Ma 7 Final Exam Solutions /3/ Name: Lecture Section: (A and B: Prof. Levine, C: Prof. Brady) Problem a) ( points) Find the eigenvalues and eigenvectors of the matrix A. A 3 5 Solution. First we find the eigenvalues. deta ri r 3 r 5 r 3 r 5 r r r 3 r r 5 r r r Clearly one root is r. Using the quadratic formula, the others are r 8 4 i For r, we solve A Iu Using elimination on the augmented matrix, we have 3 5 5 5 3 5 Thus u u 3 u 5u 3 The third component is arbitary, so any multiple of u 5
is an eigenvector. Similarly, for r i, we have the following. [The third step is an extra step of multipying the second row by i to show how this goes.] i i 5 i i 5 i i 5 i 5 5 i i i Thus u u iu 3 Again, the third component is arbitrary and any multiple of i is an eigenvector. Finally, since the entries in the matrix are all real, both eigenvalues and eigenvectors come in complex conjugate pairs and for r i, eigenvectors are multiples of i.
b) (5 points) The eigenvalues of the matrix i i are and Find a [real] general solution to x x 3 4. are i and i and the corresponding eigenvectors 3 4 x x 5t. 3 4 x Solution: First we find a real general solution to. A solution has x x the form e rt v,wherer is an eigenvalue and v is a corresponding eigenvector. We expand one of the complex solutions and take the real and imaginary parts. e it i x e t cost isint i e t cost sint i sint cost cost isint e t cost sint e t cost i e t sint cost e t sint x h c e t cost sint e t cost c e t sint cost e t sint e t cost sint e t cost e t sint cost e t sint c c Next, we find a particular solution to the given non-homogeneous equation. Since the non-homogeneous term is a polynomial of degree one, the solution must be the same. x x at b ct d Substitute into the system of d.e.s and find the coefficients. x x 3 4 x x 5t a c 3 4 at b ct d 5t a c 3a 4ct 3b 4d a ct b d 5t 3
We equate like terms on each row. The coefficients of t.are listed in the first two rows below, then the constant terms. 3a 4c 5 a c a 3b 4d c b d Thus (from the first pair of equations) a 5, c and then b 7andd 4. Combining homogeneous and particular solutions, we have a general solution. x x c e t cost sint e t cost c e t sint cost e t sint 5t 7 t 4 e t cost sin t e t cost e t sint cost e t sint c c 5t 7 t 4 4
Problem a) ( points) Evaluate the line integral F d r for Fx,y,z xi yj z k and C one turn around the spiral C r t costi sin tj tk from,, to,,. Solution: In terms of the parameter, t, we have Fx,y,z costi sintj t k d r sin ti costj k dt So C F d r costi sin tj t k sinti costj k dt sin tcost sin tcost t t dt 3 t3 8 3 3 dt 5
b)(5 points) Consider the triple integral x x y x y z dzdydx. x y i. Describe and sketch the region of integration. Solution: The region of integration is in the first octant, bounded below by the cone z x y and above by the upper portion of the sphere of radius centered at,, which has the equation x y z. r,,r..5 z..5.....5.5. x y ii. Give an equivalent triple integral in cylindrical coordinates. Solution: The cone is z r. The sphere is r z The integral is r r r z rdrd iii. Give an equivalent triple integral in spherical coordinates. Solution: In spherical coordinates, the equation of the cone is 4. algebra. We have x y z z cos cos The integral is 4 cos sinddd. 6 The sphere requires a little
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Problem 3 (5 points) Evaluate the surface integral S F ds F nds S where S is the surface of the portion of the cone z x y in the first octant and below the plane z 4 with downward normal and F yzi xzj xyk. Solution: We need to parametrize the surface and then set up the integral, in terms of the parameters that we choose, in the form S F ds F r u r v da uv. R I chose to use cylindrical coordinates, since the equation of the cone is z r. x rcos y rsin z r r 4 The computations are r xi yj zk rcosi rsinj rk r r cosi sin j k r rsin i rcosj r u r v i j k cos sin rsin rcos rcos i rsin j rk We observe that we have a normal vector pointing upward from the positive z component and use the negative, i.e r v r u. 8
S F ds 4 4 4 r sin i r cos j r cossink rcos i rsin j rk drd r 3 sin cos r 3 cossin r 3 sin cos drd r 3 sin cosdrd 44 4 64 sin / 3. sin cosd 9
Problem 4 a) ( points) Show that, if all partial derivatives of fx,y,z are continuous, curl grad fx,y,z. Solution: grad fx,y,z f x i f y j f z k curl grad fx,y,z i j k x y z f x f y f z f zy f zy i f zx f xz j f yx f xy k
b) (5 points) The figure shows the torus obtained by rotating about the z-axis the circle in the xz-plane with center,, and radius. Parametric equations for the torus are x cos coscos y sin cossin z sin,. is the usual polar angle around the z axis and is the angle around the circle in the x z plane. Find the surface area of the torus. Solution: We have that surface area, for a surface given parametrically, is given by S ds R r u r v da uv. r xi yj zk cos coscosi sin cossinj sink r coscosi cossin j sink The parametrization is given, so we proceed with the computations. r cos sin i coscosj r sin cosi sin sin j cosk r r i j k cos sin coscos sin cos sin sin cos
r r i j k i j cos sin coscos cos sin coscos sin cos sinsin cos sincos sin sin coscoscosi cossinsin k cossin cos k coscossinj cos coscosi sin sin cos k cossin j cos coscosi cossinj sink Thus r r cos cos cos cos sin sin cos cos sin cos sin cos S ds cosdd 4 d 8. sin d
Problem 5 a)(3 points) Use Stokes Theorem to evaluate curlf ds S where F z i 3xyj x 3 y 3 k and S is the part of z 5 x y above the plane z. Assume S is oriented upwards. Solution: A sketch of the surface is shown below. Stokes Theorem is curlf ds F dr S S where S is the boundary of the surface S. The boundary of S is the curve where S intersects the plane z. Thus we have 5 x y or S : x y 4, z We parametrize S by x cost,y sint, t, z so rt costi sintj k t and r t sinti costj Since F z i 3xyj x 3 y 3 k, then F t i 3costsintj cost 3 sint 3 k 3
Therefore F dr F t r tdt S sint 4sintcos t dt cost 8cos 3 t b)( points) Use the divergence theorem to evaluate F ds S where F xyi y j zk and the surface S consists of the three surfaces, z 4 3x 3y, z 4, on the top, x y, z onthesides,andz onthebottom. Solution: A sketch of the surface is shown below. The divergence theorem is where E is the volume enclosed by S. S F ds divf dv E We use cylindrical coordinates so that the limits for the variables are 4
z 4 3r r Hence S F ds divf dv E divf F y y 4 3r rdzdrd 4r 3r 3 drd r 3 4 r4 d 5 4 d 5 5
Problem 6 (5 points) Verify Green s Theorem for the line integral C y dx 3xydy where C is the closed curve consisting of the upper half of the unit circle centered at the origin oriented counterclockwise followed by the line joining, to,. Solution: The closed curve C is shown below. x y..8.6.4. -. -.8 -.6 -.4 -....4.6.8. x Green s Theorem says where R is the region enclosed by C. P y and Q 3xy so R Q x P y Pdx Qdy C R da 3y yda R Q x P y x ydydx da We now compute C y dx 3xydy y x dx x dx x x3 3 3 directly. Note the C consists of two parts, C, the semicircle of unit radius from, to,, andc the part of the x axis from, to,. 6
Thus C C : x cost,y sin t, t C : x t,y t y dx 3xydy sin t sin t 3costsintcostdt dt sin 3 t 3cos tsin t dt sint cos t 3cos tsin t dt sin t 4cos tsint dt cost 4 3 cos3 t 4 3 4 3 8 3 3 7
Problem 7 a) (3 points) The integral y y dxdy gives the area of a region R in the x,y plane. Sketch R and then give another expression for the area of R with the order of integration reversed. Do not evaluate this expression. Solution: This problem was on the Ma 7 final exam given in 4S. The curves that bound R are the parabola x y and the line x y or y x. Thus x y y 3-3 4 x - -3 The curves intersect when y y or when y y y y. Hence at, and 4,. Reversing the order of integration requires two integrals. They are x dydx x x dydx 4 x 8
b) ( points) Find the volume of the region that lies under the sphere x y z 9, above the plane z and inside the cylinder x y 5. Sketch the solid. Solution: The region is shown below. We will use cylindrical coordinates. V E dv rdzdrd E z goes from the x,y plane to the sphere. In cylindrical coordinates the sphere is z 9 x y 9 r. Thus z 9 r. The region on integration in the x,y plane is the area x y 5. Therefore r 5 and. Therefore V E dv rdzdrd E 5 9 r rdzdrd 5 r 9 r drd 9 r 3 3 3 8 7 38 3 5 d 3 4 3 9 3 d 9
Problem 8 a)(8 points) Show that the characteristic polynomial for the matrix A is Solution:The characteristic polynomial for A is pr r 3 pr det r r r r r r r r r r r r r 4 4r r r 3 8 4r r 4 4r r 4r 4r r 3 5 3r r 3 3r 3r r 3 b)(7 points) Show that pa A I 3 where A is the matrix in part a) Solution: A I Thus pa A I 3 3
so 3
c)( points) Calculate e At. Solution: e At e t e A It e t I A It A I t! e t te t e t te t te t te t te t e t te t te t te t te t e t e t
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 3