, and P D .2; 3/. Find the point Q such that PQ is equivalent to AB. Sketch PQ and AB.
""!
""!
SOLUTION The vectors AB and PQ are equivalent, therefore they have the same components. We denote the point Q by Q D
""!
""!
.a; b/, and compute the vectors AB and PQ. We get
""!
AB D h1 " 2; 4 " ."1/i D h"1; 5i
""!
PQ D ha " 2; b " 3i
Therefore
"1 D a " 2 and 5 D b " 3
or
a D 1 and b D 8
""! ""!
Hence the point Q is Q D .1; 8/. The equivalent vectors AB D PQ D h"1; 5i are shown in the figure:
1494
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
y
〈−1 , 5〉
5
x
−1
47. Find the vector with length 3 making an angle of
SOLUTION
7!
4
with the positive x-axis.
We denote the vector by v D ha; bi. v makes an angle # D
a D kvk cos # D 3 cos
b D kvk sin # D 3 sin
7!
4
with the x-axis, and its length is 3, hence,
7!
3
D p
4
2
7!
3
D "p
4
2
That is,
+
,
3
3
v D ha; bi D p ; " p :
2
2
48. Calculate 3 .i " 2j/ " 6 .i C 6j/.
SOLUTION
Using basic properties of vector algebra we have
3.i " 2j/ " 6.i C 6j/ D 3i " 6j " 6i " 36j D "3i " 42j
49. Find the value of ˇ for which w D h"2; ˇi is parallel to v D h4; "3i.
SOLUTION
If v D h4; "3i and w D h"2; ˇi are parallel, there exists a scalar % such that w D %v. That is,
h"2; ˇi D %h4; "3i D h4%; "3%i
yielding
"2 D 4%
and
ˇ D "3%
These equations imply that % D " 12 and % D " ˇ3 . Equating the two expressions for % gives
1
ˇ
3
D"
or ˇ D :
2
3
2
50. Let r1 .t/ D v1 C tw1 and r2 .t/ D v2 C tw2 be parametrizations of lines L1 and L2 . For each statement (a)–(e), provide a
proof if the statement is true and a counterexample if it is false.
(a) If L1 D L2 , then v1 D v2 and w1 D w2 .
(b) If L1 D L2 and v1 D v2 , then w1 D w2 .
(c) If L1 D L2 and w1 D w2 , then v1 D v2 .
(d) If L1 is parallel to L2 , then w1 D w2 .
(e) If L1 is parallel to L2 , then w1 D %w2 for some scalar %.
"
SOLUTION
(a) This statement is false. Consider the following lines:
L1 W
r1 .t/ D h1; 0i C th1; 1i
L2 W
r2 .t/ D h3; 2i C th2; 2i
The line L1 passes through the points P D .1; 0/ (for t D 0) and Q D .2; 1/ (for t D 1). The line L2 passes through P and Q as
well (for t D "1 and t D " 12 respectively). Therefore, L1 D L2 . However, v1 D h1; 0i, v2 D h3; 2i, w1 D h1; 1i, w2 D h2; 2i
hence v1 ¤ v2 and w1 ¤ w2 .
(b) This statement is false. Consider the following lines:
L1 W
r1 .t/ D h0; 1i C th1; 1i
L2 W
r2 .t/ D h0; 1i C th2; 2i
The line L1 passes through the points P D .0; 1/ (for t D 0) and Q D .1; 2/ (for t D 1). The line L2 passes through P and Q as
well (for t D 0 and t D 21 ). Therefore, L1 D L2 . Also v1 D v2 , but w1 ¤ w2 .
Chapter Review Exercises
1495
(c) This statement is false. Consider the following lines:
L1 W
r1 .t/ D h1; 0i C th1; 1i
L2 W
r2 .t/ D h2; 1i C th1; 1i
The line L1 passes through P D .1; 0/ and Q D .2; 1/ (for t D 1). The line L2 passes through P D .1; 0/ (for t D "1) and
Q D .2; 1/ (for t D 0). Therefore, L1 D L2 . Also, w1 D w2 but v1 ¤ v2 .
(d) This statement is false. Consider the following lines:
L1 W
r1 .t/ D h1; 1i C th1; 0i
L2 W
r2 .t/ D th2; 0i
We have w1 D h1; 0i and w2 D h2; 0i therefore w2 D 2w1 . We conclude that w1 and w2 are parallel vectors, hence the lines L1
and L2 are parallel although w1 ¤ w2 .
(e) This statement is correct. If L1 and L2 are parallel lines, the direction vectors w1 and w2 of these lines are parallel, hence they
are scalar multiples of one another.
51. Sketch the vector sum v D v1 " v2 C v3 for the vectors in Figure 5(A).
y
y
v2
v2
v3
v3
v1
v1
x
x
(A)
(B)
FIGURE 5
SOLUTION
Using the Parallelogram Law we obtain the vector sum shown in the figure.
y
v2
v3
v1 − v2 + v3
v1
v'3
x
v1 − v2
−v2
We first add v1 and "v2 , then we add v3 to v1 " v2 .
52. Sketch the sums v1 C v2 C v3 , v1 C 2v2 , and v2 " v3 for the vectors in Figure 5(B).
SOLUTION We use the definition of scalar multiple of a vector and the Parallelogram Law to sketch the vectors.
y
y
v1 + 2v2
v1 + v2 + v3
2v2
v1 + v2
v2
v2
v1
v3
v1
x
x
y
v3
v2
x
v2 − v3
1496
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
To form v2 " v3 , we draw the vector pointing from v3 to v2 and translate it back to the basepoint.
53. Use vectors to prove that the line connecting the midpoints of two sides of a triangle is parallel to the third side.
SOLUTION
Let E and F be the midpoints of sides AC and BC in a triangle ABC (see figure).
A
E
C
F
B
We must show that
""! ""!
EF k AB
Using the Parallelogram Law we have
""! "! ""! ""!
EF D EA C AB C BF
(1)
By the definition of the points E and F ,
1 "!
"!
EA D CAI
2
1 ""!
""!
BF D BC
2
We substitute (1) to obtain
1 "! ""! 1 ""! ""! 1 %"! ""!&
""!
EF D CA C AB C BC D AB C CA C BC
2
2
2
1 ""!
""! 1 %""! "!& ""! 1 "! ""! 1 ""!
D AB C BC C CA D AB C BA D AB " AB D AB
2
2
2
2
""!
""!
""!
""!
Therefore, EF is a constant multiple of AB, which implies that EF and AB are parallel vectors.
54. Calculate the magnitude of the forces on the two ropes in Figure 6.
B
A
30°
45°
Rope 1
Rope 2
P
10 kg
FIGURE 6
SOLUTION
we have
Gravity exerts a force Fg of magnitude 10g D 98 newtons. Since the sum of F1 and F2 balance the force of gravity,
F1 C F2 C Fg D 0
(1)
We resolve F1 , F2 , and Fg into a sum of a force along the ground and a force orthogonal to the ground.
F2
(F2)
F1
(F1)
30°
(F1)||
P
F1
Fg
45°
(F2)||
j
i
This gives
.F1 /k D ".kF1 k cos 30ı /i;
.F1 /? D .kF1 k cos 60ı /j
.Fg /k D 0;
.Fg /? D "10gj % "98j
.F2 /k D .kF2 k cos 45ı /i;
.F2 /? D .kF2 k cos 45ı /j
Chapter Review Exercises
1497
Substituting these forces in (1) gives
!
p
kF1 k 3
kF1 k
"
iC
j C
2
2
p
p !
kF2 k 2
kF2 k 2
iC
j " 98j D 0
2
2
"
"
p
p
1 !p
1!
2kF2 k " 3kF1 k i C
kF1 k C 2kF2 k " 196 j D 0
2
2
We now equate each component to zero, to obtain
p
p
2kF2 k " 3kF1 k
D0
2
p
kF1 k C 2kF2 k
" 98 D 0
2
)
kF1 k % 71:7
kF2 k % 87:8
We conclude that
F1 D ."71:7 cos 30ı /i C .71:7 cos 60ı /j D "62:1i C 35:9j
F2 D .87:8 cos 45ı /i C .87:8 cos 45ı /j D 62:1i C 62:1j
(As in the statement of the problems, all units are in Newtons.)
55. A 50-kg wagon is pulled to the right by a force F1 making an angle of 30ı with the ground. At the same time the wagon is
pulled to the left by a horizontal force F2 .
(a) Find the magnitude of F1 in terms of the magnitude of F2 if the wagon does not move.
(b) What is the maximal magnitude of F1 that can be applied to the wagon without lifting it?
SOLUTION
(a) By Newton’s Law, at equilibrium, the total force acting on the wagon is zero.
N
F
F1
30°
F2
F||
W
We resolve the force F1 into its components:
F1 D Fk C F?
where Fk is the horizontal component and F? is the vertical component. Since the wagon does not move, the magnitude of Fk must
be equal to the magnitude of F2 . That is,
kFk k D kF1 k cos 30ı D kF2 k
The above equation gives:
kF1 k
p
3
D kF2 k
2
)
kF1 k D
2kF2 k
p
3
(b) The maximum magnitude of force F1 that can be applied to the wagon without lifting the wagon is found by comparing the
vertical forces:
kF1 k sin 30ı D 9:8 ! 50
1
D 9:8 ! 50 )
2
56. Find the angle between v and w if kv C wk D kvk D kwk.
kF1 k !
SOLUTION
kF1 k D 9:8 ! 100 D 980 N
The cosine of the angle # between v and w is given by:
cos # D
v!w
kvkkwk
(1)
We denote by r the value kv C wk D kvk D kwk D r. To find v ! w in terms of r, we evaluate kv C wk. Using properties of the
dot product we obtain:
kv C wk2 D .v C w/ ! .v C w/ D v ! .v C w/ C w ! .v C w/
1498
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
D v ! v C v ! w C w ! v C w ! w D kvk2 C 2v ! w C kwk2
That is,
r 2 D r 2 C 2v ! w C r 2
"r 2 D 2v ! w
v!wD"
)
r2
2
2
We now substitute kvk D kwk D r and v ! w D " r2 in (1) to obtain:
2
2!
3 rad.
p
57. Find ke " 4fk, assuming that e and f are unit vectors such that ke C fk D 3.
The solution for 0 # # # ! is # D
SOLUTION
2!
3 .
"r
1
cos # D 2 D "
r !r
2
That is, the angle between v and w is
We use the relation of the dot product with length and properties of the dot product to write
3 D ke C fk2 D .e C f/ ! .e C f/ D e ! e C e ! f C f ! e C f ! f
D kek2 C 2e ! f C kfk2 D 12 C 2e ! f C 12 D 2 C 2e ! f
We now find e ! f:
3 D 2 C 2e ! f
e ! f D 1=2
)
Hence, using the same method as above, we have:
ke " 4fk2 D .e " 4f/ ! .e " 4f/
D kek2 " 2 ! e ! 4f C k4fk2 D 12 " 8e ! f C 42 D 17 " 4 D 13
Taking square roots, we get:
ke " 4fk D
p
13
58. Find the area of the parallelogram spanned by vectors v and w such that kvk D kwk D 2 and v ! w D 1.
SOLUTION
The base of the parallelogram is kvk D kwk D 2, and the height must be 2 sin #. For #, we have
then cos # D 1=4 and so sin # D
p
v ! w D kvkkwk cos # D 2 ! 2 ! cos # D 1;
p
p
p
p
1 " cos2 # D 15=16 D 15=4. Thus, the area is 2 ! 2 ! sin # D 2 ! 2 ! 15=4 D 15
In Exercises 59–64, calculate the derivative indicated.
˝
˛
59. r0 .t/, r.t/ D 1 " t; t !2
SOLUTION
60. r000 .t/,
SOLUTION
We use the Theorem on Componentwise Differentiation to compute the derivative r0 .t/. We get
E
˝
˛ D
r0 .t/ D .1 " t/0 ; .t !2 /0 D "1; "2t !3
˝
˛
r.t/ D t 3 ; 4t 2
We use the Theorem on Componentwise Differentiation to find r0 .t/:
˝
˛ ˝
˛
r0 .t/ D .t 3 /0 ; .4t 2 /0 D 3t 2 ; 8t
We differentiate r0 .t/ componentwise to find r00 .t/:
r00 .t/ D h6t; 8i
Differentiating r00 .t/ componentwise gives r000 .t/:
61. r0 .0/,
SOLUTION
˝
2˛
r.t/ D e 2t ; e !4t
r000 .t/ D h6; 0i
We differentiate r.t/ componentwise to find r0 .t/:
˝
˝
2 0˛
2˛
0
r0 .t/ D .e 2t / ; .e !4t / D 2e 2t ; "8te !4t
The derivative r0 .0/ is obtained by setting t D 0 in r0 .t/. This gives
˝
2˛
r0 .0/ D 2e 2"0 ; "8 ! 0e !4"0 D h2; 0i
Chapter Review Exercises
˝
˛
r.t/ D t !2 ; .t C 1/!1
62. r00 ."3/,
SOLUTION
We differentiate componentwise to find r0 .t/:
D
E
r0 .t/ D "2t !3 ; ".t C 1/!2
We differentiate componentwise to find r00 .t/ and evaluate at t D "3:
63.
64.
D
E
r00 .t/ D 6t !4 ; 2.t C 1/!3 ;
d t˝ ˛
e 1; t
dt
SOLUTION
+
) r00 ."3/ D
2
1
;"
27 4
,
Using the Product Rule for differentiation gives
˝
˛
r.s/ D s; 2s
d
r.cos #/,
d#
SOLUTION
1499
˝ ˛
d t˝ ˛
d ˝ ˛ % t &0 ˝ ˛
e 1; t D e t
1; t C e 1; t D e t h0; 1i C e t 1; t
dt
dt
%
˝ ˛&
˝
˛
D e t h0; 1i C 1; t D e t 1; 1 C t
We use the Chain Rule to compute the derivative. That is,
ˇ
# ˇ
$
ˇ
d
d r ˇˇ
d
ˇ
r.cos #/ D
!
.cos
#/
D
"
sin
#
!
2i
h1;
ˇ
ˇ
d#
ds sDcos "
d#
sDcos "
D " sin # ! h1; 2i D h" sin #; "2 sin #i
D " hsin #; 2 sin #i
In Exercises 65–66, calculate the derivative at t D 3, assuming that
r1 .3/ D h1; 1i ;
r01 .3/
65.
r02 .3/ D h0; 2i
D h0; 0i ;
d
.6r1 .t/ " 4 ! r2 .t//
dt
SOLUTION
66.
r2 .3/ D h1; 1i
Using Differentiation Rules we obtain:
ˇ
ˇ
d
.6r1 .t/ " 4r2 .t//ˇˇ
D 6r01 .3/ " 4r02 .3/ D 6 ! h0; 0i " 4 ! h0; 2i
dt
t D3
D h0; 0i " h0; 8i D h0; "8i
&
d % t
e r2 .t/
dt
SOLUTION
Using the Product Rule gives:
Setting t D 3 we get:
67. Calculate
Z
0
SOLUTION
3˝
&
% &0
%
&
d % t
e r2 .t/ D e t r02 .t/ C e t r2 .t/ D e t r02 .t/ C r2 .t/
dt
ˇ
&ˇ
%
&
d % t
e r2 .t/ ˇˇ
D e 3 r02 .3/ C r2 .3/ D e 3 .h0; 2i C h1; 1i/ D e 3 h1; 3i
dt
t D3
˛
4t C 3; t 2 dt.
By the definition of vector-valued integration, we have
*Z
Z 3D
Z
E
3
4t C 3; t 2 dt D
.4t C 3/ dt;
0
0
0
3
t 2 dt
+
We compute the integrals on the right-hand side:
Z
3
0
ˇ3
ˇ
.4t C 3/ dt D 2t C 3t ˇˇ D 2 ! 9 C 3 ! 3 " 0 D 27
2
0
(1)
1500
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
Z
3
0
t 2 dt D
ˇ
t 3 ˇˇ3
33
D
D9
3 ˇ0
3
Substituting in (1) gives the following integral:
68. Calculate
Z
0
SOLUTION
Z
3˝
0
!˝
˛
˛
4t C 3; t 2 dt D h27; 9i
sin #; # d# .
By the definition of vector-valued integration, we have
+Z !
Z !
Z
sin # d#;
hsin #; #i d# D
0
0
!
# d#
0
,
(1)
We compute the integrals on the right hand-side:
ˇ!
Z !
ˇ
sin # d# D " cos # ˇˇ D ".cos ! " cos 0/ D "."1 " 1/ D 2
0
0
Z
!
0
ˇ
1 2 ˇˇ!
!2
# d# D # ˇ D
2
2
0
Substituting in (1) gives the following integral:
Z
!
0
*
!2
hsin #; #i d# D 2;
2
+
˝ ˛
69. A particle located at .1; 1/ at time t D 0 follows a path whose velocity vector is v.t/ D 1; t . Find the particle’s location at
t D 2.
SOLUTION
We first find the path r.t/ by integrating the velocity vector v.t/:
+Z
, +
,
Z
Z
1
r.t/ D h1; ti dt D
1 dt; t dt D t C c1 ; t 2 C c2
2
Denoting by c D hc1 ; c2 i the constant vector, we obtain:
+
,
1
r.t/ D t; t 2 C c
2
(1)
To find the constant vector c, we use the given information on the initial position of the particle. At time t D 0 it is at the point
.1; 1/. That is, by (1):
r.0/ D h0; 0i C c D h1; 1i
or,
c D h1; 1i
We substitute in (1) to obtain:
+
,
+
,
1
1
r.t/ D t; t 2 C h1; 1i D t C 1; t 2 C 1
2
2
Finally, we substitute t D 2 to obtain the particle’s location at t D 2:
+
,
1
r.2/ D 2 C 1; ! 22 C 1 D h3; 3i
2
At time t D 2 the particle is located at the point
.3; 3/
˝
˛
2
70. Find the vector-valued function r.t/ D x.t/; y.t/ in R satisfying r0 .t/ D "r.t/ with initial conditions r.0/ D h1; 2i.
Chapter Review Exercises
SOLUTION
We rewrite the differential equation by components as:
˝ 0
˛
˝
˛
x .t/; y 0 .t/ D " hx.t/; y.t/i or x 0 .t/; y 0 .t/ D h"x.t/; "y.t/i
Equating corresponding components, we obtain:
x 0 .t/ D "x.t/
x 0 .t/
D "1;
x.t/
)
y 0 .t/ D "y.t/
y 0 .t/
D "1
y.t/
By integration we get ln.x.t// D "t C A, ln.y.t// D "t C B or:
x.t/ D ae !t
y.t/ D be !t
where
a D eA;
b D eB
By the given information r.0/ D h1; 2i. Therefore,
x.0/ D ae !0 D a D 1
y.0/ D be !0 D b D 2
)
a D 1;
bD2
We obtain the following vector:
˝
˛ ˝
˛
r.t/ D hx.t/; y.t/i D ae !t ; be !t D e !t ; 2e !t :
71. Calculate r.t/ assuming that
D
E
r00 .t/ D 4 " 16t; 12t 2 " t ;
SOLUTION
r0 .0/ D h1; 0i ;
Using componentwise integration we get:
Z D
E
r0 .t/ D
4 " 16t; 12t 2 " t dt
D
+Z
*
4 " 16t dt;
Z
12t 2 " t dt
+
t2
D 4t " 8t ; 4t "
C c1
2
2
r.0/ D h0; 1i
,
3
Then using the initial condition r0 .0/ D h1; 0i we get:
r0 .0/ D h1; 0i D c1
so then
*
+
*
+
t2
t2
2
3
r .t/ D 4t " 8t ; 4t "
C h1; 0i D 4t " 8t C 1; 4t "
2
2
0
2
3
Then integrating componentwise once more we get:
r.t/ D
D
Z *
*Z
*
t2
4t " 8t C 1; 4t "
2
2
3
2
Z
4t " 8t C 1 dt;
+
dt
t2
4t "
dt
2
+
3
+
8
t3
D 2t 2 " t 3 C t; t 4 "
C c2
3
6
Using the initial condition r.0/ D h0; 1i we have:
r.0/ D h0; 1i D c2
Therefore,
*
+
*
+
8 3
t3
8 3
t3
4
2
4
r.t/ D 2t " t C t; t "
C h0; 1i D 2t " t C t; t "
C1
3
6
3
6
2
1501
1502
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
C H A P T E R 11
˝
˛
72. Solve r00 .t/ D t 2 " 1; t C 1 subject to the initial conditions r.0/ D h1; 0i and r0 .0/ D h"1; 1i
SOLUTION
Using integration componentwise we get:
r0 .t/ D
D
D
Z D
+Z
*
E
t 2 " 1; t C 1 dt
t 2 " 1 dt;
Z
t C 1 dt
+
t3
t2
" t;
C t C c1
3
2
,
Using the initial condition r0 .1/ D h"1; 1i we get:
D
E
so then, c1 D " 13 ; " 12 and
+
,
2 3
r0 .1/ D h"1; 1i D " ;
C c1
3 2
*
+ +
+
, * 3
t3
t2
1 1
t
1 t2
1
r .t/ D
" t;
Ct C " ;" D
"t " ;
Ct "
3
2
3 2
3
3 2
2
0
Using integration componentwise once more we get:
Z *
+
t3
1 t2
1
"t " ;
Ct "
dt
3
3 2
2
r.t/ D
*Z
Z
t2
1
D
C t " dt
2
2
*
+
t4
t2
t t3
t2
t
D
"
" ;
C
"
C c2
12
2
3 6
2
2
Using the initial condition, r.1/ D h1; 0i we get:
t3
1
" t " dt;
3
3
+
+
,
3 1
r.1/ D h1; 0i D " ;
C c2
4 6
and
c2 D
+
7 1
;"
4 6
,
Therefore,
*
+ +
,
t4
t2
t t3
t2
t
7 1
r.t/ D
"
" ;
C
"
C ;"
12
2
3 6
2
2
4 6
*
+
t4
t2
t
7 t3
t2
t
1
D
"
" C ;
C
" "
12
2
3
4 6
2
2 6
73. A projectile fired at an angle of 60ı lands 400 m away. What was its initial speed?
SOLUTION Place the projectile at the origin, and let r.t/ be the position vector of the projectile.
Step 1. Use Newton’s Law
Gravity exerts a downward force of magnitude mg, where m is the mass of the bullet and g D 9:8 m/s2 . In vector form,
F D h0; "mgi D m h0; "gi
Newton’s Second Law F D mr0 .t/ yields m h0; "gi D mr00 .t/ or r00 .t/ D h0; "gi. We determine r.t/ by integrating twice:
Z t
Z t
r0 .t/ D
r00 .u/ du D
h0; "gi du D h0; "gti C v0
0
r.t/ D
Z
t
0
r0 .u/ du D
0
Z
t
0
+
,
1
.h0; "gui C v0 / du D 0; " gt 2 C tv0 C r0
2
Chapter Review Exercises
1503
Step 2. Use the initial conditions
By our choice of coordinates, r0 D 0. The initial velocity v0 has unknown magnitude v0 , but we know that it points in the
direction of the unit vector hcos 60ı ; sin 60ı i. Therefore,
* p +
˝
1
3
ı
ı˛
v0 D v0 cos 60 ; sin 60 D v0
;
2 2
* p +
+
,
1 2
1
3
r.t/ D 0; " gt C tv0
;
2
2 2
Step 3. Solve for v0 .
The projectile hits the point h400; 0i on the ground if there exists a time t such that r.t/ D h400; 0i; that is,
* p +
+
,
1 2
1
3
0; " gt C tv0
;
D h400; 0i
2
2 2
Equating components, we obtain
p
1
3
" gt 2 C
tv0 D 0
2
2
1
tv0 D 400;
2
The first equation yields t D
800
v0 .
Now substitute in the second equation and solve, using g D 9:8m/s2 :
"4:9
#
$2
C
p #
$
3 800
v0 D 0
2
v0
800
v0
$2
D
800
v0
#
p
400 3
4:9
! v "2
4:9
0
D
p % 0:00707
800
400 3
v02 D 4526:42611;
v0 % 67:279 m/s
We obtain v0 % 67:279 m/s.
74. A force F D h12t C 4; 8 " 24ti (in newtons) acts on a 2-kg mass. Find the position of the mass at t D 2 s if it is located at
.4; 6/ at t D 0 and has initial velocity h2; 3i in m/s.
SOLUTION
Recall the formula F D ma then using F D h12t C 4; 8 " 24ti and m D 2 we get:
h12t C 4; 8 " 24ti D 2a;
)
a.t/ D r00 .t/ D h6t C 2; 4 " 12ti
Then using componentwise integration,
Z
Z
D
E
v.t/ D a.t/ dt D h6t C 2; 4 " 12ti dt D 3t 2 C 2t; 4t " 6t 2 C c1
Using the initial condition v0 D v.0/ D h2; 3i, we get:
v.0/ D h2; 3i D c1
and therefore,
D
E
v.t/ D 3t 2 C 2t C 2; 4t " 6t 2 C 3
Using componentwise integration once more,
Z
Z D
E
D
E
r.t/ D v.t/ dt D
3t 2 C 2t C 2; 4t " 6t 2 C 3 dt D t 3 C t 2 C 2t; 2t 2 " 2t 3 C 3t C c2
Using the initial condition r.0/ D h4; 6i we get:
r.0/ D h4; 6i D c2
Therefore,
D
E
r.t/ D t 3 C t 2 C 2t C 4; 2t 2 " 2t 3 C 3t C 6
and the position of the mass at t D 2 is r.2/ D h20; 4i.
1504
C H A P T E R 11
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
˝
˛
75. Find the unit tangent vector to r.t/ D sin t; t at t D !.
SOLUTION
The unit tangent vector at t D ! is
T.!/ D
r0 .!/
kr0 .!/k
We differentiate r.t/ componentwise to obtain:
r0 .t/ D hcos t; 1i
Therefore,
r0 .!/ D hcos !; 1i D h"1; 1i
We compute the length of r0 .!/:
kr0 .!/k D
Substituting in (1) gives:
q
p
."1/2 C 12 D 2
+
,
"1 1
T.!/ D p ; p
2
2
(1)
AP-11
Chapter 11: Parametric Equations, Polar Coordinates, and Vector Functions
Preparing for the AP Exam Solutions
Multiple Choice Questions
1) C
7) D
13) E
19) D
2) D
8) D
14) A
20) A
3) D
9) E
15) A
4) D
10) C
16) C
5) A
11) B
17) B
6) B
12) B
18) D
Free Response Questions
1. a) a(t )  2,8e 2t , so a(3)  2,8e 6
b) The length of the velocity vector is v(0)  5,4  5 2  4 2 = 41
dy
dx
dy dy / dt 4
 4e2t and
 2t  5 , so when t  0 ,

 . An equation of the line is
dt
dt
dx dx / dt 5
4
y  2  ( x  6)
5
d) The second coordinate is always positive, so we need the first coordinate positive also. x(t )   2t  5dt
c)
= t 2  5t  C , and x(0)  6 means C  6 . Thus x(t )  t 2  5t  6  (t  6)(t  1) , which is positive
for t  6 and t  1 .
POINTS:
(a) (2 pts) 1) finds v (t ) ; 1) answer
(b) (1 pt)
dy
dx
(c) (2 pts) 1)
and
; 1) answer
dt
dt
(d) (4 pts) 1) Finds y  2e 2t ; 1) notes y(t )  0 for all t.; 1) Finds x(t )  t 2  5t  6 ; 1) answer

2. a)
1
(1  cos( )) 2 d

20

dy
d

((1  cos  ) sin  ) =
2
d d

dx
d
 sin 2   cos   cos 2  , which equals  1 when   . Also

((1  cos  ) cos  ) =
2
d d
b) The positive y-axis is given by  
, so the point is (0, 1).
AP-11
dy
dy
d = 1. An equation of the line is
 sin   2 cos  sin  which is also  1 when   . Thus

2
dx dx
d
y  x  1.

dx
d

((1  cos  ) cos  ) =  sin   2 cos  sin  =  sin  (1  2 cos  ) .
d d
1
sin   0 or cos    . When sin   0 , cos   1or –1, and the corresponding
2
c) From above , we have
Thus
dx
 0 when
d
points (in rectangular coordinates) are (2, 0) and (0, 0). When cos   
1
3
, sin   
, and the
2
2
1 3
1
3
1 3
1
3
corresponding points are ( ,
) and ( ,
) . The answer is ( ,
) and ( ,
).
4 4
4
4
4 4
4
4
POINTS:
(a) (2 pts) 1) limits of integration; 1) integrand
dy
dx
(b) (3 pts) 1)
; 1)
; 1) answer
d
d
(c) (4 pts) 1) Gets sin   0 and cos   
1
1 3
1
3
; 1) finds (2, 0) and (0, 0); finds ( ,
) and ( ,
);
2
4 4
4
4
1) answer
3. a) We have for F that x(t )  2 x(t ) , so x(t )  Ce 2t and 3  Ce 2 , thus C  3e 2 and x(t )  3e 2t  2 .
Similarly, we have y(t )  4e 2t  2 . F (t )  3e2t 2 , 4e2t 2 .
In like manner, G(t )  9e3t ,12e3t .
ln(3)  2
. We
must also have
5
ln(3)  2
ln(3)  2
. The particles are at the same point when t 
.
 12e 3t , or t 
5
5
b) F (t )  G(t ) means 3e 2t  2  9e 3t , or e 5t  2  3 ; 5t  2  ln 3 ; t 
4e 2t  2
c) F (t )  e2t 2 3, 4 ; since the range of e 2t  2 is all positive numbers, F visits all points of the form
y
4
x, x  0 . Similarly, G(t )  e3t 9,12  3e3t 3, 4 , the same set of points.
3
d)


(27e
3t
)  (36e
2
0
=
3t

) dt   (27  36 )e
2
2
0
2
6t

dt =  27  36 e
2
0
2
 3t
27 2  36 2 3t
dt = lim
e
B 
3
B
0
27  36
lim (e 3 B  1) = 27  3  36  4 = 3 9  16 = 15.
B

9
2
2
Alternatively, G goes in a straight line from 9,12 to 0, 0 in the limit, so the distance is
POINTS:
9 2  12 2 = 15.
AP-11
(a) (2 pts) 1) F (t ) ; 1) G (t )
(b) (3 pts) 1) sets either first or second coordinates equal; 1) finds t for one coordinate; 1) verifies t for other
coordinate
(c) (2 pts) 1) Finds the half-line for one; 1) Finds same half-line for the other
(d) (2 pts) 1) Set-up; 1) answer
4. a) Since x  R cos  ,
When    ,
2
b)


dx dR
2
1

cos   R sin  =
cos  
sin  .
2
d d
2  1
(2  1)
dx
2

 0 . The tangent line is not vertical.
d (2  1) 2
2
1
1
1
1 2
1
.


(
) d =
16  4 8  4
4(2 1) 
2 2  1

c) The length is given by

0
B
lim 
B 
0
(2  1) 2  4
1 2
2
2
= lim 
d >
) (
)
d

B 
2  1
(2  1) 4
(2  1) 2
0
B
(
1
(2  1) 2
B
d = lim 
d = lim ln(2 1) 0 = lim ln( 2 B  1)  ln(1) = lim ln( 2 B  1) ,
4
B 
B 
B


B
2  1
(2  1)
0
B
which is infinite.
POINTS:
(a) (3 pts) 1)
dx dR
dx
2
1
=

cos   R sin  ; 1)
cos  
sin  ; 1)
2
d d
d (2  1)
2  1
dx
2

 0 and conclusion
d (2  1) 2
(b) (3 pts) 1) limits of integration; 1) integrand; 1) answer

(c) (3 pts) 1)

0
divergent
(
1 2
2
) (
)2 d ; 1) compares integral to divergent integral; 1) shows integral is
2  1
(2  1) 2