Ma 227 Final Exam Solutions 5/9/02

Transcription

1 Ma 7 Final Exam Solutions 5/9/ Name: Lecture Section: I pledge m honor that I have abided b the Stevens Honor Sstem. ID: Directions: Answer all questions. The point value of each problem is indicated. If ou need more work space, continue the problem ou are doing on the other side of the page it is on. Youmanot use a calculator on this exam. Score on Problem # # # # #5 #6 #7 #8 Total

2 Problem a) ( points) Calculate the iterated integral x dzdxd Be sure to show all steps. Solution: x dzdxd xdxd x d d d b) ( points) Find the volume of the region R bounded below b the the x, plane and above b z 5 x. Sketch R. Solution: The region is shown in the diagram - - x The paraboloid intersects the x, plane in the circle x 5. Thus 5 V 5 5 x 5 x ddx 5 x 5 5 r rdrd 5r r 5 d 65 65

3 c) ( points) Give two integral expressions for the area of the region R bounded b x and x x. Sketch the region R. Donot evaluate the integrals. Solution: The curves intersect when x x x, thatis,at, and,. The region is shown below x - Thus x ddx x x or since x x implies that x so that x dxd dxd Problem a) ( points) Verif Green s theorem when F xi xj and C is the boundar of the rectangle x,. Solution: We must show that xdx xd C The rectangle is,,,,,,,,, rectangle x x x da x

4 Thus rectangle x x x da xdxd 7 xdx xd dx d xdx d 7 C b)(5 points) Verif Stokes Theorem is true for the vector field F x,,z x i j z k and S is the part of the paraboloid z x that lies above the x, plane and S has upward orientation. Sketch S. We must show curlf nds F dr S S curlf i j k x z x z x,,z,, Thus curlf nds S For the line integral we parametrize the boundar of S, namel the circle x inthex, plane, as x cost, sint, z t so rt costi sintj k r t sinti costj F t cos ti sin tj k Problem F dr S a) (5 points) Find the eigenvalues and eigenvectors of cos t cos tsint sin tcost dt sin t

5 A 5 Note: r 6r r 6 r r r Solution: This is example on page 559 of the text. det r r 5 r r 6r 6 r r r r Thus the eigenvalues are r,,. The sstem of equations for the eigenvectors is rx x x x rx x x x 5 rx For r we have x x x x x x x x Thus x x x x. Hence the eigenvector corresponding to r is other eigenvectors are,, 5. Similarl the. b) (5 points) Solve the initial value problem x t Axt x where A is the matrix above. Solution: This is example on page 56 of the text. Since the eigenvalues are distinct, then the general solution of the homogeneous DE is

6 xt c e t c e t c e t xt 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 Thus x c c c c c c c c c, row echelon form:,soc, c,c and xt e t e t Problem a) ( points) Give an expression for the volume of the region enclosed b the clinder x, bounded above b the paraboloid z x and bounded below b the x, plane. Sketch the region. Do not evaluate the expression. Solution: r We use clindrical coordinates. Now z goes from the x, plane, that is, to the paraboloid which is z r. The region of integration in the x, plane is the the circle x, that is r,. Hence the volume is given b r dzrdrd

7 b)( points) Solve, if possible, the sstem of equations x x x x x 5 x x 5x 6x 5x 5 x x x x Solution: 5 6 5, row echelon form: 5 so a solution does exist and x 5,x, and x 7 x x,wherex and x are arbitrar. Problem 5 a) ( points) Evaluate F nds, where S F x,,z x i xz j zk and S is the positivel oriented surface of the solid bounded b the z x and the x, plane, and n is the outward directed unit normal to S. (Hint: ou might want to consider using a theorem.) Solution: Use the Divergence Theorem. Then V b) Consider S F nds divf dv V divf x divf dv x r dv r rdzdrd V r r drd r r6 6 (5 points) (a) Sketch the region of integration. fx,dxd. d 6

8 (5 points) (b) Write the integral reversing the order of integration. x x fx,ddx (7 points) (c) Rewrite the integral in terms of polar coordinates. Solution: The limits on are clear from the sketch. Noting that the polar equation of the line x is rcos orr sec, we have / sec frcos,rsinrdrd. Problem 6 a) ( points) Find the surface area of the part of the surface z x that lies above the triangle with vertices,,,, and,. Solution: The surface projects uniquel onto the region in the x, plane as the diagram shows. x Thus we ma use the formula Surface area S where S is given b z fx,. Here f x andf so f x f da

9 where R is the triangle,,,,,,, Surface area R da Thus Surface area dxd x x ddx We use the first expression to find the surface area since this integral can be evaluated. Surface area b)( points) If dxd d 8 F x,,z x z i x z j xz k find a function f such that f F. Solution: Check that such an f exists (not required b problem) x z,x z, xz,, 6 so Then Therefore and Then f x x z f x xz h,z f x h x z h,z z gz f x xz z gz f z xz g z xz

10 so gz K, a constant. Thus f x xz z K Problem 7 a) ( points) Let A be a constant matrix and r an eigenvalue of A with corresponding eigenvector u. Show that xt t r u is a solution of the sstem tx t Axt Solution: We have that Au ru Since xt t r u,then tx t t rt r u t r ru t r Au At r u Axt b) (5 points) Solve the sstem tx t 5 xt t 5, eigenvectors:, Solution: det r r 5 r 8 6r r r r so the eigenvalues are,. rx x x 5 rx x x x x so x x and the eigenvector is. The other eigenvector is the solution is xt c t c t. From part a) Problem 8 ( points) Find a particular solution of the sstem

11 x t xt t t Solution: This is homework problem # on page 579. A, f t t t. f t consists of polnomials of degree, our guess will be a vector consisting of polnomials of degree : guess x p t at b a a t b b Now we plug in: x p t Ax p t f t a a a a t b b t t Bring all of the variables to one side: t t a a t b b a a Now equate coefficients: t : a a ; we now a sstem of equations in unknowns; solve it an wa ou like. You should get a a. Now take the constant term on both sides: b b a a b b b b, another sstem of equations in unknowns. b b. So we have x p t at b t t.