Ma 227 Final Exam Solutions 12/22/09

Transcription

1 Ma 7 Final Exam Solutions //9 Name: ID: Lecture Section: Problem a) (3 points) Does the following system of equations have a unique solution or an infinite set of solutions or no solution? Find any solutions. x x x 3 x 4 6 x x 3x 3 x 4 5 x 3x 5x 3 x 4 x 3x 4x 3 x 4 6 Solution: We use row reduction on the augmented matrix Now, with the reduced row echelon form, we can see that the system has an infinite number of solutions. x 3 can take any value. x x x 3 x 3 x 4

2 b) ( points) Show that divcurlf F. (Assume that the mixed second partial derivatives are equal.) Fx,y,z Pi Qj Rk.

3 Solution. Let F F Pi Qj Rk i j k x y x P Q R Ry Q z i R x P z j Q x P y F R yx Q zx Rxy P zy Q xz P yz 3

4 Problem a) ( points) Find the eigenvalues and eigenvectors of A. Solution: We solve deta ri. deta ri r r r r r i r i So, the eigenvalues are a complex conjugate pair. We find the eigenvector for one and take the complex conjugate to get the other. For r i, we solve A riu i i u u The second row is redundant, so iu u oru i u. Hence any multiple of i is an eigenvector for r i. Then an eigenvector corresponding to r i is b)(3 points) Find the [real] general solution to i. x x x x e t. Solution: The solution is the general solution x h to the homogeneous equation plus one [particular] solution x p to the full non-homogeneous equation. First we ll find x p. It is in the form x p c e t c e t Substituting into the d.e., we obtain 4

5 x x c e t c e t x x c e t c e t c e t c e t c e t c e t c e t c e t c e t c e t c e t c e t e t We can divide by e t (which is never zero) and move the unknowns to the left side to obtain c c x p e t For the solution to the homogeneous equation, we use one of the eigenvalues and eigenvectors found in a to write a complex solution and break it into real and imaginary parts. we ll use i. x e it i e t cost isint i e t cost ie t sin t e t sint ie t cost x h c e t cost e t sin t c e t sin t e t cost e t cost e t sint e t sint e t cost c c Finally, we add to obtain the desired solution. x e t cost e t sin t e t sint e t cost c c e t \ 5

6 Problem 3 a) ( points) Let R be the region in the first octant bounded by the plane x y 3z 6. Sketch the region R. Set up three iterated integrals for the volume with the orders of integration as specified below. dzdydx, dydxdz, dxdzdy Solution: The region is the tetrahedron bounded by the x y plane, the x z plane, the y z plane and the plane given. 6 x y 3 The integrals are V 6 3 x 3 x 3 y dzdydx 6 3z 3 x 3 z dydxdz 3 3 y 6 y 3z dxdzdy b) (3 points) The integral below represents the volume of a solid. Describe the solid. Evaluate the integral. 4 y 4 x y 4 x y dzdydx Solution: The region is a hemisphere - the right half of a sphere of radius centered at the origin and bounded by the x z plane (with y positive.) The integral is easily evaluated with a change to spherical coordinates. 4 x 4 x y dzdydx 4 x y sin ddd 3 3 sin dd d 6 3 cos d 6

7 Problem 4 a) ( points) Evaluate where S F d S S F nds F x z 3 i xyz 3 j xz 4 k and S is the surface of the box with vertices,,3. Solution: We use the divergence theorem to replace the surface integral with something that [we hope] is simpler. FdV Box S F d S F xz 3 xz 3 4xz 3 8xz 3 3 FdV 8xz 3 dzdydx 3 Box Since the inner integral yields 3 8xz 3 dz xz 4 z3 3 z 3 the result is zero. b) (5 points) Evaluate S F d S S 8x F nds where F x i xy j zk and S is the portion of the paraboloid z x y below z, oriented with the normal pointing downward. Solution: The surface is simply parametrized using x and y. r xi yj x y k r x i xk r y j yk r x r y i xk j yk i j y i k x k j 4xy k k k yj xi We observe that this is a normal pointing upward, but the other direction is required 7

8 S F d S D F r y r x da xy x y x 3 xy x y da r 3 cos r rdrd r 4 cos r 3 rdrd 5 cos 4 d. 8

9 Problem 5 (5 points) Verify Stokes Theorem for the vector field F y i x j z xk over the upper hemisphere x y z, z, oriented by the outward pointing normal n. Solution: The hemisphere is shown below x y z Stokes Theorem is S curl F nds S F nds S F d r We first evaluate F d r. (This is worth points.) S The boundary of S is the circle x y, z. We parametrize S as x cost,y sint,z, t. Then F t sinti costj costk Hence and We now evaluate S F d r rt costi sintj k r t sin ti costj k F t r t sin t cos t sin t cos t dt costsin t t costsint t costsint 3 t S curl F nds 9 3

10 (This is worth 5 points.) curlf i j k x y z i j k x y z y x x z y x x z i j k k i j j 3k i x y j y x We use spherical coordinates to parametrize the hemisphere. Thus since here x cossin, y sin sin, z cos Therefore r, cossin i sin sinj cosk r sinsini cossin j r coscosi sin cosj sin k i j k i j N r r sin sin cossin sinsin cossin coscos sin cos sin coscos sin cos cossin i sin sin cosk cos cossink sin sin j cossin i sin sin j sincosk We must check to see if N points inward or outward. Let and.thenn i,son is inner. Hence we use N cossin i sinsin j sincosk curlf N sinsin 3sincos S curl F nds sin sin 3sincos dd cos sin d 3 sincos 3sin 3

11 Problem 6 a)(3 points) It can be shown the for the force field F x,y,z e x cosyi e x sinyj k that curlf Find the work done by F on an object moving along a curve C from,, to,,3. Solution: Since curlf this means that there exists fx,y,z such that f F Hence Work F dr f,,3 f,, We have to find f f x e x cosy f y e x siny f z Holding y and z and integrating f x with respect to x leads to So Therefore g y andg hz. f e x cosy gy,z f y e x siny g y e x siny f e x cosy hz f z h z and hz z K, wherek is a constant. f e x cosy z K 6b)( points) Evaluate Work F dr f,,3 f,, C e 6 4 e arctanx y dx e y x dy where C is the path enclosing the annular region shown below, positively oriented.

12 y x Solution: We use Green s Theorem, namely, Pdx Qdy R Q x P y Where R is the region enclosed by the closed curve C. Q x x and P y y so C da arctanx y dx e y x dy x yda R 3 rcos rsinrdrd cos sin r 3 3 d sin cos 4 3

13 Problem 7 a) (3 points) Give and integral in polar coordinates for R x y into polar coordinates, where R is the part of the circle centered at, of radius to the right of the line x in the first quadrant. Be sure to sketch R. Do not evaluate this integral. Solution: The equation of the circle is x y or x y x x x da y x We must use polar coordinates to evaluate this integral. The equation of the circle in polar coordinates is r cos. The equation of the line x isrcos orr sec. r goes from the line to the circle, so sec r cos. The line x and the circle intersect at,. Thus goes from to 4, the angle that the line joining the origin and the point,, whichisy x makes. Thus b) ( points) Evaluate R x y da 4 sec cos r rdrd zdv E where E is the region within the cylinder x y 4, where z y. cos r 3 drd 4 sec Solution: The condition z y means that y, so E projects onto the semicircle y x in the first two quadrants of the x,y plane. 3

14 z y The equation of the cylinder in cylindrical coordinates is r, and the equation of the plane z y is z rsin. Sincey thismeans. The circle has radius, so r. Finally z rsin. Hence r sin zdv zrdzdrd E rsin rdrd r 3 sin drd 4 4 sin d cossin 4

15 Problem 8 a)(3 points) It can be shown that the Cauchy-Euler system tx t Axt t where A is a constant matrix, has a nontrivial solutions of the form xt t r u if and only if r is an eigenvalue of A and u is a corresponding eigenvector. Use this information to solve the system tx t 4 xt t Solution: We first find the eigenvectors of 4. 4 r 4 r r 4 r 5r r so the eigenvalues are r, 5. The system A rix is 4 rx x x rx r yields 4x x orx x. This gives the eigenvector equation x x and hence the eigenvector.thus. r 5 leadstothe xt c t c t 5 b)( points) Rewrite the system of equations where as a single scalar equation. Solution: We have A x Ax f and f cost 5

16 x x x x x 3 x 3 cost Since x t yt, x t y t, x 3 t y t so x t y t x t x t y t x 3 t x 3 t y t y y cost x x cost Thus d 3 y dt 3 dy dt y cost 6

17 Table of Integrals sin xdx cosxsinx x C cos xdx cosxsinx x C sin 3 xdx 3 sin xcosx 3 cosx C cos 3 xdx 3 cos xsinx 3 sinx C cos x sin x dt sinx C 7