Math 214 — Solutions to Assignment #5
11.1
10. Let x = t2 , y = t3 .
(a) Sketch the curve by using the parametric equations to plot points. Indicate
with an arrow the direction in which the curve is traced as t increases.
(b) Eliminate the parameter to find a Cartesian equation of the curve.
Solution.
(a)
t
x
y
−3 −2 −1 0 1 2 3
9
4
1 0 1 4 9
−27 −8 −1 0 1 8 27
y
(4,8)
0
t=2
t=1
t=−1
(4,−8)
x
t=−2
√
(b) t = ± x =⇒ y = ±x3/2 , x ≥ 0, y ∈ R. Or x = y 2/3 , x ≥ 0, y ∈ R.
For #12 and 16
(a) Eliminate the parameter to find a Cartesian equation of the curve.
(b) Sketch the curve and indicate with an arrow the direction in which the curve
is traced as the parameter increases.
12. x = 4 cos θ, y = 5 sin θ, −π/2 ≤ θ ≤ π/2.
1
Solution.
y2
x2
+
=1
(a) Since cos2 θ + sin2 θ = 1, we get (x/4)2 + (y/5)2 = 1, i.e.,
16
25
which is an ellipse with the x-intercepts x = ±4, the y-intercepts y = ±5. But since
−π/2 ≤ θ ≤ π/2, we have 0 ≤ cos θ ≤ 1 so the graph consists of only the portion
on the right side of the y-axis.
(b)
y
5 θ= π 2
4
0
x
θ =0
−5 θ =− π 2
16. x = ln t, y =
√
t, t ≥ 1.
Solution.
(a) x = ln t =⇒ t = ex =⇒ y =
(b)
√
ex = ex/2 , x ≥ 0.
y
1
0
x
22. Describe the motion of a particle with position (x, y) as t varies in the given
interval:
x = cos2 t, t = cos t, 0 ≤ t ≤ 4π.
Solution. x = y 2 is a parabola opening to the right with the vertex (0, 0).
The particle starts at the point (1, 1) (t = 0). It then moves along the parabola to
2
the point (0, 0) (t = π/2) down to (1, −1) (t = π), then back to (0, 0) (t = 3π/2)
and to (1, 1) (t = 2π). The same motion is repeated from 2π to 4π.
y
(1,1)
x
(1,−1)
11.2
8. Find an equation of the tangent to the curve x = tan θ, y = sec θ at the point
√
(1, 2) by 2 methods: (a) without eliminating the parameter and (b) by first
eliminating the parameter.
Solution. (a)
dy
dy/dθ
sec θ tan θ
tan θ
=
=
=
= sin θ,
dx
dx/dθ
sec2 θ
sec θ
√
√
π
at (1, 2), 1 = tan θ,
2 = sec θ =⇒ θ =
(or π/4 + 2nπ),
4
(since tan θ > 0 in quadrant I and III while sec θ > 0 in quad. I and IV). Hence the
√
slope of the tangent at (1, 2) is
√
π
1 or 2
0 π
y
= sin = √ =
4
4
2
2
and the equation is
y−
√
√
2
2=
(x − 1).
2
(b) Since tan2 θ + 1 = sec2 θ =⇒ x2 + 1 = y 2 , differentiating implicitly we
obtain
√
x
1
dy
=
=⇒ y 0 (1, 2) = √
2x = 2yy 0 =⇒
dx
y
2
which gives the same equation as in part (a).
16. Find dy/dx and d2 y/dx2 if x = cos 2t, y = cos t, 0 < t < π. For which values
of t is the curve concave upward?
3
Solution.
dy
dy/dt
− sin t
sin t
1
1
=
=
=
=
= sec t,
dx
dx/dt
−2 sin 2t
4 sin t cos t
4 cos t
4
d dy
1
( )
sec t tan t
d2 y
1
sin t
dt dx
4
=
=
=− ·
2
2
dx
dx/dt
−2 sin 2t
16 cos t sin t cos t
1
1
or
=−
= − sec3 t.
3
16 cos t
16
The curve is concave upward if y 00 (x) > 0 on 0 < t < π, i.e.,
1
− sec3 t > 0 ⇐⇒ sec t < 0 ⇐⇒ cos t < 0 ⇐⇒ t ∈ (π/2, π).
16
18. Find the points on the curve x = 2t3 + 3t2 − 12t, y = 2t3 + 3t2 + 1 where the
tangent is horizontal or vertical.
Solution. We find the derivative
dy
6t2 + 6t
6t(t + 1)
t(t + 1)
= 2
=
=
.
dx
6t + 6t − 12
6(t + 2)(t − 1)
(t + 2)(t − 1)
Horizontal tangents occur when y 0 = 0, i.e., t = 0, −1. So the points are
t = 0 =⇒ x = 0, y = 1, i.e., (0, 1),
t = −1 =⇒ x = −2 + 3 + 12 = 13, y = −2 + 3 + 1 = 2, i.e., (13, 2).
Vertical tangents occur at t = −2, 1 (y 0 does not exist there). Then the points are:
t = −2 =⇒ x = −16 + 12 + 24 = 20, y = −16 + 12 + 1 = −3, i.e., (20, −3),
t = 1 =⇒ x = 2 + 3 − 12 = −7, y = 2 + 3 + 1 = 6, i.e., (−7, 6).
34. Find the area of the region enclosed by the astroid x = a cos3 θ, y = a sin3 θ.
Solution. The graph of the astroid is
y
a
−a
0
−a
4
a
x
Using symmetry
A=4
Z
a
y dx = 4
0
= −12a
2
Z
0
π
2
Z
0
a sin3 θ(−3a cos2 θ sin θ) dθ
π
2
4
2
sin θ cos θ dθ = −12a
2
Z
Z
0
π
2
1
1
(1 − cos 2θ)2 · (1 + cos 2θ) dθ
4
2
0
3
(1 − 2 cos 2θ + cos2 2θ)(1 + cos 2θ) dθ
= − a2
π
2
2
Z π2 3 2
1
1
3
= a
− cos 2θ − cos 4θ + cos 2θ dθ
2
2
2
0
π2 Z π
2
1
1
3 2 1
2
θ − sin 2θ − sin 4θ +
cos
2θ} cos 2θ dθ
= a
{z
|
2
2
2
8
0
0
(1−sin2 2θ)
3 2
a
2
3
= a2 ·
2
=
1 π
1
· −0−0 +
2 2
2
π
3
= a2 π.
4
8
Z
0
0
(1 − u2 ) du (u = sin 2θ)
44. Find the length of the curve x = et + e−t , y = 5 − 2t, 0 ≤ t ≤ 3.
Solution. dx/dt = et − e−t and dy/dt = −2. So
dx
dt
2
+
dy
dt
2
∴
= (et − e−t )2 + 4 = e2t − 2 + e−2t + 4
= e2t + 2 + e−2t = (et + e−t )2
s
Z 3 2 2
Z 3
dy
dx
(et + e−t ) dt
L=
+
dt =
dt
dt
0
0
3
= et − e−t = e3 − e−3 − 1 + 1 = e3 − e−3 .
0
60. Find the area of the surface obtained by rotating the curve x = 3t−t3 , y = 3t2 ,
0 ≤ t ≤ 1 about the x-axis.
Solution.
S = 2π
Z
b
y ds = 2π
a
Z
1
3t
0
5
2
s
dx
dt
2
+
dy
dt
2
dt
where
dx
dt
2
+
dx
dt
dy
dt
dy
dt
2
2
2
= (3 − 3t2 )2 = 9 − 18t2 + 9t4
= (6t)2 = 36t2
= 9 − 18t2 + 9t4 + 36t2 = 9 + 18t2 + 9t4 = (3 + 3t2 )2
∴ S = 2π
Z
= 18π
1
2
2
3t (3 + 3t ) dt = 2π
0
Z
1
1
9t2 (1 + t2 ) dt
0
3
1
1 1
48π
t
t5 =
(t + t ) dt = 18π
+ = 18π
+
.
3
5 0
3 5
5
2
0
Z
4
66. Find the surface area generated by rotating the given curve about the y-axis.
x = et − t, y = 4tt/2 , 0 ≤ t ≤ 1
Solution. S = 2π
∴
dx
dt
2
+
dx
dt
dy
dt
∴
2
2
R1
x ds where ds =
0
t
2
2t
p
(dx/dt)2 + (dy/dt)2 dt.
t
= (e − 1) = e − 2e + 1,
dy
dt
2
= (2et/2 )2 = 4et
= e2t − 2et + 1 + 4et = e2t + 2et + 1 = (et + 1)2
S = 2π
= 2π
Z
Z
1
(et − t)(et + 1) dt
0
1
0
(e2t + et − tet − t) dt
1
t2 1 2t
t
t
t
e + e − (te − e ) − = 2π
2
2 0
1 1
1 2
e + e − e + e − − − 1 − 1 = π(e2 + 2e − 6).
= 2π
2
2 2
(Here to do
R
tet dt, integration by parts was used.)
6