Math 120 Answers for Homework 24

1.

Equation Picture Possible Reasons

(a) z = cos(

√

x2 + y2)

-8-1-8

-4-4

-0.5

00 x

y44

0

88

0.5

1

– Graph is rotationallysymmetric

– Height at all maxima is 1

– Cross section is cos graph.

(b) z = sin(y)

-4

-2-1

-40 x

-0.5

-2 2

0

0y 2

0.5

44

1

– Graph and function do notdepend on x.

– Slice when x constant is sin.

(c) z = cos2(x) sin2(y)

-3-20

-3

0.2

-1-2

0.4

-1 0

0.6

x

0.8

0 1y

1

1 2233

– Graph is always positive.

– Graph is periodic on eachx-slice and y-slice.

(d) z = sin(x2+y2)x2+y2

-3

-2

-1

0 y

3-0.2

12

0

10

2

0.2

-1x

0.4

-2 3-3

0.6

0.8

1

– Graph is rotationallysymmetric

– Maximum height decreaseswith distance from origin.

1

(e) z = cos(xy)

-3

-2

-1

0-1-3 x

-2 1-1

-0.5

0 21

0

y2 3

3

0.5

1

– Graph has height 1 whenxy = 0.

– For fixed x or y (i.e., x-slices ory-slices) graph looks like cos.

(f) z = |xy|

0 -2-2

1

-1-1

2

0

3

0 xy

4

11

22

– Slice when y constant looks likea multiple of |x|, and slice whenx constant looks like a multipleof |y|.

2. Starting with f(x, y) = xesin(y) + y2 we compute that

(a) ∂f

∂x= esin(y).

(b) ∂f

∂y= (cos(y)) · xesin(y) + 2y.

(c) ∂3f

∂x2y= 0

(the fastest way to see this is to differentiate ∂f

∂xwith respect to x to get ∂2f

∂x2 = 0,

and then differentiate once more with respect to y to get ∂3f

∂x3 = 0).

3. The tangent plane to a graph z = f(x, y) over the point (x0, y0) is the plane givenby the equation z = ax + by + c with a = fx(x0, y0), b = fy(x0, y0) and c chosen so thatthe tangent plane passes through the point (x0, y0, f(x0, y0)). This is the same as sayingthat c = f(x0, y0) − x0fx(x0, y0) − y0fy(x0, y0).

(a) f(x, y) = x3y2 − x2y3 + 4x − 5y + 6 , (x0, y0) = (3, 2).

function value at (3, 2) constantfx 3x2y2 − 2xy3 + 4 64 a = 64fy 2x3y − 3x2y2 − 5 −5 b = −5f x3y2 − x2y3 + 4x − 5y + 6 44 c = 44 − 3 · (64) − (−5) · (44) = −138

The equation of the tangent plane is

z = 64x − 5y − 138.

2

(b) g(x, y) = ex2y + sin(x) cos(y), (x0, y0) = (π, π2).

function value at (π, π2) constant

gx 2xy · ex2y + cos(x) cos(y) π2eπ3

2 a = π2eπ3

2

gy x2ex2y − sin(x) sin(y) π2eπ3

2 b = π2eπ3

2

g ex2y + sin(x) cos(y) eπ3

2 c = eπ3

2 − π · (π2eπ3

2 ) − π2· (π2e

π3

2 ) = −π3

2e

π3

2

The equation of the tangent plane is

z = (π2eπ3

2 ) x + (π2eπ3

2 ) y −π3

2e

π3

2 .

(c) h(x, y) = e(x2+y2)2 at the point (x0, y0) = (3, 4).

function value at (3, 4) constant

hx 4x(x2 + y2)e(x2+y2)2 300e625 a = 300e625

hy 4y(x2 + y2)e(x2+y2)2 300e625 b = 300e625

h e(x2+y2)2 e625 c = e625 − 3 · (300e625) − 4 · (300e625) = −2099e625

The equation of the tangent plane is therefore

z = 300e625x + 300e625y − 2099e625.

4.

(a) f(x, y) = x3 − y3 + 3xy.

f = x3 − y3 + 3xy

fx = 3x2 + 3y

fy = −3y2 + 3x

fxx = 6x

fyy = −6y

fxy = 3

∂∂x

∂∂y

∂∂x

∂∂y

∂∂x

∂∂y

3

At the point (x0, y0) = (1,−1) the value of the partial derivatives (and D) are :

Point fxx(x0, y0) fyy(x0, y0) fxy(x0, y0) D Classification(0, 0) 6 6 3 6 · 6 − 32 = 27 Min

Since D > 0 the point is either a local max or a local min. Since the secondderivative fxx is negative we see that (0, 0) is a local min by the rule for the onevariable test.

(b) g(x, y) = xy

g = xy

gx = y · xy−1

gy = ln(x) · xy

gxx = y · (y − 1) · xy−2

gyy = (ln(x))2 · xy

gxy = (1 + y ln(x)) · xy−1

∂∂x

∂∂y

∂∂x

∂∂y

∂∂x

∂∂y

At the point (x0, y0) = (1, 0) the values of the partial derivatives and D are:

Point gxx(x0, y0) gyy(x0, y0) gxy(x0, y0) D Classification(1, 0) 0 0 1 0 · 0 − 12 = −1 Saddle

By the two-variable second derivative test, the critical point (1, 0) is a saddle point.

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(c) h(x, y) = cos(x + y) − sin(x) sin(y)

h = cos(x + y) − sin(x) sin(y)

hx = − sin(x + y) − cos(x) sin(y)

hy = − sin(x + y) − sin(x) cos(y)

hxx = − cos(x + y) + sin(x) sin(y)

hyy = − cos(x + y) + sin(x) sin(y)

hxy = − cos(x + y) − cos(x) cos(y)

∂∂x

∂∂y

∂∂x

∂∂y

∂∂x

∂∂y

At the point (x0, y0) = (π2, 3π

2). the values of the partial derivatives and D are:

Point hxx(x0, y0) hyy(x0, y0) hxy(x0, y0) D Classification(π

2, 3π

2) −2 −2 −1 (−2) · (−2) − (−1)2 = 3 Max

Since D > 0 the point is either a local min or a local max. Since hxx is negative,(π

2, 3π

2) is a local max by the rule for the one variable second derivative test.

(d) f(x, y) = e2x2+3y2−5xy

f = e2x2+3y2−5xy

fx = (4x − 5y)e2x2+3y2−5xy

fy = (6y − 5x)e2x2+3y2−5xy

fxx = (4 + (4x − 5y)2)e2x2+3y2−5xy

fyy = (6 + (6y − 5x)2)e2x2+3y2−5xy

fxy = (−5 + (4x − 5y)(6y − 5x))e2x2+3y2−5xy

∂∂x

∂∂y

∂∂x

∂∂y

∂∂x

∂∂y

At the point (x0, y0) = (0, 0) the values of the partial derivatives are:

Point fxx(x0, y0) fyy(x0, y0) fxy(x0, y0) D Classification(0, 0) 4 6 −5 4 · 6 − (−5)2 = −1 Saddle

By the two-variable second derivative test, the critical point (0, 0) is a saddle point.

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