Calculus with Aalytic Geometry Fial Eam Study Guide ad Sample Problems Solutios The date for the fial eam is December, 7, 4-6:3p.m. BU Note. The fial eam will cosist of eercises, ad some theoretical questios, from the topics we covered i class from Chapters 6, 7, 8, ad 9 of the tetbook. I am listig here a umber of problems that, i my opiio, you should kow how to do. Cosider them practice. The eam as such will certaily be much shorter ad probably much easier. It could also be quite differet; the mai thig it will have i commo with this sample eam is that the skills you eed to do well with these problems are the same you eed for the fial. The agai,the fial could be just a subset of this problem list. Time will tell.. Compute the followig itegrals: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) e e π/ d l 3 () cos d t + 8 t 7t + dt 3 e d. ( + 4)( + 4 + 5) d. d 9. si 7 θ cos 4 θ dθ. 4 + + 3 6 + 8 d. 3 e d. 9 d. / / 6 d. + 6 + 9 d You ca use (for eample) Wolfram alpha to check if you computed these itegrals correctly. The itegral i (e) diverges.. The lie y = a divides the regio bouded by the curve y = ( ) ad the -ais ito two regios of equal area. Fid a. Equatig a = ( ) there is a solutio with if ad oly if a ; the it is = a. The full are bouded by the curve is ( ) d = 6.
Oe of the areas determied by the lie y = a has area We eed to have ( a)3 6 = 3. The regio bouded by the curves a (( ) a) d = ( ) thus a = 3 6. ( a)3. 6 y =, =, = 4, + is rotated about the y-ais. Draw a sketch of the regio ad calculate the volume of the solid so obtaied. Graph of the regio: By the shell method 4 V = π +, d = π l 7. 4. Let R be the regio i the first quadrat bouded by the curves y = 3 ad y =. Calculate the followig quatities. (a) The area of R. (b) The volume obtaied by rotatig R about the -ais. (c) The volume obtaied by rotatig R about the y-ais. (d) The volume obtaied by rotatig R about the lie = 3. The curves itersect at =, = ; a sketch shows that y = is above y = 3 for. The aswers are: (a) A = (b) By washers: V = π (c) By shells, V = π (d) By shells, V = π ( 3 ) d = 5. ( ( ) ( 3 ) ) d = 4π 5. ( 3 ) d = 3π 3. (3 )( 3 ) d = 3π 5. 5. The base of a solid is a square with vertices located at (, ), (, ), (, ) ad (, ). Each cross-sectio perpedicular to the -ais is a semicircle. Fid the volume of the solid. A sketch helps. The radius of the semicircular cross sectio at is r() = so that the area of the cross sectio is A() = π ad the volume is V = A() d = π d = π 3.
3 6. Fid the legth of the followig arcs. (a) y = cosh(), l 4. (b) y = l, 5. L = 4 L = + sih d = 5 4 cosh d = sih 4. + 5 d = + d. This is a asty itegral. Nothig like it will appear i the eam. The itegral is probably best computed by the substitutio = ta θ. It becomes L = arcta 5 arcta sec θcscθ dθ. This itegral ca ow be computed by itegratio by parts. (c) y = 3 5 +,. Note: There was a typo i this eercise. It should have bee y = 3 6 + L = = + ( + ( ) ) d = d = 7. + 4 4 + 4 4 d =,. The ( + ) d 7. A tak i the shape of a regular circular cylider havig a radius of meters ad a ais that is 5 meters log is layig o its side half filled with water. Figure below, stole from the web, shows the tak. The correspodece with the picture is h = m, l = 5 m. Assume water has a mass of kg/m 3 ad at sea level (where we are) a mass of kg weighs 9.8 ewtos. A pump at the top of the tak will empty the tak. Calculate the work eeded to empty it. Figure shows the frot of the tak ad a suggestio for a ais ad coordiates. Figure Figure The water eteds from y = to y =. At level y, a horizotal slab of water of thickess y has width 4 y ad legth the legth of the tak; i.e., 5 meters. Its volume is V = 4 y Dy cubic meters. The weight of such a slab is W t = V 9.8 = 98 4 y y ewtos. This
4 slab has to be lifted to the top, a distace of y (recall y < ). Thus the work eeded to lift it is W = W t ( y) = 98 4 y (y ) y Joules. Addig up over y s, lettig y (or replacig it by dy) oe gets ( W = 98 4 y ( y) dy = 98 4 y dy y ) 4 y dy. The first itegral ca be computed by the substitutio y = si t, the secod by u = 4 y. The ed result is ( W = 98 π + 8 ) joules, 3 8. Determie the ature (covergece or divergece) of the followig itegrals ad give reasos i each case. (a) (b) d. It coverges. The problem is at where the deomiator is. We have b b d = = b b ( )/ b ( = ( b ) /) =. b + 4 d. Coverges First of all the problem is at, so it will coverge if ad oly if + 4 d coverges. Sice + 4 = 3/ ad 3/ d coverges, so does d, by compariso. + 4 9. Evaluate the followig its, or declare that the correspodig sequece diverges. (a) (b) l. Thus ( ) arcta +. We have, by L Hôpital, arcta l = = =. l =. ( ) ( + = arcta ) + = arcta = π 4.
5 (c) e. L Hôpital applied directly to it is clear ad evidet (is it?, thik about it) e is a bit messy because of the square root. However, e = e ad ow two applicatios of L Hôpital show that the it is. (!) (!) (d). Hit: The ratio test applied to the series could be of some use; if you use it ()! ()! = correctly. Let s apply the ratio test: ((+)!) (+)! (!) ()! (( + )!) ()! ( + ) = ( + )!(!) == ( + )( + ) = 4 <. (e) What this tells us is that the series = (!) ()! coverges; it follows that (!) ()! =. (si ) (cos ) (cos ) = while si. The (cos si ) ( cos ). The sequece coverges to by the squeeze theorem. ( ) (f). 3 + ( ) = e 3 3 + (g) { 3 + 5 }. It should have bee { 3 + 5 (3 { ) 3 + 5 = 5{ + = 5( + ) = 5. 5. Show that the followig series coverges, ad compute its sum. k= 6k + 8k 3 6k + 8k 3 = 6(k + k 3 8 ) = 6(k 4 )(k + 3 4 ).
6 Applyig partial fractios 6k + 8k 3 = 6 (k 4 )(k + 3 4 ) = ( A 6 k 4 + B ) + 3. 4 Oe gets A =, B = ad 6k + 8k 3 = 6 ( k 4 B ) + 3 4 = 6k 4 6k +. We ow have for large k= ( 6k + 8k 3 = 4 ) ( + ) ( + 8 8 ) ( ) + + 44 6 4 6 + Sice 4 6 = 4, the series coverges ad its sum is 4. = 4 6. Determie the ature (coditioal covergece, absolute covergece, or divergece) of the followig series. Give brief reasos i each case. (a) = l(). series. si k (b) k k. k= Diverges; from 3 owards l() >, so it diverges by compariso with the harmoic The series coverges absolutely, by compariso with the p=series with p = 3/: si k k k k. 3/ (c) (d) (e) (/). l() l () = Coverges, for eample by it compariso with = l. ( ) ( ). = ( ) } Diverges, the sequece {( ) does ot coverge to (or to aythig else). ( = ). Coverges. If we apply the root test we get ( By the root test, the series coverges. ) ( ) = = e <.
7 (f) ( 6 ) + + 3. = Coverges. 6 + + 3 = ( 6 + + 3 )( 6 + + + 3 ) = 6 + + 6 6 + + + 3 6 + + + 3 = + 6 + + + + 3 3 3 ; sice coverges, so does our series by compariso.. Show that the followig series diverges: is, the series a, where a = ( ) b, b = = { /, if is odd, /, if is eve. 4 + 3 6 + 5 64 ± Eplai why the divergece of this series does ot cotradict the alteratig series covergece test. Suppose the series coverges. The oe would have ; that k = a k= = ad sice both series o the right coverge, so would the series o the left. However, the series o the left diverges. k= 4 k The divergece of this series does ot cotradict the alteratig covergece test because its terms do ot costitute a decreasig sequece. 3. Determie all the values of p for which the followig series coverges coditioally, ad all the p s for which it coverges absolutely. JUSTIFY. ( ) (l ) p. = By the itegral test, the series coverges absolutelyif ad oly if (l ) p d coverges. This itegral coverges if ad oly if p >. I omit the calculatio. The sequece decreases oce l > p ad coverges to so the coclusio is: { } (l ) p The series coverges absolutely for p >. It coverges coditioally for all values of p (icludig p < ). 4. Determie all the values of for which the followig series coverges. ( + + ). = Determie its sum for these values of. The series is geometric with ratio + +, so it will coverge for all values of (if ay) such that < + + <.
8 Now + > is true for all ; we are reduced to + < which becomes < <. So covergece occurs if ad oly if < <. The sum is ( + + ) = +. 5. Fid the radii ad itervals of covergece of the followig power series + (a).! = By the ratio test = +3 (+)! +! = + = ; (b) (c) The radius is, the iterval (, ). 3. = By the ratio test + 3 + (+) + 3 = 3( + ) ( + ) = 3. The radius is 3. For the iterval we have to test the edpoits ±3. For = ±3 the series satisfies 3 = 3/, = which is a p series with p = 3/ >, so we have (absolute) covergece at both edpoits. The iterval is [ 3, 3]. + ( 3). = By the ratio test + + ( 3)+ = + ( 3) = 3 ( + ) + = 3. The radius is. For the iterval we have to test the edpoits 3 ±. For = 3 the series becomes ( ) + which coverges by the alteratig series criterio. For = 3 + the series becomes + = which diverges (oce agai, the harmoic series). The iterval is [ 5, 7 ). =
9 (d) ( ) 3 ( + ). = This is quite similar to b or c. The radius is 3, the iterval is ( 4, ]. 6. Fid the Taylor polyomial of order 4 geerated by f() = cosh = e + e at =, ad estimate the error made whe evaluatig cosh by its fourth Taylor polyomial i the iterval (.5,.5). As a hit, or help, you may use that i that iterval sih while cosh. That is, you ca use as a boud for sih i that iterval; for cosh, if you kow what to do with it. You should ot eed both. P 4 () = + + 4 4. We have two ways of computig the error. We ca use R 4 () = f (5) (c) 5 == sih(c) 5. for <.5, usig that the also c <.5 ad the estimate for sih give above, we get R 5 () (.5)5.6. But we ca also use the fact that P 4 () = P 5 () so that we ca use R 5 9) for the error ad a much better error estimate. R 5 () cos c 7 6 7.5 6.43, 7. (a) Fid the Maclauri series for e 3. Use ay method you wat. k 3k e 3 =. k! (b) (5 poits) If f() = e 3, what is f (8) ()? f (8) () = 7 (8!). 7! 8. Oe wishes to estimate si by 6 3. What is the largest iterval of the form ( a, a) i which this ca be doe if the error is to be < 4. (There is probably o uique aswer; justify your aswer). Rather tha cosiderig 6 3 the 3rd Taylor polyomial, we ll cosider it the 4th, so the error made i the iterval ( a, a) is We have which works out to a =.4. R 4 () 9. O wishes to use a Taylor polyomial P () = the error is to be < 6. k= cos c 5 a5. a 5 < 4 if a < 5, k= k k! estimate e i the iterval (.5,.5). Determie if
By the remaider method, we fid so R () < 6 for <.5. Now R () = e c ( + )! + for some c betwee ad ; thus e c e.5 ( is a somewhat crude estimate, but will do). Thus R () By trial ad error we see that if = 6, so = 7, the we get ( + )! (.5)+ 7, so = 7 will do. ( + )! (.5)+. ( + )! (.5)+ 3 6, for = 7, so + = 8