Homework 13 Solutions
Math 21a
Spring, 2014
Due Friday, March 7th (MWF) or Tuesday, March 11th (TTh)
1. (Stewart 10.1 #26 ) Show that the curve with parametric equations x = sin t, y = cos t, z = sin2 t is the curve of
intersection of the surfaces z = x2 and x2 + y 2 = 1. Use this fact to help sketch the curve.
Solution: Here we have x2 = sin2 t = z and x2 + y 2 = sin2 t + cos2 t = 1, so the curve is contained in the intersection
of the parabolic cylinder z = x2 with the circular cylinder x2 + y 2 = 1. We get the complete intersection for 0 ≤ t ≤ 2π.
2. In this problem, we’ll use Mathematica to tackle (Stewart 10.1 #37 ).
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(a) Plot the cone z = x2 + y 2 by using the following Mathematica command:
cone = Plot3D[Sqrt[x^2+y^2], {x,-5,5}, {y,-5,5}]
Solution:
The output of Mathematica for the first three parts of this question are all shown in Figure 1
Figure 1: Mathematica output for Problem 2
(b) Plot the plane z = 1 + y by using the command:
plane = Plot3D[1+y, {x,-5,5}, {y,-5,5}, ColorFunction -> "RustTones"]
Solution:
See Figure 1.
(c) Plot the two together by using the command:
Show[cone, plane]
You should now be able to click and drag on the plot to see the curve of intersection. Try this.
Solution:
See Figure 1.
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(d) (Stewart 10.1 #37 ) Find a vector function that represents the curve of intersection of the cone z = x2 + y 2 and
the plane z = 1 + y.
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Solution: Both equations are solved for z, so we can substitute to eliminate z: x2 + y 2 = 1 + y ⇒ x2 + y 2 =
1
(1 + y)2 = 1 + 2y + y 2 ⇒ x2 = 1 + 2y ⇒ y = (x2 − 1). We can form parametric equations for the curve C of
2
1
1
1
intersection by choosing a parameter x = t, then y = (t2 − 1) and z = 1 + y = 1 + (t2 − 1) = (t2 + 1). Thus a
2
2
2
vector function representing C is
1
1
1
1
or
r(t) = t, (t2 − 1), (t2 + 1) .
r(t) = ti + (t2 − 1)j + (t2 + 1)k
2
2
2
2
(e) Plot the vector function you found in part (d) by using the command:
curve = ParametricPlot3D[{t,t^2,t^3}, {t,-5,5}, PlotStyle -> {Thickness[0.02]}]
Note: You will need to replace the vector function r(t) = ht, t2 , t3 i in this command by the vector function you
found in part (d)! Depending on the parameterization you choose, you may also need to adjust the t-interval.
Solution:
In Figure 2 we show the answer from (c) as well as the answers to (e) and (f).
Figure 2: More Mathematica output for Problem 2
(f) Plot the curve together with the two surfaces by using the command:
Show[cone, plane, curve]
Solution:
See Figure 2.
3. (Stewart 10.1 #44 ) Two particles travel along the space curves
r1 (t) = ht, t2 , t3 i
r2 (t) = h1 + 2t, 1 + 6t, 1 + 14ti
Do the particles collide? Do their paths intersect?
Solution: The particles collide if r1 (t) = r2 (t), implying that ht, t2 , t3 i = h1 + 2t, 1 + 6t, 1 + 14ti. Equating components
gives t = 1 + 2t, t2 + 1 + 6t, and t3 = 1 + 14t. The first equation gives t = −1, but this does not satisfy the other
equations, so we can determine the particles do not collide.
In order for the paths to intersect, we need to find a value of t and a value for s such that r1 (t) = r2 (s), which implies
that at some point the paths collide. Thus, ht, t2 , t3 i = h1 + 2s, 1 + 6s, 1 + 14si. Equating components gives t = 1 + 2s,
t2 = 1 + 6s, and t3 = 1 + 14s. Substituting the first equation into the second gives (1 + 2s)2 = 1 + 6s ⇒ 2s(s − 1) =
0 ⇒ s = 0 or s = 21 . From the first equation, s = 0 ⇒ t = 1 and s = 12 ⇒ t = 2. Checking, we see that both pairs of
values satisfy the third equation. Thus the paths intersect twice, once at (1, 1, 1), when s = 0 and t = 1, and at (2, 4, 8)
when s = 21 and t = 2.
4. (Stewart 10.2 #29 ) Find parametric equations for the tangent line to the curve with parametric equations
x = t cos t, y = t, z = t sin t
at the point (−π, π, 0).
Solution: Differentiating r(t) = ht cos t, t, t sin ti, we find r0 (t) = hcos t − t sin t, 1, t cos t + sin ti.
This curve passes through the point (−π, π, 0) at t = π, at which point the tangent vector is r0 (π) = h−1, 1, −πi. Thus
parametric equations of the tangent line are
x = −π − t
or
y =π+t
r(t) = h−π − t, π + t, −πti .
z = −πt
5. (Based on Stewart 10.4 #14 ) Find the velocity and position vectors of a particle that has acceleration given by
a(t) = 2i + 6tj + 12t2 k,
and initial velocity and position given by
v(0) = i
Solution:
and
r(0) = j − k.
We integrate acceleration a(t) = 2i + 6tj + 12t2 k to get velocity:
Z
v(t) = (2i + 6tj + 12t2 k) dt = 2ti + 3t2 j + 4t3 k + C
The arbitrary constant C can be found by the initial condition i = v(0) = C, so C = i and v(t) = (2t + 1)i + 3t2 j + 4t3 k.
Similarly we integrate velocity v(t) to find position:
Z r(t) =
(2t + 1)i + 3t2 j + 4t3 k dt = (t2 + t)i + t3 j + t4 k + D.
The initial condition is j − k = r(0) = D, so D = j − k and r(t) = (t2 + t)i + (t3 + 1)j + (t4 − 1)k.