Math Exam III - Fall 2011

Transcription

1 Math Exam III - Fall 211 November 16, Renato Feres NAME: STUDENT ID NUMBER: General instructions: This exam has 16 questions, each worth the same amount. Check that no pages are missing and notify your proctor if you detect any problems with your copy of the exam. Mark your ID number on the six blank lines on the top of your answer card, using one line for each digit. Print your name on the top of the card. Choose the answer that is closest to the solution and mark your answer card with a PENCIL by shading in the correct box. You may use a 3 5 card with notes and any calculator that does not have graphing functions. GOOD LUCK! 1. Find the second order partial derivative f xy (π, 1) where f(x, y) = x 2 y + cos y + y sin x. 13.3, 43 (A) π (B) 2π + 1 (C) 2π 1 (D) 2π (E) π + 3 (F) π 3 (G) π + 1 (H) π 1 (I) 2π (J) 3π 1

2 Solution: The first derivative in x is f x (x, y) = 2xy + y cos x and the mixed derivative is f xy (x, y) = 2x + cosx. Evaluating at (π, 1): f xy (π, 1) = 2π 1. 2

3 2. Evaluate w/ u at (u, v) = (, 1), where w = xy + yz + xz and x = u + v, y = u v, z = uv. 13.4, 9 (A) 4 (B) 3 (C) 2 (D) 1 (E) (F) 1 (G) 2 (H) 3 (I) 4 (J) 5 3

4 Solution: By the chain rule, The partials are: w u = w x x u + w y y u + w z z u. This gives w x = y + z, w y = x + z, w z = x + y, x u = y u = 1, z u = v. w u = (x + y)(1 + v) + 2z = 2u(1 + v) + 2uv = 2u(1 + 2v). Evaluating at (u, v) = (, 1) we get w u =. 4

5 3. Find the derivative of the function f(x, y) = 2xy 3y 2 at P (5, 5) in the direction of the vector u = 4i + 3j. 13.5, 11 (A) 4 (B) 3 (C) 2 (D) 1 (E) (F) 1 (G) 2 (H) 3 (I) 4 (J) 5 5

6 Solution: The gradient of f at P is ( f) P = 2y, 2x 6y P = 1, 2. Then the directional derivative in the direction of u (equivalently, in the direction of the unit vector u/ u = 4/5, 3/5 ) is D u = ( f) P u = 1, 2 4/5, 3/5 = 8 12 = 4. u 6

7 4. Find an equation for the tangent plane to the surface represented by z = y x at the point (x, y, z) = (1, 2, 1). 13.6, 11 (A) x y + 2z = (B) x + y + 2z = 5 (C) 2x y + z = 1 (D) 3x y + 2z = 3 (E) x 2y + z = 1 (F) x y + 2z = 3 (G) x y + 2z = 1 (H) 2x 2y + z = 1 (I) x y z = 2 (J) x 3y + 2z = 3 7

8 Solution: The gradient of f(x, y, z) = y x z at P (1, 2, 1) is 1 ( f) P = 2 y x, y x, 12, 12, 1. P = Therefore, the plane has equation 12, 12, 1 x 1, y 2, z 1 =. Evaluating the dot product and simplifying results in x y + 2z = 1. 8

9 5. Find the local maxima, minima, and saddle points, if any, of a function f(x, y) whose partial derivatives are f x = 9x 2 9, f y = 2y , 43 (A) local minimum at (1, 2), saddle at ( 1, 2) (B) local maximum at (1, 2), local maximum at ( 1, 2) (C) local maximum at (1, 2), local maximum at ( 1, 2) (D) saddle at (1, 2), local minimum at ( 1, 2) (E) saddle at (1, 2), local minimum at ( 1, 2) (F) local minimum at (1, 2), local maximum at ( 1, 2) (G) local minimum at (2, 1), local maximum at ( 2, 1) (H) local maximum at (1, 2), saddle at ( 1, 2) (I) local maximum at (2, 2), saddle at ( 2, 2) (J) local minimum at (1, 2), saddle at ( 1, 2) 9

10 Solution: To find the critical points we set 9x 2 9 =, 2y + 4 =. Therefore, the critical points are ( 1, 2), (1, 2). The determinant H involved in the second derivative test is H = f xx f xy = 18x 2 = 36x. f yx f yy At ( 1, 2) H is negative, so the function has a saddle at ( 1, 2). At (1, 2) H is positive and f xx is positive, so f has a local minimum at (1, 2). 1

11 6. Among all the points on the graph of z = 8 x 2 y 2 that lie above the plane 2x + 2y + z =, find the point farthest from the plane. 13.7, 49 (A) ( 1, 1, 6) (B) (,, 8) (C) (2, 1, 3) (D) (1, 2, 3) (E) ( 1, 2, 3) (F) (1, 2, 3) (G) (, 1, 7) (H) (1,, 7) (I) (1, 1, 6) (J) ( 1,, 7) 11

12 Solution: The distance from a point (x, y, z) to a plane through the origin with equation 2x + 2y + z = is 2x + 2y + z D(x, y, z) =. 3 If the point lies on the surface z = 8 x 2 y 2, then the distance is the following function on x and y: f(x, y) = 2x + 2y + 8 x2 y 2. 3 Where the surface lies above the plane the function f(x, y) is positive. Taking the partial derivatives and setting them equal to, f x (x, y) = (2 2x)/3 =, f y (x, y) = (2 2y)/3 =, gives x = y = 1, and z = 6. After a brief inspection of the graph we conclude that (1, 1, 6) is indeed the point of maximum. 12

13 7. Find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse x y2 9 = 1 with sides parallel to the coordinate axis. (The base is parallel to the x-axis and the height is parallel to the y-axis.) 13.8, 11 (A) base = 1 and height = 2 (B) base = 2 2 and height = 3 2 (C) base = 2 and height = 1 (D) base = 2 3 and height = 3 2 (E) base = 2 and height = 3 (F) base = 2 and height = 3 (G) base = 3 and height = 2 (H) base = 4 and height = 9 (I) base = 1 and height = 3 (J) base = 3 and height = 1 13

14 Solution: We use the method of Lagrange multipliers. Let f(x, y) = 4xy be the area of the inscribed rectangle and g(x, y) = (x/2) 2 + (y/3) 2 = 1 the constraint equation. Then the equation f = λ g yields the system of equations 4y = λx/2, 4x = 2λy/9. We first solve for λ: multiplying the respective sides of the two equations yields 4 2 xy = 9λ 2 xy. Clearly, the maximum happens for a positive value of xy, so after canceling xy on both sides and solving for λ we obtain λ = ±12. As x and y can be taken to be positive by the nature of the problem, take λ = 12. Replacing in one of the two top equations gives, x 2 = y 3. Substituting this into the constraint equation results in x = 2 and y = 3 2. Therefore, the rectangle has dimensions: base = 2 2 and height =

15 8. Find three real numbers whose sum is 6 and the sum of whose squares is as small as possible. The product of these three numbers is: 13.8, 25 (A) 4 (B) 2 (C) 2 (D) 4 (E) 6 (F) 8 (G) 1 (H) 12 (I) 14 (J) 16 15

16 Solution: We use the method of Lagrange multipliers. The function to minimize is f(x, y, z) = x 2 + y 2 + y 2 and the constraint is g(x, y, z) = x + y + z = 6. The vector equation f = λ g gives the system 2x = λ, 2y = λ, 2z = λ. Therefore, x = y = z, and as the sum is 6, then (x, y, z) = (2, 2, 2) and the product is 8. 16

17 9. Evaluate the iterated integral 1 1 y 1 + xy dxdy. 14.1, 7 (A) ln2 (B) ln 2 (C) ln 2 (D) ln2 (E) 2 + ln 2 (F) 2 1 ln2 (G) 3 ln 2 (H) 2 2 ln2 (I) ln3 (J) ln 3 17

18 Solution: For the middle integral, we use the u-substitution u = 1 + xy, so du = ydx. Then the integral I becomes I = 1 1+y 1 1 du dy = u 1 ln(1+y) dy = [(1 + y) ln(1 + y) (1 + y)] 1 = 2 ln

19 1. Integrate the function f(x, y) = y x over the triangular region cut from the first quadrant of the xy-plane by the line x + y = , 27 (A) 1/5 (B) 1/4 (C) 1/12 (D) 1/1 (E) 1/3 (F) 1/3 (G) 1/12 (H) 1/1 (I) 1/4 (J) 1/5 19

20 Solution: This integral can be written 1 1 x y x dy dx = 1 [ y 2 2 x1/2 y ] y=1 x y= dx = (1 x)2 (1 x) xdx This is easily integrated and gives the value 1/1. 2

21 11. Evaluate the integral π π x sin y y dy dx by first reversing the order of integration. 14.2, 47 (A) 2π (B) π (C) 6 (D) 4 (E) 2 (F) 2 (G) 4 (H) 6 (I) π (J) 2π 21

22 Solution: This amounts to integrating the function f(x, y) = y 1 sin y over the triangle with vertices (, ), (π, π), (, π). By reversing the order of integration, we obtain π π x sin y y dy dx = π y sin y y The last integral gives cosπ + cos = 2. dxdy = π sin y π y y dy = sin y dy. 22

23 12. The region whose area is given by the integral 1 x 1 2x dy dx x x/2 dy dx is a triangle. What are its vertices? (It is not necessary to compute the area.) 14.3, 17 (A) (, 1), (1, ), (2, 2) (B) (, ), ( 1, 2), (3, 1) (C) (, ), ( 1, 1), ( 2, 2) (D) ( 1, 1), ( 1, 2), (2, 2) (E) (, 1), ( 1, 3), (3, 2) (F) ( 1, ), ( 1, 1), (1, 2) (G) (, ), (1, ), ( 1, 2) (H) (, ), (2, 1), ( 1, 2) (I) (, ), (1, ), (, 2) (J) (, ), ( 1, ), ( 1, 2) 23

24 Solution: An inspection of the integrals shows that the triangular region is bounded by the lines y = 2x, y = x/2, and y = 1 x. The points of pairwise intersection are: (a) between y = 2x and y = x/2: x = and y = ; (b) between y = 2x and y = 1 x: x = 1 and y = 2; (c) between y = x/2 and y = 1 x: x = 2 and y = 1. Therefore, the vertices are (, ), ( 1, 2), and (2, 1). 24

25 13. Evaluate the integral ln 2 (ln 2) 2 y 2 x e 2 +y 2 dxdy by converting to polar coordinates. 14.4, 19 (A) π 2 ln 2 π (B) ( ln2) 2 (C) π (1 2 ln2) 2 (D) ( ln2)π (E) π 2 (F) 2π( ln2) (G) 1 (H) 1 (I) π/4 (J) 2π/3 25

26 Solution: x This amounts to the area integral of e 2 +y 2 = e r over the quarter disc of radius ln 2 in the positive quadrant. In polar coordinates, this is π/2 ln2 e r r dr dθ = π 2 ln2 e r r dr. The last integral can be solved by integration by parts: ln2 e r r dr = [re r ] ln 2 ln2 e r dr = 2 ln2 (2 1) = 2 ln2 1. This gives the value π (2 ln 2 1) for the original integral. 2 26

27 14. Find the average distance from a point P(x, y) in the disc x 2 + y 2 R 2 to the origin. 4.4, 35 (A) 1 2 (B) 2 3 (C) 1 4 R2 (D) 1 4 R (E) 2 5 R (F) 1 3 R2 (G) 2 3 R2 (H) 1 2 R (I) 1 3 R (J) 2 3 R 27

28 Solution: The area of the disc is πr 2 and the function to average over the disc is f(x, y) = x2 + y 2 = r. So the average value is 1 x2 + y A 2 da = 1 2π R r 2 dr dθ = 2 disc of radius R πr 2 3 R. 28

29 15. Find the volume of the region in the first octant bounded by the coordinate planes, the plane y + z = 2, and the cylinder x = 4 y , 25 (A) 1/3 (B) 2/3 (C) 5/3 (D) 1/3 (E) 2/3 (F) 2/5 (G) 7/5 (H) 4 (I) 3 (J) 2 29

30 Solution: We can obtain the volume by integrating the function z = 2 y over the region in the first quadrant of the xy-plane bounded by the y-axis and the parabola x = 4 y 2. This gives the integral 2 4 y 2 which gives 2/3. (2 y) dxdy = 2 (2 y)(4 y 2 ) dy = 2 (8 2y 2 4y + y 3 ) dy 3

31 16. Evaluate the integral y cos(x 2 ) z dxdy dz by changing the order of integration. 14.5, 41 (A) 4 sin 4 (B) 2 sin 4 (C) sin 4 (D) sin 2 (E) 2 sin 2 (F) 4 sin 2 (G) 2 cos 4 1 (H) cos 4 1 (I) 1 sin 4 (J) π 31

32 Solution: By changing the order of integration (exchanging the integrals in x and y) we obtain 4 2 x/2 cos(x 2 ) 4 2 dy dxdz = z 1/2 1 dz z 2 x cos(x2 ) dx. The value of the integral in z is 4, and the integral in x can be evaluated by a substitution u = x 2 : 2 x cos(x 2 ) dx = Therefore, the original integral has value sin 4. cosudu = 1 sin