Solutions to Selected
Homework
Week of 5/13/02
§14.3, 12. (a) Find the unit tangent and unit normal vectors T(t) and N(t). (b) Use
formula 9 to find the curvature.
r(t) = ht2 , sin t − t cos t, cos t + t sin ti, t > 0
Solution: (a) We have
r0 (t) = h2t, cos t + t sin t − cos t, − sin t + sin t + t cos ti
= h2t, t sin t, t cos ti.
Thus
||r0 (t)|| =
p
4t2 + t2 sin2 t + t2 cos2 t =
which gives us
1
T(t) = √ h2t, t sin t, t cos ti =
5t
¿
√
5t2 =
√
2 sin t cos t
√ , √ , √
5
5
5
5t,
À
.
Clearly then
√
√
T0 (t) = h0, cos t/ 5, − sin t/ 5i,
√
and so ||T0 (t)|| = 1/ 5. Therefore
√
N(t) = 5 · T0 (t) = h0, cos t, − sin ti.
(b) By formula 9,
||T0 (t)||
||r0 (t)||
√
1/ 5
1
= √
= t.
5
5t
κ(t) =
§14.4, 10. Find the velocity, acceleration and speed of a particle with the position
function
r(t) = h2 cos t, 3t, 2 sin ti.
Solution: We have
v(t) = r0 (t) = h−2 sin t, 3, 2 cos ti
a(t) = v0 (t) = h−2 cos t, 0, −2 sin ti
p
√
v(t) = ||v(t)|| = 4 sin2 t + 9 + 4 cos2 t = 13.
1
2
§15.1, 30. Match the function with its graph. Give reasons for your choices.
Solution: (a) f (x, y) = |x| + |y|. The dead giveaway on this one is that
it only touches the xy-plane at the origin. The only graph that has this
property is (VI):
4
3
2
1
0
–2
–2
–1
–1
y
0
0
x
1
1
2
2
(b) f (x, y) = |xy|. Here, we notice that the function is zero on both of the
lines x = 0, y = 0, so that the graph must touch the x and y axes. The
graph is (V):
4
3
2
1
0
–2
–2
–1
–1
y
0
0
1
1
2
2
x
3
(c) f (x, y) = 1/(1 + x2 + y 2 ). Notice that at (x, y) = (0, 0), the function
has value 1, and that as x, y → ±∞, f (x, y) → 0. Hence the graph is (I):
1
0.8
0.6
0.4
0.2
–2
–2
–1
–1
y
0
0
x
1
1
2
2
(d) f (x, y) = (x2 − y 2 )2 . Here, f equals zero when x2 − y 2 = 0, or equivalently when x = ±y. Thus the function has zeros along the lines x = ±y
in the plane. Graph (IV) has this property:
16
14
12
10
8
6
4
2
0
–2
–2
–1
–1
y
0
0
1
1
2
2
x
4
(e) f (x, y) = (x − y)2 . This function has zeros along the line (x = y) in the
plane, the graph with this property is (II):
16
14
12
10
8
6
4
2
–2
–1
x
0
1
2
–2
–1
1
0
2
y
(f) f (x, y) = sin(|x| + |y|). This is the only periodic function listed, so the
graph is (III):
1.5
1
0.5
0
–0.5
–1
–10
–5
x
0
5
10
–10
–5
5
0
y
10
5
§15.1, 38. Draw a contour map of the function showing several level curves
x+y
f (x, y) =
.
x−y
Solution: The level curves of the function are given by f (x, y) = k for
various k. Thus we have (1 + k)y = (k − 1)x. For k = −3, −2, −1, 0, 1, 2, 3
we get y = 2x, y = 3x, x = 0, y = −x, y = 0, y = x/3, and y = x/2
respectively. Here is a contour map
15
10
5
–2
–4
2
4
x
–5
–10
–15
The actual graph is
40
30
20
10
0
–10
–20
–30
–40
–4
–2
y
0
–4
–2
2
0
4
2
4
x