Math 222 Spring 2013 Exam 3 Review Problem Answers


 Sophia Daniels
 3 years ago
 Views:
Transcription
1 . (a) By the Chain ule, Math Spring 3 Exam 3 eview Problem Answers w s w x x s + w y y s (y xy)() + (xy x )( ) (( s + 4t) (s 3t)( s + 4t)) ((s 3t)( s + 4t) (s 3t) ) 8s 94st + 3t (b) By the Chain ule, w s w x x s + w y y s (t cos s) + x + y x + y (t) t cos s + t t sin s + st cos s + sin s + s Note that both parts (a) and (b) could be done by first writing w in terms of s and t and then computing w/ s directly.. We have to use the Chain ule for this one. We have dw dt w dx x dt + w dy y dt w x w (t + 3) + (5 t) y When t, we get x and y 6. Thus, when t, we have dw dt ( )(( ) + 3) + ()(5 ( )) t 6 When t, we get x 4 and y 4. Thus, when t, we have dw dt ()(() + 3) + ( 4)(5 ()) t (a) f yz e xy, xz e xy, ze xy
2 (b) This is asking for D u f(,, ), where u is the unit vector in the direction of, 3,. We have D u f(,, ) f(,, ) u e, e, e 3,, e (c) f(, 3, ) 3,, (d) f(,, ),, 5 4. (a) g y + z, x + 3z, x + 3y (b) This is asking for D u g(,, ), where u is the unit vector in the direction of 3, 4,. We have D u g(,, ) g(,, ) u, 4, 3 4,, (c) g(,, ), 4, (d) g(,, ), 4, 4 5. Let h(x, y) z.x.y. (a) Since u, is the unit vector pointing due south, we need to know the sign of D u h(5, 8), 3., 3.. Since D u h(5, 8) is positive, you will ascend if you go due south. (b) We need D u h(5, 8) where u,, which is D u h(5, 8), 3.,.. Since this is negative, you will descend if you head northwest. (c) The hill is the steepest in the direction of h(5, 8), 3. (in terms of cardinal directions, this is 7.65 south of west) 6. (a) A general form for the equation of the tangent plane at (,, ) is f x (, )(x ) + f y (, )(y ( )) (z ) Solving for f x (, ) and f y (, ), and plugging these in gives 8(x ) + 4(y + ) (z ) which is 8x + 4y z, or z 8x + 4y +. (b) The function f has horizontal tangent planes at any points where f x and f y are both. We have f x 6x + and f y y, so f x when x 3, and f y when y. Thus, f has one horizontal tangent place at the point ( 3,, f( 3, )) ( 3,, 3 ). 7. (a) A general form for the equation of the tangent plane at (3,, 3e ) is g x (3, )(x 3) + g y (3, )(y ) (z 3e ) Solving for g x (3, ) and g y (3, ), and plugging these in gives 9e (x 3) + 6e (y ) (z 3e ) which is 9e x + 6e y z 6e, or z e (9x + 6y 6).
3 (b) As in the previous problem, we look for points where both g x and g y are. We have g x (x + )e x +y and g y xye x +y. Since e x +y is never, the only way that g x can be is if x +. However, no value of x can make this equation true. So, g x is never, and hence there are no points where g has a horizontal tangent plane. 8. A general form for the linearization at (, ) is L(x, y) f x (, )(x ) + f y (, )(y + ) + f(, ) Note that f(, ). Since f x x sec (x y ) and f y y sec (x y ), we get f x (, ) and f y (, ). So, the linearization is L(x, y) (x ) + (y + ) + x + y We have tan(. (.96) ) L(.,.96). (the actual value of tan(. (.96) ) tan(.985) is about.99) 9. (a) We have f x x+4 and f y y 6, so the only critical point is (, 3). Furthermore, f xx, f yy, and f xy, so D f xx f yy f xy ] 4. At (, 3), we have D > and f xx >, so by the Second Partial Derivative Test there is a local minimum at (, 3) (b) We have g x x y + 9 and g y x + y 6, so g x when y x + 9. Plugging this into g y gives g y 3x +, so g y when x 4. Backsolving, we get y. Thus, there is one critical point at ( 4, ). At this point, g xx > and D g xx g yy g xy ] 3 >, so there is a local minimum at ( 4, ). (c) We have h x xe y/ and h y x e y/ + e y/ + yey/. Since e y/ is never, h x when x. Plugging this into h y gives h y e y/ + yey/ e y/ ( + y). So, h y when + y, i.e. when y. So, there is one critical point: (, ). Now, h xx e y/, h yy 4 x e y/ + ey/ + ey/ + 4 ey/, and h xy xe y/. At (, ), we have h xx (, ) e, h yy e, and h xy. Thus, D h xx h yy h xy ] is negative, so h has a saddle point at (, ). (d) We have k x 3x 6 and k y 4y 3 4y, so k x when x ± and k y when y,, or. This means that there are six critical points: (, ), (, ), (, ) (, ), (, ), (, ) Furthermore, k xx 6x, k yy y 4, and k xy. Hence, D 6x(y 4). The conclusions we reach are summarized in the following table Point D k xx Conclusion (, ) 4 6 saddle point (, ) 48 6 local minimum (, ) 48 6 local minimum (, ) 4 6 local maximum (, ) 48 6 saddle point (, ) 48 6 saddle point
4 . Since f x, and f y, f has no critical points. So, the absolute extrema of f occur on the boundary of S. The boundary has two pieces: y 4 x (for x ) and y (for x ). Along y 4 x, we have f(x, 4 x ) x (4 x ) x + x 4, which makes f a function of one variable. So, we need to find the extreme values of f(x) x + x 4 on, ]. We have f (x) x +, so f(x) has a critical point at x. Evaluating f(x) at this critical point and at the endpoints of, ], we have f( ) 4, f( ) 5, and f() 4. Along y, we have f(x, ) x. On the interval, ], the maximum value of x is 4 and the minimum value is 4. Comparing the extreme values of f along y 4 x with the extreme values of f along y, we see that the maximum value of f on S is 4 while the minimum value is (a) x 3xy + y da x 3xy + y dy dx (b) xy x + y da 48 ] y4 x y 3 xy + 3 y3 dx 4x 4x dx 4 3 x3 x x ] x x 3 4 xy x + y dy dx u x + y du ydy when y, u x when y 4, u x x +6 x x u du dx ] x ux +6 3 u3/ ux dx y x(x + 6) 3/ x(x ) 3/ dx x(x + 6) 3/ x 4 dx 5 (x + 6) 5/ 5 x5 ] x3 x
5 (c) π/ x sin y dx dy π/ π/ ] x 3 x3 sin y dy x 8 sin y dy 3 ] yπ/ 83 cos y y 8 3 (d) Integrating with respect to x first will require integration by parts, so we switch the order of integration: xe xy+ dx dy (e) The region can be described as either xe xy+ dy dx u xy + du x dy when y, u when y, u x + x+ e u du dx ] ux+ e u dx u e x+ e dx e x+ ex e e ] x x, y x x or y, x y so we can evaluate either or x y ln(x + ) dy dx () ln(x + ) dx dy () Integral () can be done, but we will need to use integration by parts to handle ln(x + ) dx. So, we will compute integral () (as we ll see, integration by parts
6 will also be required for (), but it will be a little easier). We have x ] yx ln(x + ) dy dx y ln(x + ) dx x ln(x + ) dx u x + y du x dx when x, u when x, u ln u du From here, we use integration by parts (with w instead of u, since we used u above): (f) The region is w ln u dw u du ln u du dv du v u ln u du u ln u ] u u ln ] ( ln ) ln u du ] u u ] ] which is horizontally simple and can be described as y, y x 8 y. So, we have 8 y ] x8 y y da y dx dy xy dy y xy 8 y 3 dy ] y 8y y4 8 y
7 (g) We can t integrate is ln y dy, so we need to switch the order of integration. The region and can be described as x ln, e x y. We want to change the way we describe the region so that y is bounded by constants. The curve y e x is the same as x ln y, so the region is Thus, ln y, x ln y e x ln y dy dx ln y x ln y ] y dx dy ln y ] xln y (h) We can t integrate e x dx, so we need to switch the order of integration. The region is y dy y x dy and can be described as y, 3y x 3. We want to change the way we describe the region so that x is bounded by constants. The curve x 3y is the same as y 3 x, so the region is x 3, y 3 x
8 Thus, 3 3y e x dx dy x 3 3 e x dy dx ye x] y 3 x dx y 3 xex dx u x du x dx when x, u when x 3, u 9 9 e u du e u ] u9 u 6 (e9 ) (i) The region is given in terms of polar coordinates, so we should integrate using polar coordinates. Using the substitutions x r cos θ and da r dr dθ, we have π/ sin θ 3x da 3(r cos θ)(r dr dθ) π/4 sin θ π/ sin θ π/4 sin θ π/ π/4 π/ π/4 π/ π/4 3r cos θ dr dθ) ] r sin θ cos θ r 3 dθ) rsin θ cos θ(8 sin 3 θ sin 3 θ) dθ 7 cos θ sin 3 θ dθ u sin θ du cos θ dθ when θ π/4, u / when θ π/, u 7u 3 du / 7 6 ] u 4 u4 u /
9 (j) The region is best described using polar coordinates: r, π θ 3π 4 Performing the substitutions r x + y and da r dr dθ, we get 3π/4 cos(x + y ) da cos(r )r dr dθ π/ 3π/4 π/ u r r cos(r ) dr dθ du r dr when r, u when r, u 4 3π/4 4 cos(u) du dθ π/ 3π/4 ] u4 sin u π/ u 3π/4 sin 4 dθ π/ ] θ3π/4 (sin 4)θ θπ/ π 8 sin 4 (k) The region is best described using polar coordinates: r 5, θ π Performing the substitutions r x + y and da r dr dθ, we get π 5 5 x y da (5 r )r dr dθ π 5 π π dθ 5r r 3 dr dθ ] r5 5 r 4 r4 dθ r 44 4 dθ 44 4 θ ] θπ θ 44π
Math 10C  Fall Final Exam
Math 1C  Fall 217  Final Exam Problem 1. Consider the function f(x, y) = 1 x 2 (y 1) 2. (i) Draw the level curve through the point P (1, 2). Find the gradient of f at the point P and draw the gradient
More informationMath 234 Final Exam (with answers) Spring 2017
Math 234 Final Exam (with answers) pring 217 1. onsider the points A = (1, 2, 3), B = (1, 2, 2), and = (2, 1, 4). (a) [6 points] Find the area of the triangle formed by A, B, and. olution: One way to solve
More informationMath 147 Exam II Practice Problems
Math 147 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also rework all examples given in lecture, all homework problems, all lab
More information1. For each function, find all of its critical points and then classify each point as a local extremum or saddle point.
Solutions Review for Exam # Math 6. For each function, find all of its critical points and then classify each point as a local extremum or saddle point. a fx, y x + 6xy + y Solution.The gradient of f is
More informationDO NOT BEGIN THIS TEST UNTIL INSTRUCTED TO START
Math 265 Student name: KEY Final Exam Fall 23 Instructor & Section: This test is closed book and closed notes. A (graphing) calculator is allowed for this test but cannot also be a communication device
More informationMath 131 Exam 2 Spring 2016
Math 3 Exam Spring 06 Name: ID: 7 multiple choice questions worth 4.7 points each. hand graded questions worth 0 points each. 0. free points (so the total will be 00). Exam covers sections.7 through 3.0
More informationx + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the
1.(8pts) Find F ds where F = x + ye z + ze y, y + xe z + ze x, z and where T is the T surface in the pictures. (The two pictures are two views of the same surface.) The boundary of T is the unit circle
More informationMath Review for Exam Compute the second degree Taylor polynomials about (0, 0) of the following functions: (a) f(x, y) = e 2x 3y.
Math 35  Review for Exam 1. Compute the second degree Taylor polynomial of f e x+3y about (, ). Solution. A computation shows that f x(, ), f y(, ) 3, f xx(, ) 4, f yy(, ) 9, f xy(, ) 6. The second degree
More informationSolutions to Exam 2, Math 10560
Solutions to Exam, Math 6. Which of the following expressions gives the partial fraction decomposition of the function x + x + f(x = (x (x (x +? Solution: Notice that (x is not an irreducile factor. If
More informationMath 226 Calculus Spring 2016 Practice Exam 1. (1) (10 Points) Let the differentiable function y = f(x) have inverse function x = f 1 (y).
Math 6 Calculus Spring 016 Practice Exam 1 1) 10 Points) Let the differentiable function y = fx) have inverse function x = f 1 y). a) Write down the formula relating the derivatives f x) and f 1 ) y).
More information1 4 (1 cos(4θ))dθ = θ 4 sin(4θ)
M48M Final Exam Solutions, December 9, 5 ) A polar curve Let C be the portion of the cloverleaf curve r = sin(θ) that lies in the first quadrant a) Draw a rough sketch of C This looks like one quarter
More informationCalculus III. Math 233 Spring Final exam May 3rd. Suggested solutions
alculus III Math 33 pring 7 Final exam May 3rd. uggested solutions This exam contains twenty problems numbered 1 through. All problems are multiple choice problems, and each counts 5% of your total score.
More informationMth Review Problems for Test 2 Stewart 8e Chapter 3. For Test #2 study these problems, the examples in your notes, and the homework.
For Test # study these problems, the examples in your notes, and the homework. Derivative Rules D [u n ] = nu n 1 du D [ln u] = du u D [log b u] = du u ln b D [e u ] = e u du D [a u ] = a u ln a du D [sin
More informationMATH 162. Midterm Exam 1  Solutions February 22, 2007
MATH 62 Midterm Exam  Solutions February 22, 27. (8 points) Evaluate the following integrals: (a) x sin(x 4 + 7) dx Solution: Let u = x 4 + 7, then du = 4x dx and x sin(x 4 + 7) dx = 4 sin(u) du = 4 [
More informationTangent Plane. Linear Approximation. The Gradient
Calculus 3 Lia Vas Tangent Plane. Linear Approximation. The Gradient The tangent plane. Let z = f(x, y) be a function of two variables with continuous partial derivatives. Recall that the vectors 1, 0,
More informationMath 180, Exam 2, Practice Fall 2009 Problem 1 Solution. f(x) = arcsin(2x + 1) = sin 1 (3x + 1), lnx
Math 80, Exam, Practice Fall 009 Problem Solution. Differentiate the functions: (do not simplify) f(x) = x ln(x + ), f(x) = xe x f(x) = arcsin(x + ) = sin (3x + ), f(x) = e3x lnx Solution: For the first
More informationMath 180, Final Exam, Fall 2012 Problem 1 Solution
Math 80, Final Exam, Fall 0 Problem Solution. Find the derivatives of the following functions: (a) ln(ln(x)) (b) x 6 + sin(x) e x (c) tan(x ) + cot(x ) (a) We evaluate the derivative using the Chain Rule.
More informationMath 2163, Practice Exam II, Solution
Math 63, Practice Exam II, Solution. (a) f =< f s, f t >=< s e t, s e t >, an v v = , so D v f(, ) =< ()e, e > =< 4, 4 > = 4. (b) f =< xy 3, 3x y 4y 3 > an v =< cos π, sin π >=, so
More informationM273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3
M7Q Multivariable alculus Spring 7 Review Problems for Exam Exam covers material from Sections 5.5.4 and 6.6. and 7.. As you prepare, note well that the Fall 6 Exam posted online did not cover exactly
More informationCalculus I Review Solutions
Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.
More informationPartial Derivatives for Math 229. Our puropose here is to explain how one computes partial derivatives. We will not attempt
Partial Derivatives for Math 229 Our puropose here is to explain how one computes partial derivatives. We will not attempt to explain how they arise or why one would use them; that is left to other courses
More informationFinal Exam 2011 Winter Term 2 Solutions
. (a Find the radius of convergence of the series: ( k k+ x k. Solution: Using the Ratio Test, we get: L = lim a k+ a k = lim ( k+ k+ x k+ ( k k+ x k = lim x = x. Note that the series converges for L
More information3 Applications of partial differentiation
Advanced Calculus Chapter 3 Applications of partial differentiation 37 3 Applications of partial differentiation 3.1 Stationary points Higher derivatives Let U R 2 and f : U R. The partial derivatives
More informationMATH 10550, EXAM 2 SOLUTIONS. 1. Find an equation for the tangent line to. f(x) = sin x cos x. 2 which is the slope of the tangent line at
MATH 100, EXAM SOLUTIONS 1. Find an equation for the tangent line to at the point ( π 4, 0). f(x) = sin x cos x f (x) = cos(x) + sin(x) Thus, f ( π 4 ) = which is the slope of the tangent line at ( π 4,
More informationPage Problem Score Max Score a 8 12b a b 10 14c 6 6
Fall 14 MTH 34 FINAL EXAM December 8, 14 Name: PID: Section: Instructor: DO NOT WRITE BELOW THIS LINE. Go to the next page. Page Problem Score Max Score 1 5 5 1 3 5 4 5 5 5 6 5 7 5 8 5 9 5 1 5 11 1 3 1a
More information1 + x 2 d dx (sec 1 x) =
Page This exam has: 8 multiple choice questions worth 4 points each. hand graded questions worth 4 points each. Important: No graphing calculators! Any nongraphing, nondifferentiating, nonintegrating
More informationSolutions to Homework 5
Solutions to Homework 5 1. Let z = f(x, y) be a twice continuously differentiable function of x and y. Let x = r cos θ and y = r sin θ be the equations which transform polar coordinates into rectangular
More informationMath 131 Final Exam Spring 2016
Math 3 Final Exam Spring 06 Name: ID: multiple choice questions worth 5 points each. Exam is only out of 00 (so there is the possibility of getting more than 00%) Exam covers sections. through 5.4 No graphing
More informationEXAM. Exam #1. Math 3350 Summer II, July 21, 2000 ANSWERS
EXAM Exam #1 Math 3350 Summer II, 2000 July 21, 2000 ANSWERS i 100 pts. Problem 1. 1. In each part, find the general solution of the differential equation. dx = x2 e y We use the following sequence of
More informationSolutions to Exam 1, Math Solution. Because f(x) is onetoone, we know the inverse function exists. Recall that (f 1 ) (a) =
Solutions to Exam, Math 56 The function f(x) e x + x 3 + x is onetoone (there is no need to check this) What is (f ) ( + e )? Solution Because f(x) is onetoone, we know the inverse function exists
More informationMath 210, Final Exam, Fall 2010 Problem 1 Solution. v cosθ = u. v Since the magnitudes of the vectors are positive, the sign of the dot product will
Math, Final Exam, Fall Problem Solution. Let u,, and v,,3. (a) Is the angle between u and v acute, obtuse, or right? (b) Find an equation for the plane through (,,) containing u and v. Solution: (a) The
More informationFinal Exam Review Quesitons
Final Exam Review Quesitons. Compute the following integrals. (a) x x 4 (x ) (x + 4) dx. The appropriate partial fraction form is which simplifies to x x 4 (x ) (x + 4) = A x + B (x ) + C x + 4 + Dx x
More informationName Date Period. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
AB Fall Final Exam Review 20020 Name Date Period MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. ) The position of a particle
More informationMath 23b Practice Final Summer 2011
Math 2b Practice Final Summer 211 1. (1 points) Sketch or describe the region of integration for 1 x y and interchange the order to dy dx dz. f(x, y, z) dz dy dx Solution. 1 1 x z z f(x, y, z) dy dx dz
More informationFall 2009 Math 113 Final Exam Solutions. f(x) = 1 + ex 1 e x?
. What are the domain and range of the function Fall 9 Math 3 Final Exam Solutions f(x) = + ex e x? Answer: The function is welldefined everywhere except when the denominator is zero, which happens when
More informationPractice problems for Exam 1. a b = (2) 2 + (4) 2 + ( 3) 2 = 29
Practice problems for Exam.. Given a = and b =. Find the area of the parallelogram with adjacent sides a and b. A = a b a ı j k b = = ı j + k = ı + 4 j 3 k Thus, A = 9. a b = () + (4) + ( 3)
More information(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)
eview Exam Math 43 Name Id ead each question carefully. Avoid simple mistakes. Put a box around the final answer to a question (use the back of the page if necessary). For full credit you must show your
More informationMAT 122 Homework 7 Solutions
MAT 1 Homework 7 Solutions Section 3.3, Problem 4 For the function w = (t + 1) 100, we take the inside function to be z = t + 1 and the outside function to be z 100. The derivative of the inside function
More informationMAS113 CALCULUS II SPRING 2008, QUIZ 5 SOLUTIONS. x 2 dx = 3y + y 3 = x 3 + c. It can be easily verified that the differential equation is exact, as
MAS113 CALCULUS II SPRING 008, QUIZ 5 SOLUTIONS Quiz 5a Solutions (1) Solve the differential equation y = x 1 + y. (1 + y )y = x = (1 + y ) = x = 3y + y 3 = x 3 + c. () Solve the differential equation
More information2015 Math Camp Calculus Exam Solution
015 Math Camp Calculus Exam Solution Problem 1: x = x x +5 4+5 = 9 = 3 1. lim We also accepted ±3, even though it is not according to the prevailing convention 1. x x 4 x+4 =. lim 4 4+4 = 4 0 = 4 0 = We
More informationc) xy 3 = cos(7x +5y), y 0 = y3 + 7 sin(7x +5y) 3xy sin(7x +5y) d) xe y = sin(xy), y 0 = ey + y cos(xy) x(e y cos(xy)) e) y = x ln(3x + 5), y 0
Some Math 35 review problems With answers 2/6/2005 The following problems are based heavily on problems written by Professor Stephen Greenfield for his Math 35 class in spring 2005. His willingness to
More informationSOLUTIONS FOR PRACTICE FINAL EXAM
SOLUTIONS FOR PRACTICE FINAL EXAM ANDREW J. BLUMBERG. Solutions () Short answer questions: (a) State the mean value theorem. Proof. The mean value theorem says that if f is continuous on (a, b) and differentiable
More informationTurn off all cell phones, pagers, radios, mp3 players, and other similar devices.
Math 25 B and C Midterm 2 Palmieri, Autumn 26 Your Name Your Signature Student ID # TA s Name and quiz section (circle): Cady Cruz Jacobs BA CB BB BC CA CC Turn off all cell phones, pagers, radios, mp3
More information1. The graph of a function f is given above. Answer the question: a. Find the value(s) of x where f is not differentiable. Ans: x = 4, x = 3, x = 2,
1. The graph of a function f is given above. Answer the question: a. Find the value(s) of x where f is not differentiable. x = 4, x = 3, x = 2, x = 1, x = 1, x = 2, x = 3, x = 4, x = 5 b. Find the value(s)
More informationSpring 2015 Sample Final Exam
Math 1151 Spring 2015 Sample Final Exam Final Exam on 4/30/14 Name (Print): Time Limit on Final: 105 Minutes Go on carmen.osu.edu to see where your final exam will be. NOTE: This exam is much longer than
More informationIntegration by Substitution
November 22, 2013 Introduction 7x 2 cos(3x 3 )dx =? 2xe x2 +5 dx =? Chain rule The chain rule: d dx (f (g(x))) = f (g(x)) g (x). Use the chain rule to find f (x) and then write the corresponding antidifferentiation
More informationDEPARTMENT OF MATHEMATICS AND STATISTICS UNIVERSITY OF MASSACHUSETTS. MATH 233 SOME SOLUTIONS TO EXAM 2 Fall 2018
DEPARTMENT OF MATHEMATICS AND STATISTICS UNIVERSITY OF MASSACHUSETTS MATH 233 SOME SOLUTIONS TO EXAM 2 Fall 208 Version A refers to the regular exam and Version B to the makeup. Version A. A particle
More informationSolutions to old Exam 3 problems
Solutions to old Exam 3 problems Hi students! I am putting this version of my review for the Final exam review here on the web site, place and time to be announced. Enjoy!! Best, Bill Meeks PS. There are
More informationSec. 14.3: Partial Derivatives. All of the following are ways of representing the derivative. y dx
Math 2204 Multivariable Calc Chapter 14: Partial Derivatives I. Review from math 1225 A. First Derivative Sec. 14.3: Partial Derivatives 1. Def n : The derivative of the function f with respect to the
More informationMath 101 Fall 2006 Exam 1 Solutions Instructor: S. Cautis/M. Simpson/R. Stong Thursday, October 5, 2006
Math 101 Fall 2006 Exam 1 Solutions Instructor: S. Cautis/M. Simpson/R. Stong Thursday, October 5, 2006 Instructions: This is a closed book, closed notes exam. Use of calculators is not permitted. You
More informationx+1 e 2t dt. h(x) := Find the equation of the tangent line to y = h(x) at x = 0.
Math Sample final problems Here are some problems that appeared on past Math exams. Note that you will be given a table of Zscores for the standard normal distribution on the test. Don t forget to have
More informationReview for Exam 1. (a) Find an equation of the line through the point ( 2, 4, 10) and parallel to the vector
Calculus 3 Lia Vas Review for Exam 1 1. Surfaces. Describe the following surfaces. (a) x + y = 9 (b) x + y + z = 4 (c) z = 1 (d) x + 3y + z = 6 (e) z = x + y (f) z = x + y. Review of Vectors. (a) Let a
More informatione x3 dx dy. 0 y x 2, 0 x 1.
Problem 1. Evaluate by changing the order of integration y e x3 dx dy. Solution:We change the order of integration over the region y x 1. We find and x e x3 dy dx = y x, x 1. x e x3 dx = 1 x=1 3 ex3 x=
More informationMath 265H: Calculus III Practice Midterm II: Fall 2014
Name: Section #: Math 65H: alculus III Practice Midterm II: Fall 14 Instructions: This exam has 7 problems. The number of points awarded for each question is indicated in the problem. Answer each question
More informationIf y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u
Section 3 4B The Chain Rule If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy du du dx or If y = f (u) then f (u) u The Chain Rule with the Power
More informationReview for the First Midterm Exam
Review for the First Midterm Exam Thomas Morrell 5 pm, Sunday, 4 April 9 B9 Van Vleck Hall For the purpose of creating questions for this review session, I did not make an effort to make any of the numbers
More informationReview For the Final: Problem 1 Find the general solutions of the following DEs. a) x 2 y xy y 2 = 0 solution: = 0 : homogeneous equation.
Review For the Final: Problem 1 Find the general solutions of the following DEs. a) x 2 y xy y 2 = 0 solution: y y x y2 = 0 : homogeneous equation. x2 v = y dy, y = vx, and x v + x dv dx = v + v2. dx =
More informationMath 152 Take Home Test 1
Math 5 Take Home Test Due Monday 5 th October (5 points) The following test will be done at home in order to ensure that it is a fair and representative reflection of your own ability in mathematics I
More informationLecture 10. (2) Functions of two variables. Partial derivatives. Dan Nichols February 27, 2018
Lecture 10 Partial derivatives Dan Nichols nichols@math.umass.edu MATH 233, Spring 2018 University of Massachusetts February 27, 2018 Last time: functions of two variables f(x, y) x and y are the independent
More informationIf y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u
Section 3 4B Lecture The Chain Rule If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy du du dx or If y = f (u) then y = f (u) u The Chain Rule
More information10550 PRACTICE FINAL EXAM SOLUTIONS. x 2 4. x 2 x 2 5x +6 = lim x +2. x 2 x 3 = 4 1 = 4.
55 PRACTICE FINAL EXAM SOLUTIONS. First notice that x 2 4 x 2x + 2 x 2 5x +6 x 2x. This function is undefined at x 2. Since, in the it as x 2, we only care about what happens near x 2 an for x less than
More informationMath 1310 Final Exam
Math 1310 Final Exam December 11, 2014 NAME: INSTRUCTOR: Write neatly and show all your work in the space provided below each question. You may use the back of the exam pages if you need additional space
More informationMath 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3
Math 201 Solutions to Assignment 1 1. Solve the initial value problem: x 2 dx + 2y = 0, y(0) = 2. x 2 dx + 2y = 0, y(0) = 2 2y = x 2 dx y 2 = 1 3 x3 + C y = C 1 3 x3 Notice that y is not defined for some
More information1. (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
1. a) 5 points) Find the unit tangent and unit normal vectors T and N to the curve at the point P 3, 3π, r t) 3 cos t, 4t, 3 sin t 3 ). b) 5 points) Find curvature of the curve at the point P. olution:
More informationReview Problems for Test 1
Review Problems for Test Math 603/06 9 9/0 007 These problems are provided to help you study The presence of a problem on this handout does not imply that there will be a similar problem on the test And
More informationx 2 y = 1 2. Problem 2. Compute the Taylor series (at the base point 0) for the function 1 (1 x) 3.
MATH 8.0  FINAL EXAM  SOME REVIEW PROBLEMS WITH SOLUTIONS 8.0 Calculus, Fall 207 Professor: Jared Speck Problem. Consider the following curve in the plane: x 2 y = 2. Let a be a number. The portion of
More informationMAY THE FORCE BE WITH YOU, YOUNG JEDIS!!!
Final Exam Math 222 Spring 2011 May 11, 2011 Name: Recitation Instructor s Initials: You may not use any type of calculator whatsoever. (Cell phones off and away!) You are not allowed to have any other
More informationfor any C, including C = 0, because y = 0 is also a solution: dy
Math 3200001 Fall 2014 Practice exam 1 solutions 2/16/2014 Each problem is worth 0 to 4 points: 4=correct, 3=small error, 2=good progress, 1=some progress 0=nothing relevant. If the result is correct,
More informationJim 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
Jim Lambers MAT 28 ummer emester 2121 Practice Final Exam olution 1. Evaluate the line integral xy dx + e y dy + xz dz, where is given by r(t) t 4, t 2, t, t 1. olution From r (t) 4t, 2t, t 2, we obtain
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More informationAnswer sheet: Final exam for Math 2339, Dec 10, 2010
Answer sheet: Final exam for Math 9, ec, Problem. Let the surface be z f(x,y) ln(y + cos(πxy) + e ). (a) Find the gradient vector of f f(x,y) y + cos(πxy) + e πy sin(πxy), y πx sin(πxy) (b) Evaluate f(,
More informationWeBWorK, Problems 2 and 3
WeBWorK, Problems 2 and 3 7 dx 2. Evaluate x ln(6x) This can be done using integration by parts or substitution. (Most can not). However, it is much more easily done using substitution. This can be written
More informationMath 21B  Homework Set 8
Math B  Homework Set 8 Section 8.:. t cos t dt Let u t, du t dt and v sin t, dv cos t dt Let u t, du dt and v cos t, dv sin t dt t cos t dt u v v du t sin t t sin t dt [ t sin t u v ] v du [ ] t sin t
More informationMath 53 Homework 7 Solutions
Math 5 Homework 7 Solutions Section 5.. To find the mass of the lamina, we integrate ρ(x, y over the box: m a b a a + x + y dy y + x y + y yb y b + bx + b bx + bx + b x ab + a b + ab a b + ab + ab. We
More informationENGI Partial Differentiation Page y f x
ENGI 3424 4 Partial Differentiation Page 401 4. Partial Differentiation For functions of one variable, be found unambiguously by differentiation: y f x, the rate of change of the dependent variable can
More informationAPPM 2350 FINAL EXAM FALL 2017
APPM 25 FINAL EXAM FALL 27. ( points) Determine the absolute maximum and minimum values of the function f(x, y) = 2 6x 4y + 4x 2 + y. Be sure to clearly give both the locations and values of the absolute
More informationSolution. This is a routine application of the chain rule.
EXAM 2 SOLUTIONS 1. If z = e r cos θ, r = st, θ = s 2 + t 2, find the partial derivatives dz ds chain rule. Write your answers entirely in terms of s and t. dz and dt using the Solution. This is a routine
More informationSolutions to Math 41 Final Exam December 10, 2012
Solutions to Math 4 Final Exam December,. ( points) Find each of the following limits, with justification. If there is an infinite limit, then explain whether it is or. x ln(t + ) dt (a) lim x x (5 points)
More informationName: Instructor: 1. a b c d e. 15. a b c d e. 2. a b c d e a b c d e. 16. a b c d e a b c d e. 4. a b c d e... 5.
Name: Instructor: Math 155, Practice Final Exam, December The Honor Code is in effect for this examination. All work is to be your own. No calculators. The exam lasts for 2 hours. Be sure that your name
More informationChapter 12 Overview: Review of All Derivative Rules
Chapter 12 Overview: Review of All Derivative Rules The emphasis of the previous chapters was graphing the families of functions as they are viewed (mostly) in Analytic Geometry, that is, with traits.
More informationFall 2013 Hour Exam 2 11/08/13 Time Limit: 50 Minutes
Math 8 Fall Hour Exam /8/ Time Limit: 5 Minutes Name (Print): This exam contains 9 pages (including this cover page) and 7 problems. Check to see if any pages are missing. Enter all requested information
More informationt 2 + 2t dt = (t + 1) dt + 1 = arctan t x + 6 x(x 3)(x + 2) = A x +
MATH 06 0 Practice Exam #. (0 points) Evaluate the following integrals: (a) (0 points). t +t+7 This is an irreducible quadratic; its denominator can thus be rephrased via completion of the square as a
More informationSolution: APPM 1350 Final Exam Spring 2014
APPM 135 Final Exam Spring 214 1. (a) (5 pts. each) Find the following derivatives, f (x), for the f given: (a) f(x) = x 2 sin 1 (x 2 ) (b) f(x) = 1 1 + x 2 (c) f(x) = x ln x (d) f(x) = x x d (b) (15 pts)
More informationMath Exam 02 Review
Math 10350 Exam 02 Review 1. A differentiable function g(t) is such that g(2) = 2, g (2) = 1, g (2) = 1/2. (a) If p(t) = g(t)e t2 find p (2) and p (2). (Ans: p (2) = 7e 4 ; p (2) = 28.5e 4 ) (b) If f(t)
More informationMath 265 (Butler) Practice Midterm III B (Solutions)
Math 265 (Butler) Practice Midterm III B (Solutions). Set up (but do not evaluate) an integral for the surface area of the surface f(x, y) x 2 y y over the region x, y 4. We have that the surface are is
More informationIntroduction to Differential Equations
Chapter 1 Introduction to Differential Equations 1.1 Basic Terminology Most of the phenomena studied in the sciences and engineering involve processes that change with time. For example, it is well known
More informationSection 5.6 Integration by Parts
.. 98 Section.6 Integration by Parts Integration by parts is another technique that we can use to integrate problems. Typically, we save integration by parts as a last resort when substitution will not
More informationChapter 7: Techniques of Integration
Chapter 7: Techniques of Integration MATH 20601: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration 7.1. Integration
More informationSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Final Exam Review Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Match the differential equation with the appropriate slope field. 1) y = x
More informationMath 121. Exam II. November 28 th, 2018
Math 121 Exam II November 28 th, 2018 Name: Section: The following rules apply: This is a closedbook exam. You may not use any books or notes on this exam. For free response questions, you must show all
More informationIntegration by Parts
Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u
More informationz 2 = 1 4 (x 2) + 1 (y 6)
MA 5 Fall 007 Exam # Review Solutions. Consider the function fx, y y x. a Sketch the domain of f. For the domain, need y x 0, i.e., y x.    0 0    b Sketch the level curves fx, y k for k 0,,,. The
More informationMath 212Lecture 8. The chain rule with one independent variable
Math 212Lecture 8 137: The multivariable chain rule The chain rule with one independent variable w = f(x, y) If the particle is moving along a curve x = x(t), y = y(t), then the values that the particle
More information= π + sin π = π + 0 = π, so the object is moving at a speed of π feet per second after π seconds. (c) How far does it go in π seconds?
Mathematics 115 Professor Alan H. Stein April 18, 005 SOLUTIONS 1. Define what is meant by an antiderivative or indefinite integral of a function f(x). Solution: An antiderivative or indefinite integral
More information1. Find and classify the extrema of h(x, y) = sin(x) sin(y) sin(x + y) on the square[0, π] [0, π]. (Keep in mind there is a boundary to check out).
. Find and classify the extrema of hx, y sinx siny sinx + y on the square[, π] [, π]. Keep in mind there is a boundary to check out. Solution: h x cos x sin y sinx + y + sin x sin y cosx + y h y sin x
More informationTest one Review Cal 2
Name: Class: Date: ID: A Test one Review Cal 2 Short Answer. Write the following expression as a logarithm of a single quantity. lnx 2ln x 2 ˆ 6 2. Write the following expression as a logarithm of a single
More informationMA261A Calculus III 2006 Fall Midterm 2 Solutions 11/8/2006 8:00AM ~9:15AM
MA6A Calculus III 6 Fall Midterm Solutions /8/6 8:AM ~9:5AM. Find the it xy cos y (x;y)(;) 3x + y, if it exists, or show that the it does not exist. Assume that x. The it becomes (;y)(;) y cos y 3 + y
More informationCHAIN RULE: DAY 2 WITH TRIG FUNCTIONS. Section 2.4A Calculus AP/Dual, Revised /30/2018 1:44 AM 2.4A: Chain Rule Day 2 1
CHAIN RULE: DAY WITH TRIG FUNCTIONS Section.4A Calculus AP/Dual, Revised 018 viet.dang@humbleisd.net 7/30/018 1:44 AM.4A: Chain Rule Day 1 THE CHAIN RULE A. d dx f g x = f g x g x B. If f(x) is a differentiable
More informationDifferential Equations: Homework 2
Differential Equations: Homework Alvin Lin January 08  May 08 Section.3 Exercise The direction field for provided x 0. dx = 4x y is shown. Verify that the straight lines y = ±x are solution curves, y
More informationMATH1013 Calculus I. Derivatives II (Chap. 3) 1
MATH1013 Calculus I Derivatives II (Chap. 3) 1 Edmund Y. M. Chiang Department of Mathematics Hong Kong University of Science & Technology October 16, 2013 2013 1 Based on Briggs, Cochran and Gillett: Calculus
More information