Math 241, Quiz 7. 10/10/11.
Name:
• Read problems carefully. Show all work. No notes, calculator, or text.
• There are 15 points total.
1. §14.5, #5 (5 points): Use the chain rule to find
w = xey/z ,
x = t2 ,
dw
when
dt
y = 1 − t,
z = 1 + 2t.
You do not have to convert x, y, z back to t in your final answer; your final answer can be in
terms of x, y, z, and t.
Solution: The chain rule in this setting has the form
∂w dx ∂w dy ∂w dz
dw
=
+
+
.
dt
∂x dt
∂y dt
∂z dt
We compute:
∂w
= ey/z ,
∂x
dx
= 2t,
dt
∂w
x
= ey/z ,
∂y
z
dy
= −1,
dt
∂w
xy
= − 2 ey/z ,
∂z
z
dz
= 2.
dt
Hence, we have
dw
x y/z
xy y/z
x 2xy
y/z
y/z
= e (2t) + e (−1) − 2 e (2) = e
2t − − 2 .
dt
z
z
z
z
√
2. §14.6, #11 (5 points): Find the directional derivative of f (x, y) = 1 + 2x y at the point
(3, 4) in the direction of ~v = h4, −3i.
Solution: We compute
√
√
∇f (x, y) = h2 y, 2x(1/2)y −1/2 i = h2 y, xy −1/2 i =⇒ ∇f (3, 4) = h4, 3/2i.
A unit vector in the direction of ~v is ~uh4/5, −3/5i. Hence, we have
D~u f (3, 4) = h4, 3/2i · h4/5, −3/5i = 16/5 − 9/10 = 23/10.
3. §14.6, #29 (5 points): Find all points at which the direction of fastest change of
f (x, y) = x2 + y 2 − 2x − 4y is ~i + ~j.
Solution: The direction of fastest change is ∇f (x, y) = (2x − 2)~i + (2y − 4)~j. It suffices to
find all points (x, y) with ∇f (x, y) k ~i + ~j. We have
∇f (x, y) k ~i + ~j ⇐⇒ ∃k with 2x − 2 = k, 2y − 4 = k
2x − 2 = 2y − 4 ⇐⇒ x − 1 = y − 2 ⇐⇒ y = x + 1.
The direction of fastest change is ~i + ~j for all points (x, y) on y = x + 1.