Exam 2 Solutions (Form A) Fall 2000.pdf

Student (Print) Section Last, First Middle Student (Sign) Student ID Instructor MATH 152 Exam 2 Fall 2000 Test Form A Solutions Part I is multiple choice. There is no partial credit. Part II is work out. Show all your work. Partial credit will be given. You may not use a calculator. 1-10 /50 11 /10 12 /10 13 /10 14 /10 15 /10 TOTAL Formulas: M n G1DG20x f x 0 G0Ex 1 2 G0Ef x 1 G0Ex 2 2 G0EG43G0Ef x nG221 G0Ex n 2 G3Blnxdx G1D xlnx G22 x G0E C T n G1D G20x 2 GA1fG9FxoGA0 G0E 2fG9Fx 1 GA0 G0E G43 G0E 2fG9Fxn G221 GA0 G0E fG9FxnGA0GA2 G3Btan xdx G1DG22ln|cos x| G0E C S n G1D G20x 3 GA1fG9FxoGA0G0E 4fG9Fx 1 GA0G0E 2fG9Fx 2 GA0G0EG43G0E 2fG9Fxn G222 GA0G0E 4fG9Fxn G221 GA0G0E fG9FxnGA0GA2 G3Bsecxdx G1D ln|secx G0E tan x| G0E C |E M | G1C KG9Fb G22 aGA0 3 24n 2 where K G1E f G55G55 G9FxGA0 for all x in GA1a, bGA2 |E T | G1C KG9Fb G22 aGA0 3 12n 2 where K G1E f G55G55 G9FxGA0 for all x in GA1a, bGA2 |E S | G1C KG9Fb G22 aGA0 5 180n 4 where K G1E f G9F4GA0 G9FxGA0 for all x in GA1a, bGA2 1 Part I: Multiple Choice (5 points each) There is no partial credit. You may not use a calculator. 1. Calculate the x-component (or coordinate) of the center of mass of a plate with uniform density G3E and whose shape is the quarter circle of radius 3 given by 0 G74 x G74 3 and 0 G74 y G74 9 G22 x 2 . a. G3D 4 b. 9G3D 4 G3E c. 9G3E d. 8 G3D e. 4 G3D correctchoice M G1D G3B 0 3 G3E 9 G22 x 2 dx G1D G3E 1 4 G3Dr 2 G1D 9G3D 4 G3E M y G1D G3B 0 3 G3Ex 9 G22 x 2 dx G1D G22G3E 1 3 G9F9 G22 x 2 GA0 3/2 0 3 G1D G3E 1 3 9 3/2 G1D 9G3E xG06G1D M y M G1D 9G3EG194 9G3DG3E G1D 4 G3D 2. Which of the following gives the Trapezoid Rule approximation to G3B 2 4 1 x dx with n G1D 4? a. 1 4 1 4 G0E 2 5 G0E 1 3 G0E 2 7 G0E 1 8 b. 1 4 1 2 G0E 2 5 G0E 1 3 G0E 2 7 G0E 1 4 c. 1 4 1 2 G0E 4 5 G0E 2 3 G0E 4 7 G0E 1 4 correctchoice d. 1 2 ln2 G0E ln 3 2 G0E ln3 G0E ln 7 2 G0E ln4 e. 1 4 ln2 G0E 2ln 3 2 G0E2ln3G0E2ln 7 2 G0Eln4 G20x G1D 4 G22 2 4 G1D 1 2 T4 G1D G20x 2 fG9F2GA0 G0E 2f 5 2 G0E 2fG9F3GA0 G0E 2f 7 2 G0E fG9F4GA0 T4 G1D 1 4 1 2 G0E 4 5 G0E 2 3 G0E 4 7 G0E 1 4 2 3. If you use the Trapezoid Rule with n G1D 4 to approximate G3B 2 4 1 x dx, which of the following is the smallest upper bound on the error |E T | in the approximation? a. 1 12 b. 1 24 c. 1 48 d. 1 96 correctchoice e. 1 192 f G55G55 G9FxGA0 G1D 2 x 3 On GA12,4GA2 the maximum is K G1D f G55G55 G9F2GA0 G1D 2 8 G1D 1 4 . So the error is bounded by |E T | G1C KG9Fb G22 aGA0 3 12n 2 G1D 1 4 G9F4 G22 2GA0 3 12G9F4GA0 2 G1D 1 96 4. An integrating factor for the differential equation dy dx G1D xsin2x G0E y tan x is IG9FxGA0 G1D a. |cos x| correctchoice b. e G22sec 2 x c. e G22|tan x| d. G22|secx| e. e |cosx| Standard form: dy dx G22 G9Ftan xGA0y G1D x sin2xPG1DG22tan x G3B Pdx G1D G22G3Btan xdx G1DG22G3B sinx cos x dx G1D ln|cosx| IG9FxGA0 G1D e G3B Pdx G1D e ln|cosx| G1D |cos x| 5. The parametric curve x G1D e t G22 t, y G1D 4e t/2 for 0 G74 t G74 2 is rotated about the x-axis. Which integral gives the area of the surface of revolution? HINT: Look for a perfect square. a. G3B 0 2 2G3DG9Fe t G22 tGA0G9Fe 2t G0E2e t G0E1GA0dt b. G3B 0 2 2G3DG9Fe t G22 tGA0G9Fe t G0E1GA0dt c. G3B 0 2 8G3De t/2 G9Fe 2t G0E 2e t G0E 1GA0dt d. G3B 0 2 8G3De t/2 G9Fe t G0E 1GA0dt correctchoice e. G3B 0 2 8G3De t/2 e t G0E 1 dt dx dt G1D e t G22 1 dy dt G1D 2e t/2 A G1D G3B 2G3Drds G1DG3B 0 2 2G3Dy dx dt 2 G0E dy dt 2 dt G1D G3B 0 2 2G3D4e t/2 G9Fe t G22 1GA0 2 G0EG9F2e t/2 GA0 2 dt G1D G3B 0 2 8G3De t/2 G9Fe 2t G22 2e t G0E 1GA0G0EG9F4e t GA0 dt G1D G3B 0 2 8G3De t/2 e 2t G0E 2e t G0E 1 dt G1D G3B 0 2 8G3De t/2 G9Fe t G0E 1GA0dt 3 6. A tank is completely filled with water. The end is a vertical semicircle with radius 5 m as shown. Which integral gives the hydrostatic force on this end of the tank? The density of water is 1000 kg m 3 and g G1D 9.8 m sec 2 . 0 5 -5 5 a. 9800 G3B 0 5 G9F5 G22 yGA0225G22y 2 dy correctchoice b. 9800 G3B 0 5 G9F5 G0E yGA0 25 G22 y 2 dy c. 9800 G3B 0 5 G9F5 G22 yGA0G9F25 G22 y 2 GA0dy d. 9800 G3B 0 5 G9F5 G0E yGA0225G22y 2 dy e. 9800 G3B 0 5 G9F5 G0E yGA0G9F25 G22 y 2 GA0dy The slice at height y is at a distance h G1D 5 G22 y below the surface. Its width is w G1D 2x where x G1D 25 G22 y 2 F G1D G3BG3Eghwdy G1D 9800 G3B 0 5 G9F5 G22 yGA0225G22y 2 dy 7. Compute G3B e G2E 1 xlnx dx. a. 0 b. 1 e c. 1 d. e e. G2E correctchoice u G1D lnxduG1D 1 x dx G3B e G2E 1 xlnx dx G1D G3B 1 u du G1D ln|u| G1D ln|lnx| e G2E G1D lim bG76G2E ln|lnb| G22 ln|lne| G1D G2E 4 8. Compute lim nG76G2E lnG9F2 G0E e n GA0 3n a. 0 b. 1 3 correctchoice c. 2 3 d. ln2 3 e. G2E Use l?Hopital?s Rule twice: lim nG76G2E lnG9F2 G0E e n GA0 3n G1D lim nG76G2E e n 2 G0E e n 3 G1D 1 3 lim nG76G2E e n 2 G0E e n G1D 1 3 lim nG76G2E e n e n G1D 1 3 OR: Since e n is much greater than 2, lim nG76G2E lnG9F2 G0E e n GA0 3n G1D lim nG76G2E lnG9Fe n GA0 3n G1D lim nG76G2E n 3n G1D 1 3 9. After separating variables, we can solve the differential equation dy dt G1D ty G0E t y by solving a. G3B 1 y G0E y dy G1D G3B tdt b. G3B y y 2 G0E1 dy G1D G3B tdt correctchoice c. G3BG9Fy 2 G0E yGA0dy G1D G3BG9Ft G0E 1GA0dt d. G3B y 2 G0E 1 y 1 dy G1D G3B t dt e. None of the above, the differential equation is not separable. dy dt G1D t y G0E 1 y G1D t y 2 G0E 1 y GAE y y 2 G0E 1 dy G1D tdt GAE G3B y y 2 G0E1 dy G1D G3B tdt 10. Find the arc length of the parametric curve x G1D 4 cos t, y G1D 4 sint for G3D 6 G74 t G74 G3D 3 . a. G3D 3 b. 2G3D 3 correctchoice c. G3D 2 d. 4G3D 3 e. 3G3D 4 dx dt G1D G224 sint dy dt G1D 4 cos t L G1D G3B G3D/6 G3D/3 dx dt 2 G0E dy dt 2 dt G1D G3B G3D/6 G3D/3 16 sin 2 t G0E 16 cos 2 t dt G1D G3B G3D/6 G3D/3 4dt G1D 4 G3D 3 G22 G3D 6 G1D 2G3D 3 OR: G3D/3 G22 G3D/6 2G3D 2G3D4 G1D 2G3D 3 5 Part II: Work Out (10 points each) Show all your work. Partial credit will be given. You may not use a calculator. 11. Determine whether each of the following sequences converges or diverges. If it converges, find the limit. Fully justify your answers. a. lim nG76G2E sin nG3D 2 Circle one: Converges Diverges Explain: sin nG3D 2 takes the values 1,0,G221,0 repeatedly. So it can never have a limit. b. lim nG76G2E cos 2 n 2 n Circle one: Converges Diverges Explain: 0 G74 cos 2 n 2 n G74 1 2 n and lim nG76G2E 1 2 n G1D 0.Bythe sandwich theorem, lim nG76G2E cos 2 n 2 n G1D 0. 12. Determine whether each of the following integrals converges or diverges. If it converges, find its value. Hint: Partial fractions a. G3B 1 G2E 1 x G0E 2x 2 dx Circle one: Converges Diverges Explain: Partial fractions gives 1 x G0E 2x 2 G1D 1 x G22 2 1 G0E 2x G3B 1 G2E 1 x G0E 2x 2 dx G1D G3B 1 G2E 1 x G22 2 1 G0E 2x dx G1D lnx G22 lnG9F1 G0E 2xGA0 1 G2E G1D ln x 1 G0E 2x 1 G2E G1D ln 1 2 G22 ln 1 3 G1D ln 3 2 b. G3B 0 1 1 x G0E 2x 2 dx Circle one: Converges Diverges Explain: G3B 0 1 1 x G0E 2x 2 dx G1D G3B 0 1 1 x G22 1 1 G0E 2x dx G1D lnx G22 lnG9F1 G0E 2xGA0 0 1 G1D GA10 G22 ln3GA2 G22 GA1G22G2E G22 0GA2 G1D G2E 6 13. A tank contains 20 lb of salt mixed with 50 gal of water. Salt water containing 2 lb of salt per gal is added to the tank at the rate of 5 gal per min. The tank is kept thoroughly mixed and drains at the same rate. a. Write out the differential equation and the initial condition for SG9FtGA0, the number of lbs of salt in the tank at time t. dS dt lb min G1D 2 lb gal 5 gal min G22 SG9FtGA0 lb 50 gal 5 gal min dS dt G1D 10 G22 1 10 SSG9F0GA0G1D20 b. Solve the initial value problem. dS dt G1D 10 G22 1 10 S G3B dS 10 G22 1 10 S G1D G3B dt G22 10ln 10 G22 1 10 S G1D t G0E C ln 10 G22 1 10 S G1D G22 t 10 G22 C 10 10 G22 1 10 S G1D Ae G22t/10 S G1D 100 G22 10Ae G22t/10 SG9F0GA0 G1D 20 G1D 100 G22 10AAG1D8SG1D100 G22 80e G22t/10 c. How much salt is in the tank after 10 hours? SG9F600GA0 G1D 100 G22 80e G22600/10 G1D 100 G22 80e G2260 The problem should have said 10 min which gives SG9F10GA0 G1D 100 G22 80e G2210/10 G1D 100 G22 80 e 14. Compute the surface area of the surface obtained by rotating the curve x G1D 1 G0E 2y 2 for 1 G74 y G74 2 about the x-axis? A G1D G3B 2G3Drds G1DG3B 1 2 2G3Dy 1G0E dx dy 2 dy G1D G3B 1 2 2G3Dy 1 G0E 16y 2 dy u G1D 1 G0E 16y 2 du G1D 32ydy du 32 G1D ydy A G1D 1 32 G3B yG1D1 2 2G3D u du G1D G3D 16 2u 3/2 3 yG1D1 2 G1D G3D 24 G9F1 G0E 16y 2 GA0 3/2 1 2 G1D G3D 24 G9F65GA0 3/2 G22 G3D 24 G9F17GA0 3/2 15. Solve the initial value problem x dy dx G0E 2y G1D cos x x with yG9FG3DGA0 G1D 1. dy dx G0E 2 x y G1D cosx x 2 P G1D 2 x I G1D e G3B Pdx G1D e G3B 2 x dx G1D e 2lnx G1D x 2 x 2 dy dx G0E 2xy G1D cos x d dx G9Fx 2 yGA0 G1D cos xx 2 yG1DG3Bcos xdx G1D sinx G0E C x G1D G3D when y G1D 1 G3D 2 G1D sinG3DG0E CCG1DG3D 2 x 2 yG1Dsinx G0EG3D 2 y G1D sinx G0EG3D 2 x 2 7