Math 233 - Exam III - Fall 2011

[Pages:32]Math 233 - Exam III - Fall 2011

November 16, 2011 - Renato Feres

NAME:

STUDENT ID NUMBER:

General instructions: This exam has 16 questions, each worth the same amount. Check that no pages are missing and notify your proctor if you detect any problems with your copy of the exam. Mark your ID number on the six blank lines on the top of your answer card, using one line for each digit. Print your name on the top of the card. Choose the answer that is closest to the solution and mark your answer card with a PENCIL by shading in the correct box. You may use a 3?5 card with notes and any calculator that does not have graphing functions. GOOD LUCK!

1. Find the second order partial derivative fxy(, 1) where f (x, y) = x2y + cos y + y sin x.

(A) (B) 2 + 1 (C) 2 - 1 (D) 2 (E) + 3 (F) - 3 (G) - + 1 (H) - - 1 (I) -2 (J) -3

13.3, 43

1

Solution: The first derivative in x is fx(x, y) = 2xy + y cos x and the mixed derivative is fxy(x, y) = 2x + cos x. Evaluating at (, 1): fxy(, 1) = 2 - 1.

2

2. Evaluate w/u at (u, v) = (0, 1), where w = xy + yz + xz and x = u + v, y = u - v, z = uv.

13.4, 9

(A) 4 (B) 3 (C) 2 (D) 1 (E) 0 (F) -1 (G) -2 (H) -3 (I) -4 (J) -5

3

Solution: By the chain rule,

The partials are:

w u

=

w x x u

+

w y

y u

+

w z

z u

.

w x

=

y

+

z,

w y

=

x

+

z,

w z

=

x

+

y,

x u

=

y u

=

1,

z u

=

v.

This gives

w u

=

(x

+

y)(1

+

v)

+

2z

=

2u(1

+

v)

+

2uv

=

2u(1

+

2v).

Evaluating at (u, v) = (0, 1) we get

w u

=

0.

4

3. Find the derivative of the function

f (x, y) = 2xy - 3y2

at P0(5, 5) in the direction of the vector u = 4i + 3j.

(A) -4 (B) -3 (C) -2 (D) -1 (E) 0 (F) 1 (G) 2 (H) 3 (I) 4 (J) 5

13.5, 11

5

Solution: The gradient of f at P0 is

(f )P0 = 2y, 2x - 6y |P0 = 10, -20 . Then the directional derivative in the direction of u (equivalently, in the direction of the unit vector u/|u| = 4/5, 3/5 ) is

u Du = (f )P0 ? |u| = 10, -20 ? 4/5, 3/5 = 8 - 12 = -4.

6

4. Find an equation for the tangent plane to the surface represented by z = y - x

at the point (x, y, z) = (1, 2, 1).

13.6, 11

(A) x - y + 2z = 0 (B) x + y + 2z = 5 (C) 2x - y + z = 1 (D) 3x - y + 2z = 3 (E) x - 2y + z = 1 (F) x - y + 2z = -3 (G) x - y + 2z = 1 (H) 2x - 2y + z = -1 (I) x - y - z = -2 (J) x - 3y + 2z = -3

7

Solution:

The gradient of f (x, y, z) = y - x - z at P0(1, 2, 1) is

(f )P0 =

2-y 1- x, 2y1- x, -1

=

P0

-

1 2

,

1 2

,

-1

.

Therefore, the plane has equation

-

1 2

,

1 2

,

-1

? x - 1, y - 2, z - 1 = 0.

Evaluating the dot product and simplifying results in

x - y + 2z = 1.

8

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