Quiz #5
Answers
February 26, 2015
1. Use the chain rule to find zs when s = 1 and t = 0 for
z = f (x, y) = x2 + 3xy 3 where x = 2s + cos ∗t) and y = 4t + e2s−2
fx = 2x + 3y 3 , fy = 9xy 2 , xs = 2, ys = 2e2s−2 .
For s = 1 and t = 0, x = 3 , y = 1, xs = 2, ys = 2, fx = 9, and fy = 27
zs = fx ∗ xs + fy ∗ ys = 9 ∗ 2 + 27 ∗ 2 = 72
2. Use f (x, y) = x2 y 3 + 4yx3 the point P (1, 2) and v = h1, 5i to answer the following.
(a) Find ∇f (x, y)
∇f (x, y) = 2xy 3 + 12yx2 , 3x2 y 2 + 4x3
(b) This problem should have said Compute Du f (1, 2)
1
5
1
v= √ ,√
∇f (1, 2) = h40, 16i and u =
|v|
26
26
5
40
80
120
1
=√ +√ =√
Du f (1, 2) = h40, 16i · √ , √
26
26
26
26
26
(c) Find the maximum rate of change of the function f at point P .
|∇f (1, 2)| = | h40, 16i | =
√
1856
Math 251.502