MATH 260 Final Exam April 30, 2013


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1 MATH 60 Finl Exm April 30, 03 Let Mpn,Rq e the spce of nyn mtrices with rel entries () We know tht (with the opertions of mtrix ddition nd sclr multipliction), M pn, Rq is vector spce Wht is the dimension of Mpn,Rq? Exhiit sis of Mp,Rq uppose A is fixed y mtrix Define the mp A : Mp,Rq Ñ Mp,Rq y A pxq AX, so the opertion A is (left) multipliction y the fixed mtrix A () how tht A is liner mp from Mp,Rq to Mp,Rq (c) how tht A B A B, tht ca c A, nd tht AB A B for y mtrices A nd B nd for ny sclr c (d) Wht is the reltionship of the determinnt of A nd the determinnt of A? Wht out the trce of A nd the trce of A? (e) uppose A is digonlizle Does it follow tht A is lso digonlizle? () ince n nyn mtrix hs n independent entries, we hve dim Mpn,Rq n A sis for Mp,Rq is " * ,,, () We hve A pαx βy q ApαX βy q αax βay α A pxq β A py q (nd the ltter is y mtrix) so A : Mp,Rq Ñ Mp,Rq is liner mp (c) First, A B pxq pa BqX AX BX A pxq B X for ll X P Mp,Rq, so A B A B Next, ca pxq pcaqx cpaxq c A pxq for ll X P Mp,Rq, so ca c A Finlly AB pxq pabqx ApBXq A p B pxqq for ll X P Mp,Rq, so AB A B (d) If A c u nd X d x v y A pxq, then u x v y cu dx cv dy,
2 so the 4y4 mtrix of A is A c 0 d 0 0 c 0 d It s esy to see tht tr A d tr A Expnding long the first row gives us det A 0 0 d 0 c 0 d 0 c c d d cd c cd pd cq pdet Aq (e) If v is n eigenvector of A corresponding to the eigenvlue λ, then the mtrix X with v in its first column nd zeros in the second column is n eigenvector of A with eigenvlue λ Likewise the mtrix X with v in the second column nd zeros in the first column is n eigenvector of A with eigenvlue λ Thus, if A is digonlizle, so is A nd it hs two eigenvectors for ech eigenvector of A note tht this gives nother proof of the ssertions out the trce nd determinnt of A Give exmples of ech of the following or explin why none cn exist: () A y mtrix A such tht A 8 I ut A k I for k P t,, 3, 4,, 6, 7u () A y mtrix A such tht A 4 I (c) A 3y3 mtrix A such tht A 4 I ¾ C (d) A vector field F, defined nd continuously differentile for ll px, yq P R, such tht F dr π, where C is the unit circle x y, trversed counterclockwise (e) A vector field F, defined nd continuously differentile for ll px, y, zq P R 3, such tht F n dσ 4π, where is the unit sphere x y z, n is the outwrdpointing norml vector to, nd dσ is the element of surfce re () Rottion through the ngle π{4 will do it, so the mtrix is () The sme mtrix s prt () will do it (c) The determinnt of the 3y3 identity mtrix is, so this cn never equl pdet Aq 4 for rel mtrix A o there is no such exmple (d) The vector field x j will do, y Green s theorem
3 3 (e) Either using the divergence theorem or the fct tht n r on the unit sphere, we see tht the vector field r x i y j z k will do 3 Find nd clssify ll of the criticl points of the function fpx, yq xy x y x We hve f x y xy nd f y xy x ince f y xpy xq we must hve either x 0 or x y t criticl point If x 0, then from f x 0 we get tht y 0, which hs no solutions If x y then we hve y nd so x 4 o there re two criticl points: p4, q nd p 4, q To clssify the criticl points, note tht f xx y, f yy x nd f xy y x Therefore H f xx f yy pf xy q 4xy px yq 4x 4xy 4y Therefore: ince Hp4, q , this is sddle point, s is p 4, q Two criticl points, oth sddles 4 Let L e liner trnsformtion from R 3 to R 3 tht stisfies Lpi jq k, Lpj kq i, Lpk iq j () how tht i j k is n eigenvector of L Wht is its eigenvlue? () Is L invertile? (c) Find the mtrix M L of L (with respect to the sis ti, j, ku (d) Find n orthogonl mtrix R such tht R T M L R is digonl () Adding the three equtions gives us Lpi j kq i j k so the eigenvlue is () ince the imge of L is ll of R 3 (since we see tht the imge contins three linerly independent vectors), L is invertile (c) We hve Lpiq Lppi jq pj kq pk iqq pk i jq, Lpjq Lppi jq pj kq pk iqq pk i jq nd Lpkq Lp pi jq pj kq pk iqq p k i jq Therefore M L (d) We lredy know one eigenvector, nmely r,, s with eigenvlue Also, it s esy to see tht M L I hs ll its entries equl to, so it s mtrix of rnk We cn tke ny orthogonl pir
4 4 of vectors perpendiculr to r,, s s our eigenvectors for the eigenvlue o we tke r,, 0s nd r,, s Then we cn tke R???6 3? 3??6? 3 0? 6 nd then R T M L R uppose f : R Ñ R is differentile function with fpxq 0 for ll x Let e the surfce of revolution in R 3 defined y revolving the grph of the function y fpxq (for x ) in R round the xxis () Using the prmetriztion x u, y fpuq cos v, z fpuq sin v, derive the formul repq» πfpxq pf pxqq dx () Apply this formul to clculte the (lterl) surfce re of the cone otined y rotting the prt of the line segment y 3 3x in the first qudrnt round the xxis () We hve r u i fpuq cospvq j fpuq sin v k, therefore r u i f puq cos v j f puq sin v k nd r v fpuq sin v j fpuq cos v k Thus dσ }r u r v } du dv }fpuqf puq i fpuq cos v j fpuq sin v k} du dv fpuq pf puqq du dv The surfce is prmetrized y u (with u ) nd v (with 0 v πq Therefore repq since u x dσ»» π 0 fpuq pf puqq dv du () ince fpxq 3 3x (for 0 x ) nd f pxq 3, we hve repq» 0» πp3 3xq? 0 dx π? 0 πfpuq pf puqq du 3x 3x 3π? 0 0» πfpxq pf pxqq dx 6 Let V e the vector field on R 3 defined (except t the origin) y V x y z () Compute V, where ρ ρ
5 () Clculte V n dσ where R is the surfce of the sphere x y z R for some R R 0 nd n is the outwrdpointing norml vector (c) Why does the result of prt () not contrdict the divergence theorem? (d) Use the divergence theorem to explin why the result of prt () doesn t depend on R (e) Clculte V n dσ where T is the surfce of the sphere px 3q py q pz 7q 4 T (nd n is the outwrdpointing norml vector) () We hve V x y z x px y z q, y 3{ px y z q, z 3{ px y z q 3{ Therefore 3 3px y z q V 0 px y z q 3{ px y z q{ () From prt (), on the sphere x nd since n rx{r, y{r, z{rs we hve which is constnt on R, so V y z R, we hve x R 3, y R 3, z R 3, V n px y z q R 4 R R V n dσ R rep Rq 4π (c) This doesn t contrdict the divergence theorem ecuse the vector field V is not defined t the origin, nd it ecomes singulr there (d) The result doesn t depend on R ecuse if we consider the solid shell Ω etween two spheres R nd R, we cn pply the divergence theorem nd conclude tht the integrl of V n is the sme over oth spheres (nd thus is independent of R) (e) The sphere in question doesn t surround the origin, so we cn pply the divergence theorem to it, nd get tht this integrl is zero 7 () Find the mximum vlue of the function fpx, y, zq ln x ln y 3 ln z on the prt of the sphere x y z r in the first octnt (where x 0, y 0 nd z 0) Be sure to explin how you know tht the criticl point you find is the mximum
6 6 () Your nswer to prt () should e in terms of r Reexpress r in terms of x, y nd z using the constrint eqution, nd write out the inequlity tht implied y your nswer to prt (), which will look like ln x ln y 3 ln z n expression in x, y nd z (c) (Extr Credit:) Multiply oth sides of the inequlity in prt () y, nd let x, y nd c z, to prove tht c c 3 7 () First off, s ny of x, y or z pproch zero, their logrithms pproch 8, so fpx, y, zq will pproch 8 t the edges of the prt of the sphere Therefore if we find only one criticl point, it will e the glol mximum We set gpx, y, zq x y z r, nd use Lgrnge multipliers The eqution f λ g gives us the three equtions x λx y λy 3 z λz We multiply these equtions y x, y nd z respectively nd dd them together to get λpx y z q 0λr using the constrint eqution Thus λ {pr q, nd we conclude tht x r y 3 r z r ince we know x, y, nd z re positive we hve only the one criticl point px, y, zq pr, r,? 3 rq Therefore, the mximum vlue of f on the domin is ln r ln r 3 lnp? 3 rq ln r 3 ln? 3 () From (), using tht r px y z q, we cn conclude tht for ll x 0, y 0 nd z 0, ln x ln y 3 ln z ln r 3 ln? 3 ln c x y z 3 ln? 3 ln x y z ln 7 (c) Multiplying y, we hve ln x ln y 3 ln z ln x y z ln 7 Now set x, y nd c z to get ln ln 3 ln c ln c ln 7
7 7 for ll 0, 0 nd c 0 Finlly, exponentite oth sides nd get c c 3 7 (exponentiting oth sides preserves the inequlity ecuse the exponentil function is monotoniclly incresing) 8 () Prove tht for ny (twice continuously differentile) vector field V on R 3, we hve p Vq 0 () Find vector field V such tht V x i 3y j z k (c) Evlute px i 3y j z kq n dσ where is the prt of the proloid z 4 x y ove the xyplne (so z 0), oriented so tht the unit norml vector points wy from the origin () Let Vpx, y, zq fpx, y, zq i gpx, y, zq j hpx, y, zq k Then V ph y g z q i pf z h x q j pg x f y q k Therefore p Vq ph y g z q x pf z h x q y pg x f y q z h xy g xz 0 f yz h xy g xz f yz () First it s n esy check tht px i 3y j z k 3 0, so there s hope Next, we need to solve h y g z x, f z h x 3y, g x f y z for f, g nd h Just to e eglitrin, let s tenttively set h xy nd g xz, so tht h y g z x p xq x, which mkes the first eqution true To mke the second eqution true, we ll then need f 4yz so tht f z h x 4y y 3y And now we cn check tht the third eqution ecomes g x f y z 4z z (tht s why we needed V 0, so tht the third eqution would e utomticlly stisfied once we rn out of choices to mke) o we ll tke V 4yz i xz j xy k (c) By tokes s theorem, we hve» px i 3y j z kq n dσ V d V dr B Now B is the circle of rdius centered t the origin in the xyplne (where z 0) And when z 0, we hve V xy k But k is perpendiculr to dr, since dr is tngent to the xyplne Therefore we end up integrting zero, nd so we cn conclude tht px i 3y j z kq n dσ 0
8 8 Just for the record, prmetrize the proloid vi x u cos v, y u sin v nd z 4 u for 0 u nd 0 v π Then r u cos v i sin v j u k nd r v u sin v i u cos v j nd so n dσ r u r v du dv pu cos v i u sin v j u kq du dv nd so px i 3y j z kq n dσ» π» 0 0 4u 3 cos v 6π 4π 80π 40π 0 6u 3 sin v p4 u qu du dv
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