Math 227 Problem Set V Solutions. f ds =

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1 Mth 7 Problem Set V Solutions If is urve with prmetriztion r(t), t b, then we define the line integrl f ds b f ( r(t) ) dr dt (t) dt. Evlute the line integrl f(x,y,z)ds for () f(x,y,z) xosz, the urve with prmetriztion r(t) tîı+t ĵj, t. (b) f(x,y,z) x+y y+z, the urve with prmetriztion r(t) ( t, 3 t3/,t ), t. Solution. () In this se r(t) tîı+t ĵj, so tht v(t) îı+tĵj nd ds dt +4t. Hene f(x,y,z)ds x(t)osz(t) ds dt dt t(os) +4t dt 8 (+4t ) 3/ 3/ (b) In this se r(t) ( t, 3 t3/,t ), so tht v(t) (,t /, ) nd ds dt +t. Hene f(x,y,z)ds x(t)+y(t) ds y(t)+z(t) dt dt t+ 3 t3/ 3 t3/ +t 53/ +tdt (+t) 3/ 8 33/ 3/ 3/. () Show tht the integrl f(x,y)ds long the urve given in polr oordintes by r r(θ), θ θ θ, is θ f ( r(θ)osθ,r(θ)sinθ ) r(θ) + ( dr (θ)) θ (b) ompute the r length of r +osθ, θ π. Solution. () The urve is r(θ) x(θ)îı+y(θ)ĵj with x(θ) r(θ)osθ, y(θ) r(θ)sinθ nd θ π. On this urve Hene v(θ) x (θ)îı+y (θ)ĵj r (θ)osθ r(θ)sinθ ] îı+ r (θ)sinθ +r(θ)osθ ] ĵj ds r (θ)osθ r(θ)sinθ ] + r (θ)sinθ +r(θ)osθ ] r (θ) +r(θ) f(x,y)ds θ f ( x(θ),y(θ) ) ds θ f ( r(θ)osθ,r(θ)sinθ ) r(θ) + ( dr (θ)) θ θ (b) In the se f(x,y), r(θ) +osθ, θ nd θ π, ds π π +osθ] + sinθ] 4os θ π (+osθ) π π os θ 4 os θ 8sin θ π 8

2 3. Find the mss of wire formed by the intersetion of the sphere x + y + z nd the plne x+y +z if the density t (x,y,z) is given by ρ(x,y,z) x grms per unit length of wire. Solution. The urve in question is the intersetion of sphere with plne through the entre of the sphere. So the urve is irle (in ft, gret irle on the sphere). A simple wy to prmetrize suh irle is to pik two unit vetors îı, ĵj tht re mutully perpendiulr nd perpendiulr to (,,) (so tht they both lie in the plne x+y+z ). For exmple îı (,,) nd ĵj (,, ). (hek the orthogonlity requirements by dotting îı nd ĵj with eh other nd with (,,).) Then r(θ) osθ îı +sinθ ĵj ( osθ, osθ, ) + ( ) sinθ, sinθ, sinθ isprmetriztionofthe irle. Asîı ndĵj remutullyperpendiulrunit vetors,r (θ) sinθîı + osθ ĵj is unit vetor nd ds. So π π mss ρds x(θ) ds osθ ] + sinθ π os θ + osθ sinθ + sin θ As π os θ π sin θ π sin θ+os θ] π nd π osθsinθ π sinθ, the mss is π + π 3 π. ] 4. Evlute F dr for () F(x,y) xyîı x ĵj long y x from (,) to (,). (b) F(x,y,z) (x z)îı + (y z)ĵj (x + y)ˆk long the polygonl urve from (,,) to (,,) to (,,) to (,,). Solution. () We my prmetrize the urve by r(t) tîı+t ĵj with t. Then v(t) îı+tĵj nd F ( x(t),y(t) ) t 3 îı t ĵj so F dr t3îı t ĵj ] îı+tĵj ] dt t 3 ] dt 4 (b) On the first segment of the urve y z so F simplifies to xîı xˆk nd dr îıdx (i.e. we n prmetrize the first segment of the urve by r(x) xîı with x ), so F dr xdx. On the seond segment of the urve x nd z so F simplifies to îı+yĵj (+y)ˆk nd dr ĵjdy, so F dr ydy. On the finl segment of the urve x y so F simplifies to ( z)îı+( z)ĵj ˆk nd dr ˆkdz, so F dr dz. So F dr xdx+ ydy + ( )dz + 5. Evlute x y dx + x 3 ydy ounterlokwise round the squre with verties (,), (,), (,) nd (,). Solution. On the first side, y nd dy. Tht is, we my prmetrize the first side by r(x) xîı with x. On the seond side x nd dx. On the third side y nd dy. We my prmetrize the third side by r(x) xîı with x running from to. On the finl side x nd dx. So x y dx+x 3 ydy x dx+ 3 ydy + x dx+ 3 ydy 3

3 . Let < b nd < d. Let r :,b] IR n be prmetriztion for some urve. Let u :,d],b] be inresing, differentible nd obey u() nd u(d) b. Set R(t) r(u(t)). It is nother prmetriztion of. () Prove tht b F ( r(t)) r (t)dt d F ( R(t)) R (t)dt (b) Set nd d. Find funtion u :,d],b] whih is inresing, infinitely differentible nd obeys u() nd u(d) b. So, it is lwys possible to prmetrize urve in suh wy tht the prmeter runs from to. () Set nd d. Find funtion u :, d], b] whih is stritly inresing, infinitely differentible nd obeys u(), u(d) b, u () nd u (d). So, it is lwys possible to prmetrize urve in suh wy tht the prmeter runs from to nd the veloity is zero t both end points. (d) When we ssign prmetriztion r :,b] IR n to urve, we re impliitly designting one of the two end points of the urve (nmely r()) s the beginning of the urve nd the other end point (nmely r(b)) s the end of the urve. This is lled ssigning n orienttion to the urve. Given urve with orienttion determined by prmetriztion r :,] IR n, define to be the urve with orienttion determined by the prmetriztion w(t) r( t). The beginning of is the end of nd vie vers. This lled reversing the orienttion. Show tht F dr F dr Solution. () Let f(u) F ( r(u)) r (u) nd g(t) F ( R(t)) R (t). By the hin rule R (t) r (u(t))u (t) so tht g(t) F ( r(u(t))) r (u(t))u (t) f ( u(t) ) u (t) nd b d F ( r(t) ) r (t)dt F ( R(t) ) R (t)dt b d f(u)du f ( u(t) ) u (t)dt Mking the hnge of vribles u u(t) in the first integrl gives the seond. (b) u(t) + b d (t ) () Define v(t) t( t). Observe tht v(), v() nd v(t) > for ll < t <. Define w(t) t v(t )dt t t3 3. Then w(), w () v(), w(), w () v() nd w is stritly inresing on t. So does the trik. (d) F dr ( u(t) +(b )w F ( w(t) ) w (t)dt Now mke the hnge of vribles t u. F dr F ( r(u) ) ( r (u) ) ( du) F dr 3 t d ) F ( r( t) ) ( r ( t) ) dt F ( r(u) ) r (u)du F ( r(u) ) r (u)du

4 7. Find the work F dr done by the fore field F (x+y)îı+(x z)ĵj+(z y)ˆk in moving n objet from (,, ) to (,,3) long ny smooth urve. Solution. The point here is tht F is onservtive, s F φ with φ x z +yx yz +. Hene for ny differentible urve prmetrized by r(t), with r(t ) (,, ) nd r(t ) (,,3), t F dr F ( r(t) ) r t (t)dt φ ( r(t) ) r t d (t)dt dt φ( r(t) )] dt t t by the hin rule. So, by the Fundmentl Theorem of lulus F dr φ ( r(t ) ) φ ( r(t ) ) φ(,,3) φ(,, ) ] + + ] 9 t 8. Evlute yexy sin(y +z)dx+e xy( xsin(y +z)+os(y +z) ) dy +e xy os(y +z)dz long the stright line segment from (,,) to (, π 4, π 4). Solution. Trik question. This F is onservtive with F ( e xy sin(y +z) ). So ye xy sin(y+z)dx+e xy( xsin(y+z)+os(y+z) ) dy+e xy os(y+z)dz e xy sin(y+z) (,π 4,π 4 (,,) ) e π/4 9. Let F be ontinuous vetor field on IR d. Prove tht the following two sttements re equivlent. (i) If nd renytwo urvesin IR d with the sme initil nd finl points then F dr F dr. (ii) If is ny losed urve in IR d (i.e. its initil nd finl points re the sme) then F dr. Solution. Assume tht property (i) holds. Let be ny losed urve. Let be the initil nd finl points of. Let be the urve with prmetriztion R(t), t. Tht is, is just single point, viewed s (degenerte) urve. Then R (t), so F dr F( R(t) ) R (t)dt. The two urves nd hve the sme initil nd finl points, so by property (i), F dr F dr nd property (ii) holds s well. Now, ssume tht property (ii) holds. Let nd be ny two urves with the sme initil nd finl points. Define to the losed urve formed by first following from its initil to its finl points nd then following bkwrds from its finl point to its initil point. This is usully denote. We n onstrut prmetriztion for s follows. Let r :,] IR d nd r :,] IR d be prmetriztions for nd respetively. Then R :,] IR d, defined by { r (t) if t R(t) r ( t) if t is prmetriztion of. (If you re wrried bout disontinuity in the veloity vetor R (t) t t, just hoose r nd r so tht r () r (). This is lwys possible by prt of the lst problem.) Then F dr F ( R(t) ) R (t)dt F ( R(t) ) R (t)dt+ F ( R(t) ) R (t)dt F dr+ F dr F dr F dr But, s is losed urve, F dr. So F dr F dr nd property (i) holds. 4

5 . Let F be ontinous vetor field on IR d. Fix ny two points Q nd Q in IR d nd ssume tht D F dr D F dr for ll urves D nd D tht strt t Q nd end t Q. Let P nd P be ny two points in IR d. Prove tht F dr F dr for ll urves nd tht strt t P nd end t P. Solution. Let P nd P be ny two points in IR d nd nd be ny two urves tht strt t P nd end t P. Fix ny urve E from Q to P nd ny urve E from P to Q. Then D E + +E nd D E + +E re urves from Q to Q. So F dr F dr D D F dr+ F dr+ F dr F dr+ F dr+ F dr E E E E F dr F dr. Is F yîı+xĵj+yˆk onservtive field? Find infinitely mny losed urves for whih F dr. Solution. y F 3 y y nd z F z x re different. So the field nnot be onservtive. On the other hnd, for ny urve tht lies in the plne z b, for ny fixed vlue of the onstnt b, dz dt so F dr ydx+xdy beuse the vetor field G yîı+xĵj (xy) is onservtive. 5