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.