Vector Calculus Solutions to Sample Final Examination #1

Transcription

1 Vector alculus s to Sample Final Examination #1 1. Let f(x, y) e xy sin(x + y). (a) In what direction, starting at (,π/), is f changing the fastest? (b) In what directions starting at (,π/) is f changing at 5% of its maximum rate? (c) Let c(t) be a flow line of F f with c() (,π/). alculate d dt [f(c(t))]. t (a) f is changing fastest in the direction of f(,π/). But f(x, y) [ye xy sin(x + y)+e xy cos(x + y)]i +[xe xy sin(x + y)+e xy cos(x + y)]j, and so f(,π/) π i. Thus f is increasing fastest in the direction i (and decreasing fastest in the direction -i). (b) If n is a unit vector, f is changing at the rate f(,π/) n π n i in the direction n. The maximum value is π/, so the rate is 5% of its maximum when π n i π 1 n i 1 This means n makes an angle θ with i where cos θ 1/, or θ ±π/3 or±6 degrees. Note that this defines two directions (if this were in space and not the plane, we would get a cone). 1

2 (c) By the chain rule, and since c (t) f(c(t)), d dt f(c(t)) f(c()) c () t ( f, π ) ( f, π ) f (, π ) π 4.. Let f : R 3 R 3 be a given mapping and write f(x, y, z) (u(x, y, z),v(x, y, z),w(x, y, z)). Let g : R 3 R 3 be defined by g(u, v, w)(u v, u + w, w + v) andlethg f. (a) Write a formula for the derivative matrix Dh. (b) Show that Dh cannot have rank 3 at any point (x, y, z). (c) Show that Dh has an eigenvalue zero at every (x, y, z). (a) By the chain rule, Dh(x, y, z) Dg(u, v, w) Df(x, y, z) v v 1 1 w w v x x x x v x x v v z x w z v z z z z θ z z (b) We claim that this matrix has determinant zero. Its determinant is the product of the determinants of the two factors. But Thus, Dh(x, y, z) is not invertible, so it must have nullity, so rank Dh(x, y, z) 3. (c) Since nullity some non-zero vector must get sent to zero; this is an eigenvector with eigenvalue zero. 3. Extremize f(x, y, z) x subject to the constraints x + y + z 1 and x + y + z 1.

3 We are to extremize f(x, y, z) xsubject to g 1 x + y + z 1 and x+y+z 1. Using the method of Lagrange multipliers, this means f λ 1 g 1 + λ g g 1 g Subtracting () and (3) gives 1 λ 1 x + λ (1) λ 1 y +λ () λ 1 z +λ (3) x + y + z 1 (4) x+y +z 1. (5) (6) λ 1 (y z) and so either λ 1 oryz.butλ 1 is not consistent with (1) and (). Hence, λ 1 and so y z. Thuswehave Substituting (9) in (8) gives 1 λ 1 x+λ (7) λ 1 y +λ (8) x +y 1 (9) x+y 1 (1) (11) (1 y) +y 1 1 4y +4y +y1 y +3y. 3

4 Thus, either y,ory/3.if y thenx1(andλ,λ 1 1/) and if y /3thenx 1/3. Therefore, the solutions are (1,, ) and ( 1/3, /3, /3) The former maximizes f while the latter minimizes it. 4. (a) Evaluate D exp[(x + y + z ) 3/ ] dx dy dz where D is the region defined by 1 x + y + z andz. (b) Sketch or describe the region of integration for 1 x y and interchange the order to dy dx dz. (a) We use spherical coordinates exp[(x + y + z ) 3/ ]dxzdydz D f(x, y, z)dz dy dx, π π/ π ϕ θ π/ ϕ ρ1 ( π 1 3 exp(ρ3 ) π 3 (e 8 e). exp(ρ 3 ) ρ sin ϕdρdθdϕ ) exp(ρ 3 )ρ dρ sin ϕdρ ρ1 π/ ( cos ϕ) 1 (b) The region for 1 x y f(x, y, z)dzdydx is that under the plane z y and over the region in the plane bounded by the x-axis, the line x y and the line x 1. (The student should draw a figure here). 4

5 (c) In the order dydxdz, the integral is easiest to write down by consulting the figure drawn in the previous part; one gets z x f(x, y, z)dydxdz. 5. Let G(x, y) (xe x +y +xy)i +(ye x +y + x )j. (a) Show that G f for some f; find such an f. (b) Use (a) to show that the line integral of G around the edge of the triangle with vertices (, ), (, 1), (1, ) is zero. (c) State Green s theorem for the triangle in (b) and a vector field F and verify it for the vector field G above. (a) If G(x, y) Pi+Qj,P xe x +y +xy, Q ye x +y + x, note that and so G is a gradient. Writing we see that P +y y xyex +x Q x, P f f, and Q f (x, y) e x +y +x y. (b) Let be the boundary of the triangle T (the student should draw a figure of T.) Since the integral of a gradient around any closed curve is zero in general, it is zero in this particular case. (c) Green s Theorem states, in this case that We computed that Q P Pdx+Qdy ( Q x P y ) dx dy above, so the right hand side is zero as well. 6. Let W be the three dimensional region under the graph of f(x, y) exp(x +y )and over the region in the plane defined by 1 x + y. (a) Find the volume of W. (b) Find the flux of the vector field F (x xy)i yj + yzk out of the region W. 5

6 (a) The volume of W is exp(x + y )dxdy 1 x +y π θ r1 e r rdrdθ (b) The flux of F is, by Gauss theorem F ds W W W W π 1 (e e) πe(e 1). divfdxdydz ( y 1+y)dxdydz dx dy dz π(e 4 e). 7. Let be the curve x + y 1 lying in the plane z 1. LetF(z y)i+yk. (a) alculate F. (b) alculate F ds using a parametrization of and a chosen orientation for. (c) Write S for a suitably chosen surface S and, applying Stokes theorem, verify your answer in (b). (d) onsider the sphere with radius and center the origin. Let S be the part of the sphere that is above the curve ( lies in the region z 1), and has as boundary. Evaluate the surface integral of F over S. Specify the orientation you are using for S. (a) We evaluate the curl by writing out the expression for the curl as a cross product of and F : i j k F z i + j + k z y y (b) Parameterize by x cosθ, y sinθ, z 1, θ π, where the orientation is counter-clockwise as viewed from above. Then π π F ds (1 sin θ)( sin θ)dθ sin θdθ π 6

7 (since the average of sin θ is 1/). (c) Let us choose S to be the disk x + y 1,z 1. Then F ds F ds k kdx dy π. S S (d) Let the orientation of S be given by the outward normal. The student should draw a figure that shows S. Then, ( F) ds F ds π. S 7