Math 217: Some Assignment 4 Solutions
14.4 # 14:√
f (x, y) = x + e4y is differentiable at (3, 0) since it is a composition of differentiable
functions (linear, exponential, and square root (where its argument is positive)). We have
fx = (1/2)(x + e4y )−1/2 → fx (3, 0) = 1/4,
fy = (1/2)(x + e4y )−1/2 4e4y → fy (3, 0) = 1.
So the linearization is
1
5
1
L(x, y) = 2 + (x − 3) + (y − 0) = x + y + .
4
4
4
14.5 # 16: With x(r, s) = 2r − s, y(r, s) = s2 − 4r,
gr (1, 2) = fx (0, 0)xr (1, 2) + fy (0, 0)yr (1, 2) = 4(2) + 8(−4) = −24,
gs (1, 2) = fx (0, 0)xs (1, 2) + fy (0, 0)ys (1, 2) = 4(−1) + 8(2)(2) = 28.
14.5 # 40: Since I = V /R,
dI
Vt
V
−0.01 (400)(0.08)
0.0124
=
− 2 Rt =
−
(0.03) = −
= −0.000031 A/s
2
dt
R
R
400
(400)
400
14.5 # 58: The equation F (x, y, z) = 0 is supposed to implicitly define, for example,
z as a function z(x, y) of x and y. Thus we have
0 = F (x, y, z(x, y)),
and differentiating this with respect to x yields, by the chain rule,
0 = Fx + Fz
∂z
∂x
so
∂z
= −Fx /Fz .
∂x
Now do the same thing, but with the variables permuted, to find
∂x
= −Fy /Fx ;
∂y
∂y
= −Fz /Fy .
∂z
1
Thus
∂z ∂x ∂y
= (−Fx /Fz )(−Fy /Fx )(−Fz /Fy ) = −1.
∂x ∂y ∂z
Bonus. Prove the following: if a function f (x, y) is differentiable at (a, b), then it is
continuous at (a, b):
We have
f (x, y) − f (a, b) =
#
"
f (x, y) − f (a, b) − fx (a, b)(x − a) − fy (a, b)(y − a) p
p
(x − a)2 + (y − b)2
(x − a)2 + (y − b)2
+fx (a, b)(x − a) + fy (a, b)(y − b).
As (x, y) → (a, b), the first term tends to zero by the definition of differentiability of f at
(a, b) (the extra factor only helps – it also tends to zero). The remaining two terms also
clearly tend to zero. Thus
lim
f (x, y) = f (a, b).
(x,y)→(a,b)
That is, f is continuous at (a, b).
Oct. 2, 2013
2