Homework Problem Solution

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 f_y and surface b is f_x.

19.

∂z

∂x = 10(2x + 3y)⁹· 2 = 20(2x + 3y)⁹

∂z

∂y = 10(2x + 3y)⁹· 3 = 30(2x + 3y)⁹

21.

∂f

∂x = 1 y

∂f

∂y = −x y²

29.

∂F

∂x = cos(e^x)

∂F

∂y = − cos(e^y) by the Fundamental Theorem of Calculus, part 1.

1

43.

∂

∂yf (x, y, z) = 1 · (x + y + z) − y(1) (x + y + z)²

= x + z (x + y + z)² f_y(2, 1, −1) = 2 − 1

(2 + 1 − 1)² = 1 4

45. By definition,

f_x(x, y) = lim

h→0

f (x + h, y) − f (x, y) h

= lim

h→0

(x + h)y²− (x + h)³y − xy²+ x³y h

= lim

h→0

hy² − 3x²hy + 3xh²y − h³y h

= y²− 3x²y f_y(x, y) = lim

h→0

f (x, y + h) − f (x, y) h

= lim

h→0

x(y + h)²− x³(y + h) − xy²+ x³y h

= lim

h→0

2hxy + xh²− x³h h

= 2xy − x³

50.

Differentiating both sides with respect to x:

y∂z

∂x + ln y = 2z∂z

∂x

⇒∂z

∂x = ln y 2z − y Differentiating both sides with respect to y:

1 · z + y∂z

∂y + x

y = 2z∂z

∂y

⇒∂z

∂y = z +^x_y

2z − y = yz + x y(2z − y) 2

56.

v_x = y(x − y) − xy(1)

(x − y)² = −y² (x − y)² v_y = x(x − y) − xy(−1)

(x − y)² = x² (x − y)²

vxx = −y²· (−2)(x − y)⁻³ = 2y² (x − y)³ v_xy = −2y(x − y)²+ y²· 2(x − y)

(x − y)⁴ = −2xy

(x − y)³ v_yx = 2x(x − y)²− x²· 2(x − y)

(x − y)⁴ = −2xy

(x − y)³ v_yy = −2x²(x − y)⁻³(−1) = 2x²

(x − y)³

79.

u_x = f⁰(x + at) + g⁰(x − at) uxx = f⁰⁰(x + at) + g⁰⁰(x − at) u_t= a[f⁰(x + at) − g⁰(x − at)]

u_tt = a²[f⁰⁰(x + at) + g⁰⁰(x − at)] = a²u_xx

101.

(a)

3

(b) For (x, y) 6= (0, 0)

fx = (3x²y − y³)(x²+ y²) − (x³y − xy³) · 2x

(x²+ y²)² = x⁴y + 4x²y³− y⁵ (x²+ y²)² and by symmetry

fy = x⁵− 4x³y²− xy⁴ (x²+ y²)² (c) Using equations 2 and 3,

fx(0, 0) = lim

h→0

f (h, 0) − f (0, 0)

h = lim

h→0

0 h² = 0 f_y(0, 0) = lim

h→0

f (0, h) − f (0, 0)

h = 0

(d) By definition,

f_xy(0, 0) = lim

h→0

f_x(0, h) − f_x(0, 0)

h = lim

h→0

−h⁵ h⁴ − 0

h = −1

f_yx(0, 0) = lim

h→0

f_y(h, 0) − f_y(0, 0)

h = lim

h→0 h⁵ h⁴ − 0

h = 1 (e) For (x, y) 6= (0, 0),

f_xy(x, y) = x⁶+ 9x⁴y²− 9x²y⁴− y⁶ (x² + y²)³

As (x, y) → (0, 0) along the x-axis, f_xy(x, y) → 1 while along the y-axis f_xy(x, y) → −1. Thus f_xy 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.

4