1. (30 points) In the x-y plane, find and classify all local maxima, local minima, and saddle points of the function. f(x, y) = 3y 2 2y 3 3x 2 + 6xy.

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1 APPM 35 FINAL EXAM FALL 13 INSTUTIONS: Electronic devices, books, and crib sheets are not permitted. Write your name and your instructor s name on the front of your bluebook. Work all problems. Show your work clearly. Note that a correct answer with incorrect or no supporting work may receive no credit, while an incorrect answer with relevant work may receive partial credit. 1. (3 points) In the x-y plane, find and classify all local maxima, local minima, and saddle points of the function Solution: Using the second derivative test it follows that f(x, y) 3y y 3 3x + 6xy. f x 6x + 6y Setting the partial derivatives equal to and solving gives f y 6y 6y + 6x x y y This implies that the possible critical points are (, ) and (, ). Then the determinant of the derivative matrix is D f xx f xy f yx f yy y 7y 7 Plugging in the critical points gives D(, ) 7 (, ) is a saddle point and (, ) is a local max.. (35 points) onsider the function f(x, y) exp(xy + y). D(, ) 7, f xx (, ) < (a) If one moves from (1, 1) to (1.1, 1.), estimate the change in the value of f. Solution: The change in f can be approximated by f f f x + x y ye xy+y (1,1) (.1) + (x + 1)e xy+y (1,1) (.) e (.1) + e (.) Therefore, the estimate for the change in the value of f is (b) From (1, 1), in what direction should one move to observe the greatest rate of increase in the value of f. (Your answer should be an explicit vector.) Solution: The greatest rate of increase in the value of f is given by the gradient evaluated at (1, 1). The gradient of f is f F x, f ye xy+y, (x + 1)e xy+y e F (1,1) e, e

2 (c) If one moves away from (1, 1) in the direction of the vector A 3 i + 4 j, what is df/ds, where s is the arc length. Solution: f s f u where u 3/5, 4/5. f s 3 5 e e 11 5 e (d) If one moves away from (1, 1) in the direction of A at the constant speed V v, what is df/dt? Solution: Note that df dt f v and u v v Hence, 3. (3 points) onsider the function f(x, y) x exp(xy). (a) Determine the value of the double integral f t f s v 11 5 e v y f(x, y) dx dy. Solution:Switching the order of integration, it follows that y x e xy dx dy x x x e xy dy dx x exy x dx xe x dx x dx Let u x. Then 1 xe x dx x dx 1 e u du 1 eu 1 x 1 e 1 1 e 1 (b) Determine the value of f, the average value of f over the region of integration,. Hint: f f da (Area of region ). Solution: The region in question is a triangle with base and height of length 1, so the area of the region is 1/. f 1 ( e ) 1 1 x dx

3 4. (3 points) onsider the force field F (y + cxz) i + y(bx + cz) j + (y + cx ) k. (a) For what values of b and c will F be a potential field? Solution: In order for F to be a potential field the curl of F must be. It follows that i j k F x z y + cxz y(bx + cz) y + cx (y cy)i + (cx cx)j + (by y)k c, b (b) Using your values for b and c, determine the work done by moving from (1,, ) to (1,, π) along on the helical path r(t) cos t i + sin t j + t k. Solution: Note that v sin(t), cos(t), 1 and F sin (t) + 4t cos(t), sin(t)(cos(t) + t), sin (t) + cos (t). Then Then the work is given by F v sin 3 (t) 4t cos(t) sin(t) + sin(t) cos(t)(cos(t) + t) + sin (t) + cos (t) sin 3 (t) t sin(t) + sin(t) cos (t) + sin (t) + cos (t) π W F v dt sin 3 (t) + sin(t) cos (t) dt + ( [ + 3π t cos(t) ] π ) π cos(t) + dt 3π + π 4π π sin (t) + cos (t) dt π t sin(t) dt 5. (35 points) onsider the point (1,, 3) in the plane described by 3x + y + z 1. Around this point, and in the plane, is a circular path of radius. Determine the value of the circulation of the field F 7y i + 5z j 1x k around the circular path. Solution: Using Stoke s Theorem it follows that F T ds where ( F) k da F 5i + 1j 7k Note that k 3,, 1. Then the circulation is given by 5, 1, 7 3,, 1 da ( ) da da 4π

4 6. (4 points) onsider the portion of( a solid sphere of radius 1 in the first octant. In this object the temperature q ) (1 distribution is given by T (x, y, z) z ) where q is a constant. You will be working with the vector F κ T, which is traditionally called the heat flux vector, where κ is the thermal conductivity. (For this problem, you may assume that κ 1.) (a) In words, clearly describe the level surfaces of the temperature distribution T (x, y, z). Solution: The level surfaces are described by q (1 z ) k where k is a constant. Therefore the level surfaces are quarter discs parallel to thex-y axis. (b) alculate the heat flux vector F κ T. Solution: The gradient of T is given by T x,,,, qz z F,, qz (c) The amount of heat passing crossing any surface can be found by evaluating the surface integral representing the outward flux of F over that surface. ompute the outward flux across each of the four bounding surfaces (bottom, left, back and curved top) of the object. learly label each of the four flux values, flux bottom, flux left, and so on. Solution: The flux through the bottom is given by FdA qda q π 4 Since F is perpendicular to both i and j the outward flux through the left and right surfaces is. The flux through the top is given by where and It follows that π/4 π/ π/4 π/ F (r φ r θ ) dφ dθ r φ r θ sin (φ) cos(θ), sin (φ) sin(θ), sin(φ) cos(φ) F (r φ r θ ) dφ dθ F (r(φ, θ)),, q cos(φ) π/4 π/ q π 4 π q π q π 1 j + + q sin(φ) cos (φ) dφ dθ 1 sin(φ) + sin(3φ) dφ 4 (d) Based on your results in part (c), determine the total amount of heat leaving the object. Solution: The outward flux is equal to the sum of all the sources minus the sum of all the sinks inside the volume. Therefore, the total amount of heat leaving the object is q π 4 q π 1 q π 6

5 (e) Now, if possible, verify your answer to part (d) using an appropriate alculus III theorem. If this is not possible, write BUMME. Solution: Using the Divergence Theorem it follows that F dv q dv q qπ 6 ( 4π 3 ) ( 1 8 ) ENJOY YOU BEAK!

6 Projections and distances proj A B ( ) A B A d P S v A A v Arc length, frenet formulas, and tangential and normal acceleration components dt ds κn ds v dt T dr ds v v db ds τn κ a a N N + a T T dt ds v a v 3 a T d v dt N dt/ds dt/ds dt/dt dt/dt f (x) 1 + (f (x)) a N κ v d n P S n B T N ẋÿ ẏẍ 3/ ẋ + ẏ 3/ τ db ds N a a T The Second Derivative Test Suppose f(x, y) and its first and second partial derivatives are continuous in a disk centered at (a, b), and f x(a, b) f y(a, b). Let D f xxf yy f xy. 1. If D > and f xx < at (a, b), then f has a local maximum at (a, b).. If D > and f xx > at (a, b), then f has a local minimum at (a, b). 3. If D < at (a, b), then f has a saddle point at (a, b). 4. If D at (a, b), then the test is inconclusive. Directional derivative, discriminant, and Lagrange multipliers df ds ( f) u fxxfyy (fxy) f λ g, g Taylor s formula (at the point (x, y )) [ ] f(x, y) f(x, y ) + (x x )f x(x, y ) + (y y )f y(x, y ) + 1 [ ] (x x ) f xx(x, y ) + (x x )(y y )f xy(x, y ) + (y y ) f yy(x, y )! + 1 [ (x x ) 3 f xxx(x, y ) + 3(x x ) (y y )f xxy(x, y ) 3! ] + 3(x x )(y y ) f xyy(x, y ) + (y y ) 3 f yyy(x, y ) + Linear approximation error E(x, y) M ( x x + y y ) {, where max fxx, f xy, f } yy M! Polar coordinates x r cos θ y r sin θ r x + y da dx dy r dr dθ ylindrical and spherical coordinates ylindrical to ectangular Spherical to ylindrical Spherical to ectangular x r cos θ r ρ sin φ x ρ sin φ cos θ y r sin θ z ρ cos φ y ρ sin φ sin θ z z θ θ z ρ cos φ dv dx dy dz dz r dθ dr ρ sin φ dρ dφ dθ Substitutions in multiple integrals f(x, y) dx dy f ( x(u, v), y(u, v) ) J(u, v) du dv where J(u, v) x G u v x u v x/ u / u x/ v / v Mass, moments, and center of mass Mass M δ da Moments M x y δ da M y x δ da enter of mass x M y/m ȳ M x/m Green s Theorem in the x-y plane (The curve is traversed counterclockwise, and F(x, y) M(x, y) i + N(x, y) j.) ( N irculation F T ds ( F) k da x M ) da ( M Outward Flux F n ds F da x + N ) da Surface area of level surface g(x, y, z) c S S Stokes Theorem F T ds ( F) n dσ S Divergence Theorem of Gauss F n dσ S Fond memories sin x 1 cos x and dσ F dv D cos x 1 + cos x g g p da