Homework 7 Solutions
Problem 1.
(a) A time t, the black line has z = sin t and θ = t. If we let r = u and t = v, we get
~
X(t)
= (u cos v, u sin v, sin v)
for u ∈ R and 0 < v < 2π.
(b) Observe that x2 + y 2 = u2 . Then y 2 = u2 sin2 t = (x2 + y 2 )z 2 , so the answer is
y 2 = (x2 + y 2 )z 2
Problem 2.
At time t, the circle has center (0, t, 0), and it lies in a plane parallel to the vectors
(cos t, 0, sin t) and (0, 1, 0). If we let u = t and v be the angle along the circle, then
~
X(u,
v) = (0, u, 0) + (cos v)(cos u, 0, sin u) + (sin v)(0, 1, 0)
where u ∈ R (though only 0 < u < π is shown in the animation) and 0 < v < 2π.
Problem 3.
We simply draw a line segment between ~x(t) and ~x(t + π) for every value of t. Note that
~x(t + π) = (2 sin 2t + sin t, 2 cos 2t − cos t, − sin 3t).
If we let u = t and v be the coordinate along the line segment, we get the parameterization
~
X(u,
v) = (1 − v)~x(u) + v~x(u + π)
which simplifies to
~
X(u,
v) = (2 sin 2u, 2 cos 2u, 0) + (1 − 2v)(− sin u, cos u, sin 3u)
The line segment is in the same place when u = 0 and u = π, and each line segment uses
only values of v between 0 and 1, so we need 0 < u < π and 0 < v < 1 .
Problem 4.
Let ~x(t) = (t, t2 , 0) be a point on the parabola. Then ~x(t) will travel along a circle when
rotated, whose center is the projection ~c(t) of ~x(t) onto the line. The normal vector to the
1
line on the xy-plane is ~n = √ (1, −1, 0), so the distance from ~x(t) to the line is
2
r(t) =
1
t2 − t + 1
√
(1, 0, 0) − ~x(t) · ~n(t) = √ (1 − t, −t2 , 0) · (1, −1, 0) =
2
2
This will be the radius of the circle. The center is
~c(t) = ~x(t) + r(t)~n = (t, t2 , 0) +
t2 − t + 1
1
(1, −1, 0) = (t2 + t + 1, t2 + t − 1, 0).
2
2
Note that the circle is in a plane parallel to ~n and (0, 0, 1). If we let u = t and v be the angle
along the circle, then the parameterization is
2
cos
v
1
u
−
u
+
1
2
2
~
√
√ (1, −1, 0) + sin v (0, 0, 1)
X(u,
v) = (u + u + 1, u + u − 1, 0) +
2
2
2
where u ∈ R and 0 < v < 2π.