Homework Problem Solution
§14.3
9.
First we note that surface a is even in both x and y, surface b is odd in both
x and y, and surface c is even in x but odd in y. Since the derivative of
an odd function is even and the derivative of an even function is odd, we
may conclude that surface c is function f . Next, we may note that along
x = 3 the function is similar to the sine function, whose derivative is the
cosine function. On the other hand, along a fixed y value, say 1.5, f is like
a parabola, whose derivative is a straight line. With these hints, we may
conclude that surface a is fy and surface b is fx .
19.
∂z
= 10(2x + 3y)9 · 2 = 20(2x + 3y)9
∂x
∂z
= 10(2x + 3y)9 · 3 = 30(2x + 3y)9
∂y
21.
1
∂f
=
∂x
y
∂f
x
=− 2
∂y
y
29.
∂F
= cos(ex )
∂x
∂F
= − cos(ey )
∂y
by the Fundamental Theorem of Calculus, part 1.
1
43.
∂
1 · (x + y + z) − y(1)
f (x, y, z) =
∂y
(x + y + z)2
x+z
=
(x + y + z)2
1
2−1
=
fy (2, 1, −1) =
(2 + 1 − 1)2
4
45. By definition,
f (x + h, y) − f (x, y)
h→0
h
2
(x + h)y − (x + h)3 y − xy 2 + x3 y
= lim
h→0
h
hy 2 − 3x2 hy + 3xh2 y − h3 y
= lim
h→0
h
2
2
= y − 3x y
f (x, y + h) − f (x, y)
fy (x, y) = lim
h→0
h
2
x(y + h) − x3 (y + h) − xy 2 + x3 y
= lim
h→0
h
2hxy + xh2 − x3 h
= lim
h→0
h
3
= 2xy − x
fx (x, y) = lim
50.
Differentiating both sides with respect to x:
∂z
∂z
+ ln y = 2z
∂x
∂x
∂z
ln y
=
⇒
∂x
2z − y
y
Differentiating both sides with respect to y:
∂z x
∂z
+ = 2z
∂y y
∂y
x
z+y
∂z
yz + x
⇒
=
=
∂y
2z − y
y(2z − y)
1·z+y
2
56.
y(x − y) − xy(1)
−y 2
vx =
=
(x − y)2
(x − y)2
x(x − y) − xy(−1)
x2
vy =
=
(x − y)2
(x − y)2
2y 2
vxx = −y · (−2)(x − y) =
(x − y)3
2y(x − y)2 + y 2 · 2(x − y)
−2xy
vxy = −
=
4
(x − y)
(x − y)3
−2xy
2x(x − y)2 − x2 · 2(x − y)
=
vyx =
4
(x − y)
(x − y)3
2x2
vyy = −2x2 (x − y)−3 (−1) =
(x − y)3
−3
2
79.
ux = f 0 (x + at) + g 0 (x − at)
uxx = f 00 (x + at) + g 00 (x − at)
ut = a[f 0 (x + at) − g 0 (x − at)]
utt = a2 [f 00 (x + at) + g 00 (x − at)] = a2 ux x
101.
(a)
3
(b) For (x, y) 6= (0, 0)
fx =
(3x2 y − y 3 )(x2 + y 2 ) − (x3 y − xy 3 ) · 2x
x4 y + 4x2 y 3 − y 5
=
(x2 + y 2 )2
(x2 + y 2 )2
and by symmetry
fy =
x5 − 4x3 y 2 − xy 4
(x2 + y 2 )2
(c) Using equations 2 and 3,
0
f (h, 0) − f (0, 0)
= lim 2 = 0
h→0 h
h→0
h
f (0, h) − f (0, 0)
=0
fy (0, 0) = lim
h→0
h
fx (0, 0) = lim
(d) By definition,
fx (0, h) − fx (0, 0)
= lim
fxy (0, 0) = lim
h→0
h→0
h
fy (h, 0) − fy (0, 0)
fyx (0, 0) = lim
= lim
h→0
h→0
h
−h5
h4
−0
= −1
h
h5
−0
h4
=1
h
(e) For (x, y) 6= (0, 0),
fxy (x, y) =
x6 + 9x4 y 2 − 9x2 y 4 − y 6
(x2 + y 2 )3
As (x, y) → (0, 0) along the x-axis, fxy (x, y) → 1 while along the y-axis
fxy (x, y) → −1. Thus fxy is not continuous at (0,0) and Clairaut’s Theorem
does not apply, so there is no contradiction. The graphs of fxy and fyx are
identical except at the origin, where we observe the discontinuity.
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