Practice final exam solutions


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1 University of Pennsylvni Deprtment of Mthemtics Mth 26 Honors Clculus II Spring Semester 29 Prof. Grssi, T.A. Asher Auel Prctice finl exm solutions 1. Let F : 2 2 be defined by F (x, y (x + y, x y. If F denotes force field, then show tht the work done by this field to move prticle long the curve α : [, b] 2 defined by α(t (f(t, g(t, only depends on the vlues f(, f(b, g(, g(b. Solution. The work done is clculted by the line integrl F dα b b b f(b f( F (α(t α (t dt (f(t + g(t, f(t g(t (f (t, g (t dt ( f(tf (t + g(tf (t + f(tg (t g(tg (t dt ( u du + g(tf(t b b g (tf(t dt + b f(tg (t dt 1 2 (f(b2 f( 2 + f(bg(b f(g( 1 2 (g(b2 g( ((f(b2 + 2f(bg(b g(b 2 (f( 2 + 2f(g( g( 2 g(b using integrtion by prts nd the substitutions u f(t nd v g(t. This nswer might led one to guess tht there s potentil function, i.e. tht F φ, for some φ : 2. Indeed, letting φ(x, y 1 2 (x2 + 2xy y 2 works! b Find the mount of work done when f( 1, f(b 2, g( 3, g(b 4. Solution. The mount of work done is 1 2 (( ( g( v dv 2. A sphere is inscribed in cylinder. The sphere is sliced by two prllel plnes tht re perpendiculr to the xis of the cylinder. Show tht the portions of the sphere nd cylinder lying between these plnes hs the sme surfce re. Solution. We might s well ssume tht the sphere is centered t the origin, hs rdius, nd tht the cylinder (with rdius hs xis long the zxis. Now the sphere hs prmeteriztion r : [, 2π] [ π/2, π/2] 3 given by r(θ, ϕ ( cos(θ cos(ϕ, sin(θ cos(ϕ, sin(ϕ Let s sy tht the plnes re z b nd z c where b c. Then The portion S of the sphere between the two plnes is prmeterized (vi r by the subset [, 2π] 1
2 [sin 1 (b/, sin 1 (c/]. Using the prmeteriztion formul for surfce integrls, we hve 2π sin 1 (c/ 1 det ( (D (θ,ϕ r t D (θ,ϕ r dϕdθ S θ ϕsin 1 (b/ 2π sin 1 (c/ θ ϕsin 1 (b/ 2 cos(ϕ dϕdθ. The re of the portion C of the cylinder between the two plnes is Check for yourself tht these gree. 2π(c b. 3. Let D 2 be the solid dimondshped region in the plne bounded by the points (,, (1, 1, (2,, (1, 1. Compute D (x2 + y 2 using the chnge of coordintes (u, v (x y, x + y. Solution. Define f : 2 by f(x, y x 2 + y 2 nd Φ : 2 2 by Φ(x, y (x y, x + y. To use the chnge of vribles formul (integrl prmeteriztion formul, we wnt to find nice [ subset S] 2 so tht Φ( D, i.e. Φ 1 (D. Note tht Φ is liner mp given by the mtrix 1 1, so the function theoretic inverse Φ is lso liner mp given by the inverse mtrix Since Φ nd Φ 1 re liner mps, they tke prllelogrms to prllelogrms. Then 1 1 Φ 1 (D is the solid region in the plne bounded by the point (, Φ 1 (,, (1, Φ(1, 1, (1, 1 Φ(2,, (, 1 Φ(1, 1, i.e. [, 1] [ 1, ] is the squre in the fourth qudrnt of side length 1 with one corner t the origin. So Φ : D is prmeteriztion of D by nice region in which we cn use strightforwrd iterted Fubini integrl. Now we ll consider the prmeterizion fctors in the integrl. We hve Also (f Φ(x, y (x y 2 + (x + y 2 2(x 2 + y 2. D (x,y Φ since Φ is liner (so equls its own totl derivtive nd so det ( (D (x,y Φ t D (x,y Φ det D (x,y Φ 2. Finll we cn use the prmeteriztion formul f (f Φ det ( (DΦ t DΦ D x 1 x y 1 (x 2 + y 2 dydx (x dx By the wy, if you wnted to compute the originl integrl with n iterted integrl, it would be 1 x x y x (x 2 + y 2 dydx + 2 x1 2 x y 2+x (x 2 + y 2 dydx which you cn check for yourself gives the sme nswer, lbeit fter long ugly clcultion. Try it!
3 4. Find the volume of the solid tetrhedrl region bounded in the first octnt by the plne 3x + 4y + 2z 12. Set up the complete iterted integrl. Solution. The solid region we re considering is S {(x, y, z 3 : x, y, z, 3x + 4y + 2z 12}. First note tht the intersection of the plne with the positive x, y, nd zxes is 4, 3, nd 6, respectively. In prticulr, x hs the freedom to go from to 4 in S. Next, we ll consider the freedom of y for ech fixed x. For this, projecting the region to the xyplne (i.e. setting z, we get the eqution 3x + 4y 12. Solving for y in terms of x gives y 3 3 4x. This shows tht for fixed x, y hs the freedom to move from to 3 3 4x. Finlly, we consider the freedom of z for ech fixed x nd y. Solving the eqution of the plne for z in terms of x nd y yields z x 2y. This shows tht for fixed x nd y, z hs the freedom to move from to z 6 3 2x 2y in the region. Thus the iterted integrl is x y 4 x x 2y z 1 dzdydx. 5. Using Green s theorem, evlute the line integrl F dr, where F (x, y (x3 2y 3, x 3 +2y 3 nd D {(x, y 2 : x 2 + y 2 2, x, y }, for some fixed positive rel number. Solution. Since the region D is contined in the xyplne, we cn use version (1 of Green s theorem. In this cse, F dr (x 3 2y 3 dx + (x 3 + 2y 3 dy ( (x 3 + 2y 3 (x3 2y 3 D x y (3x 2 + 6y 2. D Now to compute this integrl, we use the chnge of coordintes (prmeteriztion formul to use circulr coordintes. Indeed, let Φ : 2 2 be defined by Φ(r, θ (r cos(θ, r sin(θ. Then Φ : [, ] [, π/2] D gives prmeteriztion of the region D in terms of nice box. As usul with circulr coordintes, the prmeteriztion fctor is given by det ( (D (r,θ Φ t D (r,θ Φ det D (r,θ Φ cos(θ r sin(θ det r. sin(θ r cos(θ Thus we use the prmeteriztion formul F dr 3 (x 2 + 2y D r r r π/2 θ (r 2 cos 2 (θ + 2r 2 sin 2 (θ r dθdr π/2 r 3 (1 + sin 2 (θ dθ dr θ r 3 ( π 2 + π 4 dr 9π4 16. Do you think this is esier thn computing the originl line integrl? First, there re three pieces to the curve, γ 1 (t ( cos(t, sin(t, t π/2, γ 2 (t (, t, t, γ 3 (t (t,, t,
4 so the line integrl breks into three integrls F dr π/2 + ( 3 cos 3 (t 2 3 sin 3 (t, 3 cos 3 (t sin 3 (t γ 1(t dt ( 2t 3, 2t 3 γ 2(t dt + (t 3, t 3 γ 3(t dt π/2 4 ( cos 3 (t sin(t + 2 sin 4 (t + cos 4 (t + 2 sin 3 (t cos(t dt 2 t 3 dt + Then you will need to do freshmn clculus tricks to compute this. Wht do you think is esier? t 3 dt. 6. Using Stokes Theorem, compute the surfce integrl S F n where F (x3 z, xz 2, 3 nd where S is the solid region {(x, y, z 3 : x 2 + y 2 4z 2, z 1} nd n is some choice of unit norml. NOTE: there s chnge of nottion from the originl sttement of the problem. Solution. By Stokes Theorem we hve F n S S 1 z divf 2z y 2z 4z 2 y 2 x 4z 2 y 2 3x 2 z dxdydz constructing the iterted Fubini integrl just s in problem 4. But this integrl looks hrd, but doble with pproprite trig substitutions. Insted, let s chnge to cylindricl coordintes. Define Φ : 3 3 by φ(r, θ, z (r cos(θ, r sin(θ, z. We see tht the region {(r, θ, z 3 : z 1, θ 2π, r 2z} prmeterizes S vi Φ, i.e. Φ : S. The prmeteriztion fctor is cos(θ r sin(θ det D (r,θ,z Φ det sin(θ r cos(θ 1 r. Finlly, using the prmeteriztion formul, we hve F n divf S S (divf Φ r π 1 θ 2π θ 2π θ 2π θ z 2z cos 2 (θ cos 2 (θ r 1 r 2 cos 2 (θz r drdzdθ z 1 z [ r 4 z 4 cos 2 (θ dθ 2π. ] 2z z 5 dzdθ dzdθ
5 7. Let S 2 {(, } nd define F : S 2 by ( y F (x, y x 2 + y 2, x x 2 + y 2. Compute the line integrl C F dr for ny closed curve C 2 not enclosing the origin. Solution. Since the problem is sking to compute line integrl for pretty much rbitrry curve, the nswer probbly doesn t depend on the curve. So let s just sy we hve ny curve C no enclosing the origin nd let 2 be the region it bounds. Then by Green s theorem version (1 (which we cn pply, since F is nice in the region, C F dr y x 2 + y 2 dx + ( x x x 2 + y 2 y. x x 2 + y 2 dy y x 2 + y 2 8. Find bsis for the vector spce of 3 3 symmetric mtrices. Here symmetric mens A t A. Solution. A bsis is given by e ij for ll 1 i j 3, where e ij is the 3 3 mtrix of ll zeros except for 1 in the ij nd ji spots. If i j then the mtrix e ii just hs 1 in the ii spot. For exmple e 11 1, e In totl, there re 6 such mtrices, so the vector spce of 3 3 symmetric mtrices hs dimension Let V n be the vector spce of polynomils of degree n. Consider the liner trnsformtion T : V 1 V 3 defined by T (p(x x 2 p(x 1. Compute bsis of ker(t nd im(t. Solution. Choose bses 1, x nd 1, x, x 2, x 3 for V 1 nd V 3, respectively. Then T (1 x 2 T (x x 2 (x 1 x 3 x 2 so the mtrix of T with respect to these bses is In prticulr, we see tht the rnk of T is 2, so by the rnknullity theorem, the kernel of T hs dimenion, i.e. ker(t {} nd T is injective. The imge of T is spnned by x 2 nd x 3 x 2, which re linerly independent, so form bsis. A nicer bsis is x 2 nd x 3, which is gotten by tking liner combintions. 1. Let A b c where ±b. Compute the dimensions of ker(a nd im(a. b c
6 Solution. The condition ±b implies, in prticulr, tht nd b re both not zero. So A is never the zero mtrix, so cn never hve rnk. Since A is 2 3 mtrix, dim ker(a 3, dim im(a 2, nd dim ker(a + dim im(a [ 3. Let s [ consider the imge, which is the column spn of A. We b lredy see tht the vectors nd re linerly independent. Indeed, b] ] b det b 2 b 2, which is never zero s long s ±b. In prticulr, the imge of A hs dimension t lest 2, thus is must hve dimension 2. By the rnknullity theorem, the dimension of the kernel is 1. By the wy, bsis for the kernel is given by the (column vector ( c +b, c +b, 1t. 11. Let A be n n n rel mtrix. Wht re the possible vlues of det A if A is symmetric. Solution. Any rel number. Indeed, for ny rel number, the n n digonl mtrix with digonl consisting of one nd n 1 1 s (nd ll the offdigonl entries zero is symmetric nd hs determinnt. b Wht re the possible vlues of det A if A is invertible. Solution. Any nonzero rel number. We know tht n n n invertible mtrix hs nonzero determinnt. Now we ll show tht the determinnt chieves ll nonzero vlues. Given ny nonzero rel number, the digonl mtrix considered bove is invertible nd hs determinnt. 12. For ll (x, y (,, define f(x, y sin(x2 + y 2 x 2 + y 2. Cn f(, be ssigned to mke f continuous t the origin? Solution. It s wellknown from clculus (when you find the derivtive of sin tht nd tht the function sin sin(z ( lim cos( 1, z z g(z { sin(z z for z 1 for z is continuous. Thus we should set f(, 1. To prove tht f, s defined, is continuous, let γ : 2 be continuous curve in the plne, with γ( (,. Note tht r : 2 defined by r(x, y x 2 +y 2 is continuous. Finlly, we now relize f s composition of functions f g r, which re ech continuous everywhere, mking f continuous everywhere, with f(, g(r(, g( Show tht the function f : 2 defined by f(x, y xy, is not differentible t the origin (,. Solution. A good necessry criterion for function f to be differentible t p is tht the difference quotient, f(p + h f(p, h
7 is bounded s h p. In our sitution, unfortuntely our difference quotient is f(u, v f(, uv (u, v u 2 + v, 2 for (u, v (,, which is bounded (by 1 2. So we must use different strtegy. We wnt to sk if liner mp D (, f exists so tht the limit f(h + D (, f(h lim. h h Now long the pth h (, b with b, we hve + D (, f(1, 1 lim lim (,b (, 1 + D (,(1, 1, b which gives limit of either 1+D (, f(1, 1 or 1 D (, f(1, 1 depending on whether the pther is pproching from the first or third qudrnt in the plne. Now both of these vlues cn t be, so the originl limit cn t equl. Thus f is not differentible t the origin.
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