Score - SJSU

[Pages:7]San Jose? State University

Math 32, Fall 2008

Final Exam Solutions

December 17, 2008

Name: Granwyth Hulatberi

Score 1 20 2 20 3 20 4 20 5 20 6 20 Total 120

Explain your work

1. (20 points) Let a function f be defined by

f (x, y) =

x4+x2y2+y4 x4+x3y+x2y2+xy3+y4

1

if (x, y) = (0, 0) if (x, y) = (0, 0).

(a) Is f continuous at the point (x, y) = (2008, e2008)? (b) Is f continuous at the origin?

Justify your answers.

Solution: (a) f is continuous at (2008, e2008) because f is a rational function and the denominator of f is non-zero at that point.

(b) No. Since

3x4 3 3 f (x, x) = = ,

5x4 5 5

and f (x, 0) = 1 1, as x 0, the limit of f at (0, 0) doesn't exist and is therefore not

equal f (0, 0) = 1.

2. (20 points) Suppose that f is a differentiable function. If

z = y + f (x2 - y2),

compute Solution: By the Chain Rule,

yzx + xzy.

Therefore,

zx = f (x2 - y2)2x and zy = 1 + f (x2 - y2)(-2y).

yzx + xzy = y[f (x2 - y2)2x] + x[1 + f (x2 - y2)(-2y)] = x.

3. (20 points) Show that the ellipsoid 3x2 + 2y2 + z2 = 9 and the sphere

x2 + y2 + z2 - 8x - 6y - 8z + 24 = 0

have a common tangent plane at the point (1, 1, 2).

Solution: Let

f (x, y, z) = 3x2 + 2y2 + z2 = 9

and g(x, y, z) = x2 + y2 + z2 - 8x - 6y - 8z + 24.

We have

f (x, y, z) = 6x, 4y, 2z

and g(x, y, z) = 2x - 8, 2y - 6, 2z - 8 ,

so f (1, 1, 2) = 6, 4, 4 and g(1, 1, 2) = -6, -4, -4 . Since f (1, 1, 2) = -g(1, 1, 2), the tangent planes of the surfaces f = 0 and g = 9 have the same normal, hence coincide with each other.

4. (20 points) Find the maximum and minimum of f (x, y) = e-xy

on the region D = {(x, y) : x2 + 4y2 1}. Solution: Interior of D: Since

fx = -ye-xy and fy = -xe-xy, the only critical point of f in the interior of D is (0, 0). Note that f (0, 0) = 1.

Boundary of D: Let g(x, y) = x2 + 4y2. Since gx = 2x and gy = 8y, the Lagrange equations are

-ye-xy = 2x,

(1)

-xe-xy = 8y,

(2)

x2 + 4y2 = 1.

(3)

Multiplying (1) by x and (2) by y yields 2x2 = 8y2 or (x2 - 4y2) = 0. If = 0, then (1) and (2) imply x = y = 0, which contradicts (3). Therefore, = 0, so x2 = 4y2.

Substituting into (3) yields 8y2 = 1,

which means y = ? 1 . Since x2 = 4y2, this implies x2 = 1/2, so x = ?1/ 2. This 22

gives us four points:

1 f( ,

1

11

) = f (- , -

) = e1/4

and

1 f (- ,

1

11 ) = f( ,-

) = e-1/4.

22 2

2 22

22 2

2 22

Since e-1/4 < 1 < e1/4, it follows that e-1/4 is the minimum and e1/4 is the maximum of f on D.

5. (20 points) Compute the double integral

x cos y dA,

D

where D is the region bounded by the curves y = 0, y = x2, and x = /2.

Solution: Since D is a region of type I, by Fubini's theorem we have:

/2 x2

x cos y dA =

x cos y dy dx

D

0

0

/2

=

(x sin y)|yy==0x2 dx

0

/2

=

x sin x2 dx

0

=

-

1 2

cos

x2|0

/2

=

cos -

2

-

cos 0

2

1 =.

2

6. (20 points) Let H be the solid above the xy-plane bounded by the unit sphere x2 + y2 + z2 = 1 and the xy-plane.

(a) Sketch H. (b) Compute the triple integral

xyz dV.

H

Solution: (a) The upper half of the unit ball.

(b) The projection of H onto the xy-plane is the unit disk D : x2 + y2 1. Using

Fubini's theorem, we obtain:

1-x2-y2

xyz dV =

xyz dz dA

H

D0

z2 z= 1-x2-y2

= xy

dA

D

2 z=0

1 =

xy(1 - x2 - y2) dA.

2D

Here we use polar coordinates (r, ) in which D is described by

0 r 1, 0 2.

Thus the last integral equals:

2 1

2

1

r2 cos sin (1 - r2)r dr d = cos sin d ? r3(1 - r2) dr

00

0

0

= 0,

since

2

1 2

cos sin d =

sin 2 d

0

20

= 0.

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