Math 53 Worksheet Solutions- Vector Functions and Derivatives
1. At what points does the helix r(t) = hsin t, cos t, ti intersect the sphere
x2 + y 2 + z 2 = 5?
Solution. We have
sin2 t + cos2 t + t2 = 5,
so t2 = 4 and t = ±2 are where we have intersection. The points are then
P1 = r(2) = hsin 2, cos 2, 2i,
P2 = r(−2) = h− sin 2, cos 2, −2i.
2. Find a vector function that represents the curve of intersection of the paraboloid
z = 4x2 + y 2 and the parabolic cylinder y = x2 .
Solution. We have r(t) = hx(t), y(t), z(t)i. Set x(t) = t. Then y(t) = x(t)2 = t2 . And
z(t) = 4x(t)2 + y(t)2 = 4t2 + t4 . Then
r(t) = ht, t2 , 4t2 + t4 i.
3. Find parametric equations for the tangent line to the curve
√
x = ln t, y = 2 t, z = t2 ,
at the point (0, 2, 1).
√
Solution. The position vector is r(t) = hln t, 2 t, t2 i, so the tangent vector is
r0 (t) = h 1t , √1t , 2ti. The point (0, 2, 1) corresponds to t = 1, since x(t) = ln t = 0 implies
that t = 1. Then the tangent vector at the point (0, 2, 1) is
v = r0 (1) = h1, 1, 2i.
So the parametric equation of the line is
r(t) = h0, 2, 1i + th1, 1, 2i = ht, 2 + t, 1 + 2ti.
4. If u(t) = r(t) · [r0 (t) × r00 (t)], show that
u0 (t) = r(t) · [r0 (t) × r000 (t)].
Solution. Laws of derivatives for cross products and dot products. We have
d
d
(r(t) · [r0 (t) × r00 (t)]) = r0 (t) · [r0 (t) × r00 (t)] + r(t) · [r0 (t) × r00 (t)]
dt
dt
00
00
0
= 0 + r(t) · (r (t) × r (t) + r (t) × r000 (t))
= r(t) · (r00 (t) × r00 (t)) + r(t) · (r0 (t) × r000 (t))
= r(t) · (r0 (t) × r000 (t))
1
5. Reparametrize the curve
r(t) = e2t cos 2ti + 2j + e2t sin 2tk
with respect to arc length measured from the point where t = 0 in the direction of
increasing t.
Solution. First we find s(t), the arc length of the curve between t = 0 and t = t. We
calculate
r0 (t) = h2e2t cos 2t − 2e2t sin 2t, 0, 2e2t sin 2t + 2e2t cos 2ti.
Then
Z t
s(t) =
|r0 (u)| du
Z0 t q
=
(2e2u cos 2u − 2e2u sin 2u)2 + (2e2u sin 2u + 2e2u cos 2u)2 du
0
Z t q
e2u (cos 2u − sin 2u)2 + (sin 2u + cos 2u)2 du
=2
Z0 t p
=2
e2u cos2 2u + sin2 2u − 2 cos 2u sin 2u + sin2 2u + cos2 2u + 2 sin 2u cos 2u du
Z0 t √
=2
e2u 2 du
√ 0 2u u=t
= 2 e u=0
√
= 2 e2t − 1 .
Now we want to invert this to find t = t(s). So we solve for t. Thus
s
√ + 1 = e2t ,
2
and then
1
s
t = ln √ + 1 .
2
2
Substitute this in to r(t) and call the new parametrization r0 (s). We have
s
s
s
s
r0 (s) = r(t(s)) = h √ + 1 cos ln √ + 1 , 2, √ + 1 sin ln √ + 1 i,
2
2
2
2
after reducing with basic logarithm rules.
6. What force is required so that a particle of mass m has the position function
r(t) = t3 i + t2 j + t3 k?
Solution. We know F = ma.
a(t) = r00 (t) = h6t, 2, 6ti,
so
F(t) = ma(t) = h6mt, 2m, 6mti.
2
7. (Challenge Problem) A cable has radius r and length L and is wound around a spool
with radius R without overlapping. What is the shortest length along the spool that is
covered by the cable?
Solution. We look at points in the center of the bottom of the cable. These points
form a helix. In fact, the parametrization can readily be determined from a simple
picture. It is
t
,
x = (R + r) cos t, y = (R + r) sin t, z = h
2π
where R + r comes from the distance from the center of the cable to the point at the
center of the bottom of the cable, which is one large radius and one small radius. Here
h is the vertical distance between consecutive loops; as the cable makes one loop in 2π
units of t, h is the distance traveled upwards in one loop. As soon as we find h, we can
find the arc length.
Draw a picture. The width of the small cable is 2r. The height change of one full
rotation in h. That same height change can be achieved by moving 2π(r + R) units
ahead on the cable. Drawing these distances as straight lines that create triangles, a
similar triangle argument produces
√
2r
h2 − 4r2
=
,
2π(r + R)
h
and so cross-multiplying and solving for h gives us
2π(r + R)r
h= p
.
π 2 (r + R)2 − r2
Now, the length of one cycle, `, around the main axis is
Z 2π p
`=
x0 (t)2 + y 0 (t)2 + z 0 (t)2 dt
0
s
2
Z 2π
h
(R + r)2 +
dt
=
2π
0
s
(r + R)2 r2
= 2π (R + r)2 + 2
π (r + R)2 − r2
s
π 2 (r + R)2 − r2 + r2
= 2π(R + r)
π 2 (r + R)2 − r2
2π 2 (R + r)2
.
=p
π 2 (r + R)2 − r2
3
The number of complete cycles is n = floor(L/`), where the floor function gives the
largest integer less than the given number, in this case L/`. The shortest length is
then along the length before the final cycle closes, and this length is
!
!
p
L π 2 (r + R)2 − r2
2π(r + R)r
· floor
.
total length = hn = p
2π 2 (R + r)2
π 2 (r + R)2 − r2
4