Math 231E. Homework 15. Solutions.
Problem 1. In each of the following cases, you are given a parametrically-defined pair of
equations. First, plot the curve in the (x, y)-plane; make sure you identify which direction
the curve is traversed! Then, determine whether it is possible to algebraically eliminate
the parametric variable to obtain a relation between x and y; if it is, write down this
relationship, and if it is not, say why it is not.
a. x = t2 + 1,
y = t2 − 1,
−2 ≤ t ≤ 3.
b. x = cos(t),
y = sin(t),
0 ≤ t ≤ 2π.
2
c. x = cos (t),
2π.
2
y = sin (t),
d. x = cos(2t),
e. x = e2t ,
0 ≤ t ≤
f. x =
y = sin(t),
y = et − 1,
√
e2t − 1,
y=
√
0 ≤ t ≤ 2π.
t ∈ (−∞, ∞)
e2t + 1,
t≥0
Solution:
a. If we subtract the two equations, we see the t2 term cancels, and we have
x − y = 2.
This means that x and y live on a line, in fact the line y = x − 2. However, notice that parts
of the curve get traversed twice. Note that the range of x values is [1, 10] and the range of y
values is [−1, 8], but the starting location is (5, 3). So in fact the curve moves from (5, 3),
down to (1, −1), and then back up to (10, 8).
y
8
6
4
2
2
4
6
8
10
x
b. We discussed this one in lecture. Using the well-known trig identity
sin2 t + cos2 t = 1,
gives x2 + y 2 = 1, so we have the unit circle.
y
1.0
0.5
-1.0
0.5
-0.5
1.0
x
-0.5
-1.0
c. Using the trig identity from part (b), we then have x + y = 1, so the plot lies on the line
y = 1 − x. The range of x (and y) is [0, 1], and the line is actually covered four times, as the
plot goes back and forth twice.
y
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
x
d. Here we have to be a little clever with trig identities. Recall that
cos(2t) = cos2 t − sin2 t = (1 − sin2 t) − sin2 t = 1 − 2 sin2 t,
so we have
x = 1 − y2.
This gives a parabola, opening to the left. It actually starts at (1, 0) when t = 0. Then at
t = π/2, it has gone to (−1, 1), then at t = π it has returned to (1, 0). For t ∈ [π, 2π] it
double-covers the lower branch of the parabola.
y
1.0
0.5
-1.0
0.5
-0.5
1.0
x
-0.5
-1.0
e. If we solve for t in the y equation we obtain
t = ln(y + 1).
Plugging this into the x equation gives
2
x = e2 ln(y+1) = eln(y+1) = (y + 1)2 .
This is the equation for a parabola that opens to the right and has apex at (x, y) = (0, −1).
However, notice that y = et − 1 is always at least −1, so that it is only the top branch of the
parabola that we hit.
y
1.5
1.0
0.5
0.5 1.0 1.5 2.0 2.5 3.0 3.5
x
-0.5
-1.0
f. Noting that
x2 = e2t − 1,
y 2 = e2t + 1,
we have
y 2 − x2 = 2.
(1)
√
This is a graph of a hyperbola that intersects the y-axis at (0, 2). Notice, however, that
x, y ≥ 0 by definition, and so we only obtain the right half of the top branch of the hyperbola:
y
6
4
2
2
4
6
x
Problem 2. For each of the following parametric systems, determine dy/dx and d2 y/dx2 .
Determine where the function is increasing and decreasing. Determine where the function is concave up or concave down. Identify all places where the first and/or second
derivatives do not exist.
a. x = t2 + 1, y = t
d. x = 2 sin(θ), y = 5 cos(θ)
b. x = s2 + 1, y = s2 − 1
e. x = 2 sin(t), y = 5 sin(t)
c. x = cos(t), y = sin(t)
f. x = ecos(t) , y = esin(t)
Solution:
a. We have
dy
dy/dt
1
1
=
=
= .
dx
dx/dt
2t
2y
Here we see that there is a vertical tangency at t = y = 0, so at the point (1, 0). The curve
is increasing when y is positive and decreasing when y is negative. Moreover, we have
d dy
d 1
dy
1
1
= dt dx = dt 2t = − 3 = − 3 .
2
dx
dx
2t
2t
2y
dt
2
We see that the second derivative has the opposite sign as that of y: therefore it is concave
down for y > 0 and concave up for y < 0:
y
3
2
1
-1
-2
-3
2
4
6
8
10
Also notice that x = y 2 + 1 so this is a parabola that opens right.
x
b. We have
dy
2s
=
= 1,
dx
2s
so the slope is always 1. The second derivative is thus always zero. See [1a].
c. We have
dy
cos t
x
=
=− .
dx
− sin t
y
This will have a horizontal tangency whenever x = 0 and a vertical one when y = 0. We
also have
d
d dy
(− cot t)
d2 y
csc2 t
1
1
dt dx = dt
=
= − 3 = − 3.
=
2
dx
dx
− sin t
− sin t
y
sin t
dt
2
2
Thus the sign of d y/dx is the opposite as the sign of y: the graph is concave down for y > 0
and concave up for y < 0. See [1b].
d. We have
5 sin θ
5x
dy
=
=− .
dx
−2 cos θ
2y
This will have a horizontal tangency whenever x = 0 and a vertical one when y = 0. Recall
[2c], and we will have
d2 y
5 1
= − 3.
2
dx
2y
Thus the sign of d2 y/dx2 is the opposite as the sign of y: the graph is concave down for y > 0
and concave up for y < 0. In fact, this will give the graph of an ellipse centered at the origin:
for example, we see that
5x2 + 2y 2 = 10 sin2 θ + 10 cos2 θ = 10,
which is the equation of an ellipse.
y
4
2
-2 -1
-2
-4
1 2
x
e. This is the same as [2d] but with a different independent variable.
f. We have
dy
cos t · esin t
y ln x
=
.
=
cos
t
dx
− sin t · e
x ln y
This has a horizontal tangency whenever x = 1 or y = 0, but notice that y > e−1 so is never
zero. So there is a horizontal tangency whenever x = 1.
This has a vertical tangency whenever x = 0 or y = 1, but notice that x > e−1 so is never
zero. So there is a vertical tangency whenever y = 1.
We also have
d dy
d2 y
dt dx = esin(t)−2 cos(t) cot2 (t) + cot(t) − csc3 (t) ,
=
dx
dx2
dt
which is fun. Behold!
y
2.5
2.0
1.5
1.0
0.5
0.5 1.0 1.5 2.0 2.5
x
Problem 3. In each of the following problems, you are given the description of a curve
in either Cartesian coordinates (x, y) or polar coordinates (r, θ). Whichever one you are
given, convert your equation to the other form (i.e. if it is in Cartesian, put it in polar, and
vice versa). In all cases, sketch the curve.
a. y = 2
d. θ = π/2
b. r = 2 sin θ
e. r = 3 sin2 θ
c. y = 2x
f. xy = 6
Solution:
a. Writing y = r sin θ we obtain the equation
r sin θ = 2.
b. Writing y = r sin θ, we have
2y
,
r
or r2 = 2y, which we can then write as x2 + y 2 − 2y = 0.
r=
c. We write
r sin θ = 2r cos θ,
or
tan θ = 2.
d. This is the positive y-axis.
e. Write
3y 2
= 3 sin2 θ,
r2
so this is r3 = 3y 2 , or (x2 + y 2 )3/2 − 3y 2 = 0.
(2)
f. We write
(r cos θ)(r sin θ) = 6,
or
r2 cos θ sin θ = 6.
We can simplify this a bit by using the trig identity sin(2θ) = 2 sin θ cos θ, giving
r2 sin 2θ = 3.
Problem 4. In each case, you are given an equation in polar coordinates. Sketch the curve,
then compute the area contained inside the curve.
a. r = cos θ
c. r = 3 + 4 sin θ
b. r = 2 sin θ
d. r = 1 − 5 cos(2θ)
Problem 5. Find an interesting parametric or polar curve formula from the Internet. Put
its equation and its plot here.
Solution:
From the internet:
Problem 6. In each of the following cases, you will be asked to write down a family of
parametric curves that have the property that at t = 1, we have x0 (t) = y 0 (t) = 0, but the
slope of the curve has the property listed here:
a. Horizontal tangent.
c. A slope of 4.2.
b. Vertical tangent.
d. Slope of the curve is undefined.
Solution:
a. x = (t − 1)31 , y = (t − 1)33 . It is clear that x0 (1) = y 0 (1) = 0. Moreover, if we compute
dy
33(t − 1)32
33
=
= (t − 1)2 .
30
dx
31(t − 1)
31
At t = 1, this is zero!
b. x = t33 , y = t31 . Consider part [a] with x, y flipped.
c. x = t3 ,
d. Cycloid!
y = 4.2t3 . Here we have y = 4.2x so the slope is 4.2.