Chapter 10
1.
Consider the curve represented by x = 4 sin θ and y = 3 cos θ , 0 ≤ θ ≤ 2π .
a.
Identify the parameter and determine whether the curve is traced
clockwise or counterclockwise as the parameter varies. Justify your
conclusion.
b.
Eliminate the parameter and find the corresponding rectangular equation.
Identify the resulting curve
c.
Find the area of the region traced out as θ varies from 0 to π .
1
Chapter 10
1.
2
Consider the curve represented by x = 4 sin θ and y = 3 cos θ , 0 ≤ θ ≤ 2π .
a. Identify the parameter and determine whether the
curve is traced clockwise or counterclockwise as the
parameter varies. Justify your conclusion.
+1: correct parameter
The parameter is θ ; the curve is traced clockwise. When
+1: clockwise
θ = 0 , the point (0, 3) is on the graph. When θ =
+1: justification
point is (2,
when θ =
π
6
, the
3
π
3 3
). When θ = , the point is ( 2 3 , );
2
3
2
π
, the point is (4, 0).
2
b. Eliminate the parameter and find the corresponding
rectangular equation. Identify the resulting curve.
y
x
y
+2: cos θ = and
and sin θ =
3
4
3
2
2
x
⎛ x⎞ ⎛ y⎞
sin θ =
Since, sin 2 θ + cos 2 θ = 1, ⎜ ⎟ + ⎜ ⎟ = 1 .
4
⎝4⎠ ⎝ 3⎠
x2 y2
+1: ellipse
+
= 1 . The curve is an ellipse.
Therefore,
16 9
c. Find the area of the region traced out as θ varies from
0 to π .
+2: integrand
3
16 2
+1: limits, answer
∫−3 16 − 9 y dy = 18.850
cos θ =
Chapter 10
2.
3
Consider the curve represented by x = t and y = t − 1 .
a.
Complete the table:
0
1
t
x
y
2
3
4
b.
Plot the points and indicate orientation of the graph.
c.
Find the rectangular equation by eliminating the parameter. Identify the
curve and its domain.
Chapter 10
Consider the curve represented by x = t and y = t − 1 .
2.
a.
Complete the table:
0
1
t
0
1
x
y
b.
4
−1
0
2
2
1
3
3
2
4
2
+1: (0, −1) & (1, 0)
3
+1: ( 2 ,1), ( 3 ,2) & (2,3)
Plot the points and indicate orientation of the graph.
+1: correct points
+1: orientation
c. Find the rectangular equation by eliminating the
parameter. Identify the curve and its domain.
t = y +1 ⇒
x=
y +1 ⇒
x2 = y +1 ⇒ y = x2 −1
This curve is a parabola with domain x ≥ 0 .
+2: t = y + 1
+2: solve for y
+1: parabola with domain
Chapter 10
3.
5
Consider the curve represented by x = t and y = t 2 − 2 .
dy
for the given curve.
dx
a.
Find
b.
Find the slope of this curve at the point (2, 14).
c.
Find concavity at (1, −1).
Chapter 10
3.
a.
6
Consider the curve represented by x = t and y = t 2 − 2 .
Find
dy
for the given curve.
dx
b.
dy
dy
= dt =
dx dx
dt
3
2t
= 4t 2
1 −12
t
2
Find the slope of this curve at the point (2, 14).
c.
x= t =2⇒t =4
3
dy
= 4(4) 2 = 4(8) = 32
dx
Find concavity at (1, −1).
d ⎛ dy ⎞
d y dt ⎜⎝ dx ⎟⎠
=
=
dx
dx2
dt
1
3
4 ⎛⎜ ⎞⎟ t 2
⎝ 2⎠
= 12t
1 t−12
2
dy
dy
+2:
= dt
dx dx
dt
+1: answer
+2: determines t = 4
+1: slope = 32
2
At (1, −1), t = 1 . Therefore,
concave up.
+2: finds
d2y
dx 2
+1: answer
d2y
= 12 . So concavity is
dx 2
Chapter 10
4.
7
Consider the curve represented by x = t 2 and y = t 3 − 3t .
dy
for the given curve.
dx
a.
Find
b.
Find the slope of this curve at the point (4, 2).
c.
Identify points on the curve where the tangent line is horizontal.
Determine concavity at these points.
Chapter 10
8
Consider the curve represented by x = t 2 and y = t 3 − 3t .
4.
a.
Find
dy
for the given curve.
dx
dy dy dt 3t 2 − 3
=
=
dx dx dt
2t
b. Find the slope of this curve at the point (4, 2).
+2: uses chain rule
+1: answer
x = t 2 = 4 ⇒ t = ±2
+1: determines t = 2
y = (2) − 3 (2) = 2
+1: answer
3
y = ( −2 ) − 3 ( −2 ) ≠ 2 so t = 2
3
dy 3 ( 2 ) − 3 9
=
=
dx
2 (2)
4
2
c. Identify points on the curve where the tangent line is
horizontal. Determine concavity at these points.
+1: t = ±1
2
3t − 3
= 0 ⇒ 3t 2 − 3 = 0 ⇒ t 2 = 1∴ t = ±1
2t
The points on the curve where the tangent line is
horizontal are (1, −2) and (1, 2).
d ⎛ dy ⎞
d y dt ⎜⎝ dx ⎟⎠ 3 t 2 + 1
Since
, at the point where t = 1 ,
=
=
dx
dx 2
4t 3
dt
the graph is concave up. At the point where t = −1 , the
graph is concave down.
2
(
)
+2: points
+1: concave up at (1, −2)
and concave down at (1, 2)
Chapter 10
5.
Consider the sphere generated by rotating the curve given by x = 3 cos t and
y = 3 sin t , 0 ≤ t ≤ π .
a.
b.
described.
c.
Use the parametric equations to determine the surface area of this sphere.
Eliminate the parameter to find the rectangular equation for the circle
Determine any points of horizontal tangency.
9
Chapter 10
10
5.
Consider the sphere generated by rotating the curve given by x = 3 cos t and
y = 3 sin t , 0 ≤ t ≤ π .
a. Use the parametric equations to determine the surface
area of this sphere.
dx
dy
= −3 sin t and
= 3 cos t ;
dt
dt
Surface area =
Since
π
∫ 2π (3)sin t (− 3 sin t ) + (3 cos t )
2
2
+2: integrand
+1: limits, answer
dt =
0
π
6π ∫ sin t 9 sin 2 t + 9 cos 2 t dt =
0
π
18π ∫ sin t dt = 18π [− cos t ] 0 = 36π
π
0
b. Eliminate the parameter to find the rectangular
equation for the circle described.
+2: identifies cos t and sin t
x
y
cos t = and sin t = . Therefore,
3
3
2
2
⎛ x⎞ ⎛ y⎞
2
2
⎜ ⎟ + ⎜ ⎟ =1⇒ x + y = 9
⎝3⎠ ⎝ 3⎠
c. Determine any points of horizontal tangency.
dy
3 cos t
=
= − cot t
dx − 3 sin t
− cot t = 0 ⇒ cot t = 0 ⇒ t =
+1: rectangular equation of
circle
+2: finds
π
2
Therefore, (0, 3) is a point of horizontal tangency.
dy
dx
+1: point (0, 3)
Chapter 10
6.
11
Consider the curve represented by x = t 3 , y = t 2 , 0 ≤ t ≤ 1 .
a.
Use the parametric equations to find the surface area of the surface
obtained by rotating this curve about the x-axis.
b.
Eliminate the parameter and find the rectangular equation for this curve.
c.
Determine any points of vertical tangency.
Chapter 10
6.
12
Consider the curve represented by x = t 3 , y = t 2 , 0 ≤ t ≤ 1 .
a. Use the parametric equations to find the surface area
of the surface obtained by rotating this curve about the xaxis.
1
S = ∫ 2π ⋅ t
0
2
1
(3t ) + (2t ) dt = 2π ∫ t
2 2
2
2
9t 4 + 4t 2 dt =
+2: integrand
+1: answer
0
1
2π ∫ t 3 9t 2 + 4dt = 4.936
0
b. Eliminate the parameter and find the rectangular
equation for this curve.
t=x
1
3
⇒ y = ⎛⎜ x
⎝
1
3
⎞⎟
⎠
2
+1: determines t = x
1
3
+1: solves for y
2
c.
y=x 3
Determine any points of vertical tangency.
Points of vertical tangency occur when the slope is
undefined. Therefore,
dy 2t
2
= 2 =
dx 3t
3t
3t = 0 ⇒ t = 0
(0, 0) is the only point of vertical tangency.
+2: finds
dy
dx
+1: finds the point (0, 0)
Chapter 10
7.
13
Consider the curve whose polar equation is r = 1 − cos θ .
a.
Write this equation in parametric form.
b.
Determine the slope of the tangent line when θ =
c.
Identify the values of θ at which horizontal tangent lines to this curve
occur.
π
6
.
Chapter 10
Consider the curve whose polar equation is r = 1 − cos θ .
7.
a.
14
Write this equation in parametric form.
x = r cos θ = (1 − cos θ ) cos θ = cos θ − cos θ
y = r sin θ = (1 − cos θ )sin θ = sin θ − cos θ sin θ
Determine the slope of the tangent line when
+1: finds x
2
b.
θ=
π
6
+1: finds y
.
dy
=
dx
dy
dx
dθ
dθ
=
cos θ − [cos θ cos θ + sin θ (− sin θ )]
=
− sin θ − 2 cos θ (− sin θ )
cos θ − cos 2 θ + sin 2 θ
− sin θ + 2 cos θ sin θ
2
Therefore, when θ =
π
6
⎛ 3 ⎞
3
⎟
− ⎜⎜
⎟
2
⎝ 2 ⎠
⎛ 3
−1
+ 2 ⎜⎜
2
⎝ 2
⎛ 1 ⎞
+ ⎜ ⎟
⎝ 2 ⎠
⎞⎛ 1 ⎞
⎟⎜ ⎟
⎟⎝ 2 ⎠
⎠
+2: finds
dy
dx
+1: answer
2
=
3
1
−
2
2 = 1
−1
3
+
2
2
c.
Identify the values of θ at which horizontal
tangent lines to this curve occur.
Horizontal tangent lines occur whenever
dy
= 0 as long
dθ
dx
does not also equal 0.
dθ
Therefore,
cos θ − cos 2 θ + sin 2 θ = cos θ − cos 2 θ + 1 − cos 2 θ = 0
as
(
⇒ 2 cos 2 θ − cos θ − 1 = 0
⇒ (cos θ − 1)(2 cos θ + 1) = 0
cos θ = 1 ⇒ θ = 0
2π
4π
−1
cos θ =
⇒θ =
or
2
3
3
Horizontal tangent lines occur when θ =
2π
4π
or
.
3
3
)
+2: sets
dy
=0
dθ
+1: solves for θ
+1: eliminates θ = 0
Chapter 10
8.
15
A curve is represented by the polar equation r = 2 − sin θ .
a.
Write this equation in parametric form.
b.
Determine the slope of the tangent line when θ =
c.
Identify the values of θ at which horizontal tangent lines occur.
π
2
Chapter 10
16
8.
A curve is represented by the polar equation r = 2 − sin θ .
a.
Write this equation in parametric form.
x = r cos θ = (2 − sin θ ) cos θ = 2 cos θ − sin θ cos θ
b.
θ=
y = r sin θ = (2 − sin θ )sin θ = 2 sin θ − sin 2 θ
Determine the slope of the tangent line when
π
2
dy
=
dx
+1: finds x
+1: find y
+2: finds
dy
dx
dθ
dθ
=
2 cos θ − 2 sin θ cos θ
=
− 2 sin θ − [sin θ (− sin θ ) + cos θ (cos θ )]
2 cos θ − 2 sin θ cos θ
− 2 sin θ + sin 2 θ − cos 2 θ
dy
dx
+1: answer
2(0) − 2(1)(0)
=0
2 − 2(1) + 1 − 0
c. Identify the values of θ at which horizontal tangent
lines occur.
Therefore, when θ =
π
,
Horizontal tangent lines occur whenever
as
dx
does not also equal 0.
dθ
dy
= 0 as long
dθ
π 3π
2
,
1 − sin θ = 0 ⇒ θ = π
2
2
Horizontal tangent lines occur when θ = π
dy
=0
dθ
+1: solves for θ
Therefore, 2 cos θ − 2 sin θ cos θ = 2 cos θ (1 − sin θ ) = 0
2 cos θ = 0 ⇒ θ =
+2: sets
2
or 3π
2
+1: determines that both
values of θ are valid
Chapter 10
9.
17
The graph of a cardioid is represented by f (θ ) = r = 2 + 2 cos θ .
a.
Find the area of the region above the x-axis and under the graph of this
curve.
b.
Determine the length of the arc from θ = 0 to θ = 2π .
c.
Write, but do not solve, the integral used to find the surface area of the
solid generated when the region in part (a) is rotated about the polar axis.
Chapter 10
18
The graph of a cardioid is represented by f (θ ) = r = 2 + 2 cos θ .
9.
a. Find the area of the region above the x-axis and under
the graph of this curve.
+2: integral
Area =
π
π
2
2
∫ (2 + 2 cosθ ) dθ = 12 ∫ (4 + 8 cosθ + 4 cos θ )dθ =
1
2
0
1
2
+1: answer
0
[4θ + 8 sin θ + 4(
1
2
θ + sin 2θ )] 0 =
π
1
4
(2θ + 4 sin θ + θ + 12 sin 2θ ) π0
= 3π
Determine the length of the arc from θ = 0 to
θ = 2π .
b.
Note: f ′(θ ) = −2 sin θ Therefore, arc length s is
2π
∫ (2 + 2 cosθ ) + (− 2 sin θ )
2
2
dθ =
+2: integral
+1: answer
0
2π
∫
0
2π
8(1 + cos θ ) dθ = 2 2 ∫ 1 + cos θ dθ = 16
0
c. Write, but do not solve, the integral used to find the
surface area of the solid generated when the region in part
(a) is rotated about the polar axis.
+2: integrand
+1: limits, constant
π
S = 2π ∫ (2 + 2 cos θ )sin θ
0
(2 + 2 cos θ )2 + (− 2 sin θ )2 dθ
Chapter 10
10.
Consider the polar equation f (θ ) = r = 3 cos θ .
a.
Find the area of the region above the x-axis and below the graph of this
curve.
b.
Find the length of the arc from θ = 0 to θ = π .
c.
Find the area of the region that lies inside both the curve f (θ ) and the
curve g (θ ) = r = sin θ
19
Chapter 10
20
Consider the polar equation f (θ ) = r = 3 cos θ .
10.
a. Find the area of the region above the x-axis and
below the graph of this curve.
π
Area =
1
2
∫(
)
2
3 cos θ dθ =
0
3
2
( 12 θ + 14 sin 2θ ) π0 = 3 π
+2: integrand
+1: limits, constant, answer
4
Find the length of the arc from θ = 0 to θ = π .
b.
π
s=∫
0
π
∫
(
3 cos θ
3 dθ = 3θ
0
) + (−
2π
0
)
2
2
3 sin θ dθ =
+2: integral
+1: answer
= 3π
c. Find the area of the region that lies inside both the
curve f (θ ) and the curve g (θ ) = r = sin θ
+2: integrand
Area =
π
1
2
π
3
2
∫ sin θ dθ + 12
0
[
1 1
2 2
∫(
π
2
)
2
+1: limits, constant, answer
3 cos θ dθ =
3
π
π
θ − 14 sin 2θ ] 0 3 + 32 [12 θ + 14 sin 2θ ] π 2 =
3
5
3
π−
24
4