Math 212 Spring 2008: Solutions: HW #7
Instructor: S. Cautis
1. section 4.1, #2
so
√
v(t) = c′ (t) = (sin t + t cos t, cos t − t sin t, 3)
a(t) = v ′ (t) = (cos t + cos t − t sin t, − sin t − sin t − t cos t, 0)
√
Hence at t = 0 the velocity vector is v(0) = (0, 1, 3) and the acceleration
vector is a(0) = (2, 0, 0). The tangent line is
√
(0, 0, 0) + s(0, 1, 3)
2. section 4.1, #14
Let
g(t) = ||r(t)|| = r(t) · r(t)
At a local maximum or minimum we have g ′ (t) = 0. But
g ′ (t) = r′ (t) · r(t) + r(t) · r′ (t)
so we get r(t) · r′ (t) = 0 which means r(t) is perpendicular to r′ (t).
3. section 4.1, #18
Suppose c(t) = (c1 (t), c2 (t), c3 (t)). The fact it has zero acceleration means
c′′ (t) = 0 so that c′′i (t) = 0 for i = 1, 2, 3. Integrating c′′i (t) = 0 we get
c′i (t) = ai (some constant). Integrating again we get ci (t) = ai t + bi . Thus
c(t) = (a1 , a2 , a3 ) + t(b1 , b2 , b3 )
which is a straight line.
4. section 4.2, #8
c′ (t) = (1, sin t + t cos t, cos t − t sin t) so that
x21 +x22 +x23 = 1+(sin2 t+2t sin t cos t+t2 cos2 t)+(cos2 −2t sin t cos t+t2 sin2 t) = 2+t2
so that the length of the arc is
Z
π
0
p
2 + t2 dt
5. section 4.2, #12
(a) We are given that T (t) · T (t) = 1 so differentiating gives T ′ (t) · T (t) +
T (t) · T ′ (t) = 0 which means T ′ (t) · T (t) = 0.
(b) We have
T ′ (t) =
c′′ (t)
2c′ (t) · c(t)
′
−
c
(t)
||c′ (t)||
||c′ (t)||2
6. section 4.3, #16
We have
F(
1
3
1 t 1
, e , ) = (− 4 , et , − 2 )
t3
t
t
t
Meanwhile
1
3 t
,e ,− 2)
4
t
t
Since these are equal c(t) is a field line for F (x, y, z).
c′ (t) = (−
7. section 4.3, #18
To show h(t) = V (c(t)) is decreasing we just need to show that h′ (t) ≤ 0.
Now by the chain rule
h′ (t) = ∇V (c(t)) · c′ (t)
Since c(t) is a vector field for −∇V this means −∇V (c(t)) = c′ (t). Thus
h′ (t) = −c′ (t) · c′ (t) = −||c′ (t)||2 ≤ 0
8. section 4.4, #8
The flow lines tend towards the origin but not in a straight line. We have
∇ · F = −3 − 1 = −4
This is consistent with the description of the flow lines since both calculations suggest that the gas is contracting.
9. section 4.4, #14
∇×F = (∂y (xy) − ∂z (xz), −∂x (xy) + ∂z (yz), ∂x (xz) − ∂y (yz)) = (x−x, −y+y, z−z) = (0, 0, 0)
10. section 4.4, #26
If F were a gradient field then ∇ × F = 0. On the other hand,
∇×F = (∂y (0)−∂z (−2xy), −∂x (0)+∂z (x2 +y 2 ), ∂x (−2xy)−∂y (x2 +y 2 )) = (0, 0, −4y)
which is not zero!
11. section 4.4, #28
a) ∇ · (F + G) = ∇ · F + ∇ · G = 0 so F + G must have zero divergence.
b) Take F = (y, 0, 0) and G = (0, z, 0). Clearly both have zero divergence.
On the other hand, F × G = (0, 0, yz) which has divergence y. So this
example shows that F × G need not have zero divergence.
12. section 4.4, #32
a) We have
∇×F = (∂y (0)−∂z (x3 +y 3 ), −∂x (0)+∂z (3x2 y), ∂x (x3 +y 3 )−∂y (3x2 y)) = (0, 0, 0)
b) We want a function f (x, y, z) such that ∂z (f ) = 0 so that f should only
depend on x and y. We also want that ∂x (f ) = 3x2 y so that if we integrate
with respect to x we find that f = x3 y + φ(y) where φ(y) is any function
of y. Finally, we want ∂y (f ) = x3 + y 3 which means x3 + φ′ (y) = x3 + y 3
which means φ(y) = y 4 /4. So the function we seek is
f = x3 y + y 4 /4.
One can also do this by trial and error as the problem suggests. This
involves making an educated guess, testing it, and then refining your guess.