Winter 2007 - Takeda's Class - Practice Final Exam

Winter 2007 - Takeda's Class - Practice Final Exam - Math...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 20C, Practice Final Exam Solutions March 18, 2007 Name : PID : TA : Sec. No : Sec. Time : This exam consists of 11 pages including this front page. Ground Rules 1. No calculator is allowed. 2. Show your work for every problem. A correct answer without any justification will receive no credit. 3. You may use two 4-by-6 index cards, both sides. 4. You have two hours for this exam. Score 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total 100 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
1. (a) Let f ( x, y ) = x y - ln | x + y 2 | . What is f x ? f x = yx y - 1 - 1 x + y 2 . (b) Compute the double integral Z 1 0 Z x 0 e y dy dx Z 1 0 Z x 0 e y dy dx = Z 1 0 e y ± y = x y =0 dx = Z 1 0 e x - 1 dx = ( e x - x ) ± 1 0 = e - 1 - 1 = e - 2 . 2
Background image of page 2
2. (a) Let f ( x, y ) = xe y . Find the directional derivative of f at (1 , 0) in the direction of u = h 3 5 , - 4 5 i . First note that f ( x, y ) = e y i + xe y j . Hence D u ( f )(1 , 0) = f (1 , 0) · u = h 1 , 1 i · h 3 5 , - 4 5 i = 3 5 - 4 5 = - 1 5 . (b) Let a = 3 i - 1 j + 2 k and b = - i + 3 k . Compute a × b . a × b = ± ± ± ± ± ± i j k 3 - 1 2 - 1 0 3 ± ± ± ± ± ± = - 3 i - 11 j - k . 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Let S be the tangent plane to the surface defined by z = x 2 - 3 y 3 at (2 , 1 , 1). Find the symmetric equation of the line which is perpendicular to S and goes through the point (1 , 2 , 3). Let f ( x, y ) = x 2 - 3 y 3 . Then f x = 2 x and f y = - 9 y 2 . Then the equa- tion of S is z - 1 = f x (2 , 1)( x - 2) + f y (2 , 1)( y - 1) = 4( x - 2) - 9( y - 1) . Hence the normal vector of S is h 4 , - 9 , - 1 i . So the symmetric equation of the line is x - 1 4 = y - 2 - 9 = z - 3 - 1 . 4
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.

Page1 / 13

Winter 2007 - Takeda's Class - Practice Final Exam - Math...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online