Midterm II 2004 Solutions

Midterm II 2004 Solutions - MATH 52 SAMPLE MIDTERM II...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 52 SAMPLE MIDTERM II November 13, 2004 Name: Numeric Student ID: Instructor’s Name: I agree to abide by the terms of the honor code: Signature: Instructions: Print your name, student ID number and instructor’s name in the space provided. During the test you may not use notes, books or calculators. Read each question carefully and show all your work ; full credit cannot be obtained without sufficient justification for your answer unless explicitly stated otherwise. Underline your final answer to each question. There are 5 questions. You have 120 minutes to do all the problems. Question Score Maximum 1 20 2 10 3 10 4 10 5 10 Total 60 Question 1 of 5, Page 2 of 9 Solutions 1. Evaluate the following line integrals: (a) Z C F · d r where F ( x, y, z ) = h xy, yz, xz i and C : r ( u ) = h u, u 2 , u 3 i from (- 1 , 1 ,- 1) to (1 , 1 , 1). Solution: According to the given parametrization, F ( r ( u )) = h u 3 , u 5 , u 4 i . Hence F · d r = u 3 + 5 u 6 and we integrate from u =- 1 to u = 1. This gives 10/7. Question 1 of 5, Page 3 of 9 Solutions (b) Z C F · T ds where F ( x, y, z ) = h cos x, sin y, yz i and C is the line segment from (0 , , 0) to (2 , 3 ,- 1). Solution: Again this is another notation for the integral of F · d r , the work done over the curve C by F . Here we are not given the parametrization, so we must determine one for the line from (0 , , 0) to (2 , 3 ,- 1). There are lots of ways we can do this. Remember that a line in space is given by a point P on the line and a direction vector v so we may write x ( t ) = 0 + 2 t, y ( t ) = 0 + 3 t, z ( t ) = 0- t In general, given any two points ( a , b , c ) and ( a 1 , b 1 , c 1 ) we can parametrize the line between them by x ( t ) = a 1 t + a (1- t ) , y ( t ) = b 1 t + b (1- t ) , z ( t ) = c 1 t + c (1- t ) You might think about why this is true given the equation for a line in terms of a point and a vector. In any case, F ( r ( t )) = h cos 2 t, sin 3 t,- 3 t 2 i and F ( r ( t )) · d r = 2 cos 2 t + 3 sin 3 t + 3 t 2 and so the integral, now from t = 0 to t = 1 according to the parametrization, is 2 + sin 2- cos 3. Question 1 of 5, Page 4 of 9 Solutions (c) Compute the work done by F ( x, y, z ) = h 2 x ln y- yz, x 2 y- xz,...
View Full Document

This note was uploaded on 04/16/2008 for the course MATH 52 taught by Professor Demanet,l;wieczorek,w during the Winter '08 term at Stanford.

Page1 / 9

Midterm II 2004 Solutions - MATH 52 SAMPLE MIDTERM II...

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