APPM 35 FINAL EXAM PING 15 INTUTION: 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. how 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) You are playing the video game Qbert. Qbert jumps around on the top surfaces of unit cubes (dimensions 1 1 1 ) arranged as seen in the pictures below. Assume the center of the surface Qbert stands on, A, is located at the origin, that is, the point A is in the x-y plane and hence is located at (,, ). The coordinate axes are parallel to the edges of the cubes. Qbert can jump from A to the center of adjacent cube surfaces located at points B,, D, or E. To make things more interesting, there are evil coilys in space trying to kill[ Qbert as it jumps around. When Qbert stands at point A, the likelihood of coilys killing it is given by P (x, y, z) exp 1 ( (y 4) + (x 3) (z 5) )]. Be sure to read this function carefully! (a) Find the coordinates of the points where Qbert can jump, specifically the coordinates of points B,, D, and E. olution: Let s take point B as an example. To move there from point A (the origin), Qbert needs to move up (along the Z axis) the length of 1 cube (which is defined to be 1 unit), and along the negative Y axis the length of 1 cube. Thus, the point B is located at (, 1, 1). This same approach can be taken for the remaining points, yielding ( 1,, 1), D (1,, 1), and E (, 1, 1). (b) If Qbert could pick any direction in space, in what direction should it jump to minimize its chance of being killed? olution: The gradient of a function points in the direction of greatest increase, which implies that the negative of the gradient points in the direction of greatest decrease. ince we want to find the direction of greatest decrease of the function P (x, y, z), we must calculate P and evaluate it at the point (,, ): P (,,) (3 x)p (x, y, z), (4 y)p (x, y, z), (z 5)P (x, y, z) (,,) 3, 4, 5 Direction of maximum decrease P (,,) 3, 4, 5 (c) If Qbert is restricted to jumping to one of the four locations B,, D, or E, where should it jump to to minimize its chance of being killed? Warning: don t simply plug in coordinates, use alculus 3 concepts to support your answer. Hint: you might find it helpful to note things like 1 14. olution: To determine how the function P (x, y, z) changes with respect to the direction chosen, we can utilize the directional derivative, which states that P û dp ds. We found in part (b) that P (,,) 3, 4, 5, and we know that dp ds represents how P (x, y, z) changes with respect to distance traveled (this is what we want to minimize). Thus, we can calculate 4 different û corresponding to the directions from A to B,, D, and E, and pick the one that makes dp ds the least. A B : û 1 1, 1, 1 dp 1 ( 4 5) 9 ds 1 A : û 1 1,, 1 dp 1 ( 3 5) 8 ds A D : û 3 1 1,, 1 dp 1 (3 + 5) 8 ds 3 A E : û 4 1, 1, 1 dp 1 (4 + 5) 9 ds 4 Thus, moving from point A to point B would minimize Qbert s chances of being killed.
. (3 points) What values of a and b will maximize the value of the integral a < b. Be sure to support your answer using alculus 3 concepts. b a ( 3x + 4x 45) dx? Assume that olution: Note that the integral is a function fo a and b that we want to maximize. Note also that F (a, b) b ( 3x + 4x 45) dx a a b Using the Fundamental Theorem of alculus, we have ( 3x + 4x 45) dx F a 3a 4a + 45 and F b 3b + 4b 45 We find the critical values of F by setting the partials equal to zero. We have 3 ( a 8a + 15 ) 3 (a 3) (a 5) a 3, a 5 3 ( b 8b + 15 ) 3 (b 3) (b 5) b 3, b 5 ince we re told that a < b we know that the only valid critical point is (a, b) (3, 5). We should still use the second derivative test to check that this point is a maximum. We have F aa 6a 4, F ab, F bb 6b + 4 At the critical point we have F aa 6 < F ab F bb 6 < F aa F bb F ab 36 > thus the point (a, b) (3, 5) is a maximum.
3. (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 conservative field? olution: A vector field F is conserivative if F. We have F learly F if b c. i j k x y z y + cxz bxy + cyz y + cx (y cy) i + (by y) k (b) Using your values for b and c, determine the work done by moving on the helical path given by r(t) cos t i + sin t j + t k from (1,, ) to (1,, π). olution: ince this field is conservative we know that there exists a potential function f such that F f. We ll determine this potential function and then use the Fundamental Theorem of Line Integrals to find the work. We have Integrating, we have F y + 4xz, xy + yz, y + x f x, f y, f z f x y + 4xz f xy + x z + g (y, z) f y xy + yz f xy + y z + h (y, z) f z y + x f y z + x z + k (y, z) By inspection we see that f xy + y z + x z Then, by the FTLI we have W f(1,, π) f(1,, ) 4π 4π
4. (3 points) onsider the point (1, 1, 3) in the plane x + y + z 7. Around this point, and in the plane, is a circular path of radius. alculate the value of the circulation of the vector field F y i+3z j x k around the circular path. olution: Note that the path is not an ellipse, but an actual circle in the plane. We could potentially parameterize the path and use a line integral to calculate the circulation, but this path is pretty difficult to parameterize. Instead we ll use tokes Theorem. We have irculation ( F) n dσ where is any surface that has as its boundary curve. We ll take to be the plane that contains the curve. Assuming the curve is oriented counterclockwise when viewed from above, the upward oriented unit normal vector of the plane is n g,, 1 g 3 The curl of the vector field is F i j k x y z y 3z x 3, 1, Thus we have irculation ( F) n dσ 3, 1,,, 1 3 dσ dσ The surface integral expression is just the surface area of the circular cutout of the plane. Thus we have irculation dσ π
5. (4 points) onsider the path r(t) formed by the intersection of the two surfaces z x + y and z 1 + y, and the vector field given by F y i + z j + y k. (a) Determine a parametrization of the closed path r(t). Be sure to specify the values of t. olution: To find the parameterization of the surface of integration we substitute the expression for z in the first surface for the z in the second surface. This gives x + y 1 + y x + y 1 Thus on the intersection of the two surfaces the x and y components lie on a circle of radius 1. This gives r(t) cos t, sin t, z(t) t π To determine the z-component of the parameterization we can plug x(t) cos t and y(t) sin t into the equation for either of the surfaces and solve for z. Plugging into the second surface equation gives r(t) cos t, sin t, 1 + sin t t π (b) Determine the value of the circulation of F around r(t) if one moves in a counter-clockwise direction. olution: The circulation of a vector field around a line segment is given by F dr. Plugging in the parameterization of the curve, we have F sin t, 1 + sin t, sin t and dr sin t, cos t, 4 sin t cos t dt Then π sin t, 1 + sin t, sin t sin t, cos t, 4 sin t cos t dt π sin t + cos t + sin cos t + 4 sin t cos t dt π sin t + cos t + 6 sin cos t dt π (c) If it is possible to verify your result in part (b) with another technique, do so. Either way, be sure to justify your reasoning. olution: We ll verify our result from part (b) using tokes Theorem. The curve of intersection is a boundary curve for either of the surfaces involved, but we ll choose to integrate over the second surface : z 1 + y. From tokes Theorem we have The curl of our vector field F can be calculated as follows: ( F) n dσ î ĵ ˆk curl(f ) F x y z y z y î(1 1) ĵ( ) + ˆk( 1) ˆk
We ll project the surface into the xy-plane since we know from the parameterization of the bounding curve that this will just be the unit circle x + y 1 which we denote. We choose g z y 1 and p k. We then have We have ( F) n dσ ( F) g, 4y, 1 and g k 1 ± g g p da ince the gradient of g points in the positive z direction the outward pointing normal vector is correct with respect to the counterclockwise orientation of the boundary curve. Thus we choose + g. We then have ( F) ± g g p da,, 1, 4y, 1 da da π where the last step follows from the fact that the area integral is just the area of the unit circle,. 6. (4 points) onsider the finite sized object bounded by the surfaces z 1 x y and z, and the vector field given by F ( xz sin(yz) + x 3) i + cos(yz) j + ( 3zy exp(x + y ) ) k. Determine the value of the total outward flux of F across the bounding surface of the object. olution: alculating the total flux out of the object requires the use of the Divergence Theorem: F ˆn dσ F d Therefore, the total flux can be written as follows: Total flux F d x, y, z xz sin(yz) + x 3, cos(yz), 3zy e x +y d z sin(yz) + 3x z sin(yz) + 3y d 3 x + y d To evaluate this integral, it is easiest to use cylindrical coordinates. Note that the bounds on z go from to 1 x y 1 r. 3 π 1 1 r 3 (r )(r) dzdrdθ 3 π π 1 ( 1 4 r4 1 ) 1 6 r6 dθ 3 (1 r )(r 3 ) drdθ 3 π π 1 ( 1 4 1 ) dθ 6π 6 1 π r 3 r 5 drdθ ENJOY YOU UMME!
Projections and distances proj A B ( ) A B A d P 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 n B T N ẋÿ ẏẍ 3/ ẋ + ẏ 3/ τ db ds N a a T The econd Derivative Test uppose 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 pherical to ylindrical pherical 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 d dx dy dz dz r dθ dr ρ sin φ dρ dφ dθ ubstitutions 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 y G u v y x u v x/ u y/ u x/ v y/ 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 y ( M Outward Flux F n ds F da x + N ) da y urface area of level surface g(x, y, z) c tokes Theorem F T ds Divergence Theorem of Gauss F n dσ Fond memories sin x 1 cos x and dσ g g p da ( F) n dσ F dσ F d D cos x 1 + cos x