Practice Problems for Exam 3 (Solutions) 1. Let F(x, y) = xyi+(y 3x)j, and let C be the curve r(t) = ti+(3t t 2 )j for 0 t 2. Compute F dr.

Similar documents
Math 233. Practice Problems Chapter 15. i j k

Sections minutes. 5 to 10 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed.

Math Exam IV - Fall 2011

Jim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt

One side of each sheet is blank and may be used as scratch paper.

MATH 52 FINAL EXAM SOLUTIONS

1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is

In general, the formula is S f ds = D f(φ(u, v)) Φ u Φ v da. To compute surface area, we choose f = 1. We compute

Math 11 Fall 2016 Final Practice Problem Solutions

MLC Practice Final Exam

M273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3

Page Problem Score Max Score a 8 12b a b 10 14c 6 6

(b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (0, 0)

Problem Points S C O R E

Name: SOLUTIONS Date: 11/9/2017. M20550 Calculus III Tutorial Worksheet 8

Name: Date: 12/06/2018. M20550 Calculus III Tutorial Worksheet 11

Solutions for the Practice Final - Math 23B, 2016

e x2 dxdy, e x2 da, e x2 x 3 dx = e

MATH 228: Calculus III (FALL 2016) Sample Problems for FINAL EXAM SOLUTIONS

MATH 332: Vector Analysis Summer 2005 Homework

1. If the line l has symmetric equations. = y 3 = z+2 find a vector equation for the line l that contains the point (2, 1, 3) and is parallel to l.

Calculus III. Math 233 Spring Final exam May 3rd. Suggested solutions

Math 350 Solutions for Final Exam Page 1. Problem 1. (10 points) (a) Compute the line integral. F ds C. z dx + y dy + x dz C

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the

MATHS 267 Answers to Stokes Practice Dr. Jones

Final exam (practice 1) UCLA: Math 32B, Spring 2018

Contents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9

Disclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.

Math 23b Practice Final Summer 2011

f. D that is, F dr = c c = [2"' (-a sin t)( -a sin t) + (a cos t)(a cost) dt = f2"' dt = 2

7a3 2. (c) πa 3 (d) πa 3 (e) πa3

Jim Lambers MAT 280 Fall Semester Practice Final Exam Solution

Divergence Theorem December 2013

Practice problems **********************************************************

SOLUTIONS TO THE FINAL EXAM. December 14, 2010, 9:00am-12:00 (3 hours)

Math 3435 Homework Set 11 Solutions 10 Points. x= 1,, is in the disk of radius 1 centered at origin

Page Points Score Total: 210. No more than 200 points may be earned on the exam.

(a) 0 (b) 1/4 (c) 1/3 (d) 1/2 (e) 2/3 (f) 3/4 (g) 1 (h) 4/3

Divergence Theorem Fundamental Theorem, Four Ways. 3D Fundamental Theorem. Divergence Theorem

Without fully opening the exam, check that you have pages 1 through 12.

MAC2313 Final A. (5 pts) 1. How many of the following are necessarily true? i. The vector field F = 2x + 3y, 3x 5y is conservative.

Math 31CH - Spring Final Exam

MAT 211 Final Exam. Spring Jennings. Show your work!

Final Review Worksheet

e x3 dx dy. 0 y x 2, 0 x 1.

Math 6A Practice Problems II

SOME PROBLEMS YOU SHOULD BE ABLE TO DO

No calculators, cell phones or any other electronic devices can be used on this exam. Clear your desk of everything excepts pens, pencils and erasers.

Solutions to old Exam 3 problems

PRACTICE PROBLEMS. Please let me know if you find any mistakes in the text so that i can fix them. 1. Mixed partial derivatives.

Solutions to the Final Exam, Math 53, Summer 2012

McGill University April 16, Advanced Calculus for Engineers

McGill University April 20, Advanced Calculus for Engineers

Ma 1c Practical - Solutions to Homework Set 7

Math 20C Homework 2 Partial Solutions

Math Review for Exam 3

(a) The points (3, 1, 2) and ( 1, 3, 4) are the endpoints of a diameter of a sphere.

Final Exam. Monday March 19, 3:30-5:30pm MAT 21D, Temple, Winter 2018

Vector Calculus, Maths II

Practice problems. m zδdv. In our case, we can cancel δ and have z =

1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r (t) = 3 cos t, 0, 3 sin t, r ( 3π

MTH 234 Exam 2 November 21st, Without fully opening the exam, check that you have pages 1 through 12.

Answers and Solutions to Section 13.7 Homework Problems 1 19 (odd) S. F. Ellermeyer April 23, 2004

ARNOLD PIZER rochester problib from CVS Summer 2003

Math 212-Lecture 20. P dx + Qdy = (Q x P y )da. C

Peter Alfeld Math , Fall 2005

Practice Final Solutions

Math 11 Fall 2007 Practice Problem Solutions

Without fully opening the exam, check that you have pages 1 through 12.

Review problems for the final exam Calculus III Fall 2003

MA 441 Advanced Engineering Mathematics I Assignments - Spring 2014

MATH 2203 Final Exam Solutions December 14, 2005 S. F. Ellermeyer Name

Mathematics (Course B) Lent Term 2005 Examples Sheet 2

Summary of various integrals

APPM 2350 Final Exam points Monday December 17, 7:30am 10am, 2018

Arnie Pizer Rochester Problem Library Fall 2005 WeBWorK assignment VectorCalculus1 due 05/03/2008 at 02:00am EDT.

Solutions to Sample Questions for Final Exam

Final exam (practice 1) UCLA: Math 32B, Spring 2018

Green s, Divergence, Stokes: Statements and First Applications

G G. G. x = u cos v, y = f(u), z = u sin v. H. x = u + v, y = v, z = u v. 1 + g 2 x + g 2 y du dv

Without fully opening the exam, check that you have pages 1 through 10.

Practice problems ********************************************************** 1. Divergence, curl

Archive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma

(0,2) L 1 L 2 R (-1,0) (2,0) MA4006: Exercise Sheet 3: Solutions. 1. Evaluate the integral R

Math 11 Fall 2018 Practice Final Exam

MULTIVARIABLE CALCULUS

Stokes Theorem. MATH 311, Calculus III. J. Robert Buchanan. Summer Department of Mathematics. J. Robert Buchanan Stokes Theorem

MAT 211 Final Exam. Fall Jennings.

Math 265H: Calculus III Practice Midterm II: Fall 2014

D = 2(2) 3 2 = 4 9 = 5 < 0

Exercises for Multivariable Differential Calculus XM521

Review Questions for Test 3 Hints and Answers

Math 234 Exam 3 Review Sheet

Note: Each problem is worth 14 points except numbers 5 and 6 which are 15 points. = 3 2

Math 32B Discussion Session Week 10 Notes March 14 and March 16, 2017

MATH H53 : Final exam

Direction of maximum decrease = P

Created by T. Madas LINE INTEGRALS. Created by T. Madas

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 10 (Second moments of an arc) A.J.Hobson

Vector Fields and Line Integrals The Fundamental Theorem for Line Integrals

Transcription:

1. Let F(x, y) xyi+(y 3x)j, and let be the curve r(t) ti+(3t t 2 )j for t 2. ompute F dr. Solution. F dr b a 2 2 F(r(t)) r (t) dt t(3t t 2 ), 3t t 2 3t 1, 3 2t dt t 3 dt 1 2 4 t4 4. 2. Evaluate the line integral x 2. x 3 y ds, where is the portion of the graph y x2 /2 for Solution. Note that this is the line integral of a scalar function with respect to arc length. We can parametrize as r(t) t, 1 2 t2 for t 2. Then so that for f(x, y) x 3 /y: x 3 y ds b a 2 5 r (t) 1, t and r (t) 1 + t 2, f(r(t)) r (t) dt 2 t 3 1 2 t2 1 + t2 dt 2t 1 + t 2 (use substitution u 1 + t 2, du 2t dt) 2 u du 5 2 3 (t3/2 1). 3 u3/2 1 3. Let F(x, y) (ax 2 y + y 3 + 1)i + (2x 3 + bxy 2 + 2)j be a vector field, where a and b are constants. (a) Find the values and a and b for which F is conservative. Solution. P ax 2 y + y 3 + 1 and Q 2x 3 + bxy 2 + 2 so P y ax 2 + 3y 2 and Q x 6x 2 + by 2. Thus, P y Q x if and only if a 6 and b 3. (b) For these values of a and b, find f(x, y) such that F f. Math 257 1

Solution. f x 6x 2 y + y 3 + 1 f 2x 3 y + xy 3 + x + g(y). Therefore, f y 2x 3 +3xy 2 +g (y). Setting this equal to Q gives 2x 3 +3xy 2 +g (y) 2x 3 +3xy 2 +2 so g (y) 2 and hence g(y) 2y + K where K is a constant. Hence, f(x, y) 2x 3 y + xy 3 + x + 2y + K. (c) Still using the values of a and b from part (a), compute such that x e t cos t, y e t sin t, t π. F dr along the curve Solution. The curve starts at (1, ) and ends at ( e π, ), so by the fundamental theorem for line integrals (page 175), F dr f( e π, ) f(1, ) e π 1. 4. Let F(x, y, z) (cos x + 2y 2 + 5yz)i + (4xy + 5xz)j + (5xy + 3z 2 )k on R 3. Show that F is conservative, and find a potential function for F. Solution. Writing F P i + Qj + Rk, the necessary and sufficient condition for F to be conservative is that curl(f). This means that R y Q z, R x P z, and Q x P y. But R y 5x Q z, R x 5y P z, and Q x 4y + 5z P y. Thus F is conservative. To find a potential function f(x, y, z) such that f F, first set f x P and integrate with respect to x to get f(x, y, z) sin x + 2xy 2 + 5xyz + g(y, z). Now differentiate with respect to y and set the result equal to Q to get f y 4xy + 5xz + g y 4xy + 5xz. Thus g y so g(y, z) h(z) and f(x, y, z) sin x + 2xy 2 + 5xyz + h(z). Now differentiate the result with respect to z and set the result equal to R to get f z 5xy + h (z) 5xy + 3z 2. Thus, h (z) 3z 2 so integrating gives h(z) z 3 + K where K is a constant. Thus, f(x, y, z) sin x + 2xy 2 + 5xyz + z 3 + K. 5. Verify that the vector field F(x, y) (y 2 y cos x)i + (2xy sin x + 1)j is conservative. Then compute F dr, where is the curve x t3 t, y 1 + t 2, 1 t 2. Math 257 2

Solution. To verify that F is conservative, compute P y y (y2 y cos x) 2y cos x, Q x (2xy sin x + 1) 2y cos x. x Since the two partial derivatives are equal, F is conservative. We can thus evaluate the line integral by the fundamental theorem for line integrals. We must first find a potential function f with f F. Thus, we must have f x y 2 y cos x and f y 2xy sin x + 1. Integrating f x with respect to x gives f(x, y) xy 2 y sin x + g(y). Differentiate this with respect to y to get f y 2xy sin x + g (y). Setting this equal to the expression for f y found earlier gives g (y) 1 so g(y) y + K. Thus, our potential function is f(x, y) xy 2 y sin x + y + K. The fundamental theorem for line integrals then gives F dr f dr f(r(2)) f(r(1)) f(6, 5) f(, 2) (6(25) 5 sin(6) + 5 + K) ( + 2 + K) 153 5 sin(6). 6. Let be the circle x 2 + y 2 1, oriented counterclockwise. Find (2xy + e x ) dx + (x 2 sin y + 3x) dy. Solution. Since is a simple closed curve, Green s Theorem applies. First compute Q x P y x (x2 sin y + 3x) y (2xy + ex ) (2x + 3) (2x) 3. We can then calculate, using Green s Theorem (note that the inside of is D {x 2 + y 2 1}, the disc of radius 1 centered at the origin): ( Q P dx + Q dy x P ) da 3 da y D 3(Area(D)) 3 π(1) 2 3π. 7. Let be the ellipse x2 a + y2 1. Evaluate the line integral 2 b2 x a i + y 2 b j. 2 D F dr where F(x, y) Math 257 3

Solution. Parametrize by x a cos t, y b sin t, t 2π. Then 2π ( ) a cos t b sin t F dr i + j ( a sin ti + b cos tj) dt a 2 b 2 2π ( ) a 2 cos t sin t + b2 sin t cos t dt a 2 b 2 2π 3 dt. 8. Let be the positively oriented closed curve formed by the parabola y x 2 running from (, 1) to (1, 1) and by a straight line running back from (1, 1) to (, 1). Evaluate the line integral y 2 dx + 4xy dy in two ways: (a) directly, and Solution. Let 1 + 2, where 1 is the curve running along the parabola and 2 is the horizontal segment. Parametrize 1 by x(t) t, y(t) t 2, t 1. Then dx dt and dy 2t dt, so the line integral over 1 is 1 y 2 dx + 4xy dy 1 ((t 2 ) 2 + 4tt 2 (2t)) dt 1 9t 4 dt 9 5 t5 1 18 5. Parametrize 2 by x(t) t, y(t) 1, t 1. Then dx dt and dy, so the line integral over 2 is 2 y 2 dx + 4xy dy 1 dt t 1 2. Putting all this together we get y 2 dx + 4xy dy y 2 dx + 4xy dy + y 2 dx + 4xy dy 18 1 2 5 2 8 5. (b) by using Green s theorem. Solution. y 2 dx + 4xy dy 2y da 1 1 R x 2 2y dy dx. Math 257 4

The inner integral is Thus, the outer integral is 1 x 2 2y dy y 2 1 x 2 1 x 4. 1 (1 x 4 ) dx ) (x x5 1 8 5 5. 9. Let F(x, y, z) yi+zj 2xk and let be a simple closed curve in the plane x+y+z 1. Show that F dr. Solution. Apply Stokes Theorem. Let S be the part of the plane x + y + z 1 inside. Then Stokes theorem states that F dr (curl F) n ds, S where n is the unit normal vector to the surface S. Since S is the plane x + y + z 1, the normal vector is n i + j + k. The curl of F is 3 i j k curl F x y z i + 2j k. y z 2x Thus, and hence (curl F) n ( i + 2j k) i + j + k 3 F dr (curl F) n ds. S 1. Let Σ be the unit sphere centered at the origin, with the outward normal orientation, and let F(x, y, z) x 3 i + y 3 j + z 3 k. ompute the integral: F n ds. Hint: Use the divergence theorem. Σ Solution. Let B be the unit ball, so that Σ is the boundary of B. The divergence of F is div F 3x 2 + 3y 2 + 3z 2. Then the divergence theorem gives F n ds div F dv 3(x 2 + y 2 + z 2 ) dv. Σ B Math 257 5 B

This last integral can be evaluated in spherical coordinates ρ, φ, θ as 2π π 1 3ρ 2 ρ 2 sin φ dρ dφ dθ 12 5 π. 11. Let f be a scalar field (i.e., function) and F a vector field. Determine if each of the following expressions is meaningful. If it is meaningful, indicate whether it is a scalar field or a vector field. Expression Meaningful Scalar Vector f Yes x div F Yes x curl f No curl( f) Yes x F No div( F) Yes x Math 257 6