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 n-y-n 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 n-y-n 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 4-y-4 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 3-y-3 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 outwrd-pointing 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 3-y-3 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 x-xis () 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 x-xis () 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 outwrd-pointing 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 outwrd-pointing 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 Re-express 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 xy-plne (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 xy-plne (where z 0) And when z 0, we hve V xy k But k is perpendiculr to dr, since dr is tngent to the xy-plne 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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