{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Winter 2007 - Takeda's Class - Practice Final Exam

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

This preview shows pages 1–6. Sign up to view the full content.

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

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

View Full Document
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 1 0 x 0 e y dy dx 1 0 x 0 e y dy dx = 1 0 e y y = x y =0 dx = 1 0 e x - 1 dx = ( e x - x ) 1 0 = e - 1 - 1 = e - 2 . 2
2. (a) Let f ( x, y ) = xe y . Find the directional derivative of f at (1 , 0) in the direction of u = 3 5 , - 4 5 . First note that f ( x, y ) = e y i + xe y j . Hence D u ( f )(1 , 0) = f (1 , 0) · u = 1 , 1 · 3 5 , - 4 5 = 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

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

View Full Document
3. 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 4 , - 9 , - 1 . So the symmetric equation of the line is x - 1 4 = y - 2 - 9 = z - 3 - 1 . 4
4. Compute the double integral 1 0 2 2 x e y 2 dy dx. We need to change the order of integration. 1 0 2 2 x e y 2 dy dx = 2 0 y/ 2 0 e y 2 dxdy = 2 0 xe y 2 x = y/ 2 x =0 dy = 2 0 y 2 e y 2 dy = 1 4 e y 2 2 0 = 1 4 e 4 - 1 4 .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern