Solutions to Math 41 Final Exam December 12, 2011

Transcription

1 Solutions to Mth Finl Em December,. ( points) Find ech of the following its, with justifiction. If there is n infinite it, then eplin whether it is or. ( ) / ln() () (5 points) First we compute the it: [ ( ) ] / ln() ln ln (/) ln() ln ( ) ln() ln() ln() Now since f() e ep() is continuous function, we compute ( ) [ ( / ln() ( ) )] / ln() ep ln [ ( ( ) )] / ln() ep ln ep( ) e /e (b) (5 points) As goes to 3 from the right, 6 + goes to 3 8 +, nd + 3 goes to (by direct substitution, since polynomils re continuous everywhere). Therefore the given it hs form constnt over, nd so it involves infinity. Becuse we re tking bsolute vlues, the it must be positive infinity:

2 Mth, Autumn Solutions to Finl Em December, Pge of 5. ( points) In ech prt below, use the method of your choice (tht works) to find the derivtive. Show the steps in your computtions. () Find dz in terms of z nd t given tht tn(zt) t ln(z) dt (6 points) Differentite both sides of the eqution with respect to t, nd solve for z dz/dt: tn(zt) t ln(z) sec (zt)(z t + z) t z z + ln(z) z t sec (zt) z t z ln(z) z sec (zt) z ln(z) z sec (zt) t sec (zt) t/z (b) Find dy if y [cos()]sin() d (6 points) Use logrithmic differentition: tke the nturl log of both sides, then differentite implicitly nd solve for y. Epress the nswer in terms of. y cos() sin() ln(y) sin() ln(cos()) ( ) y y (sin ) ( sin ) + cos() ln(cos ) cos y y [ cos() ln(cos()) tn() sin() ] y cos() sin()[ cos() ln(cos()) tn() sin() ]

3 Mth, Autumn Solutions to Finl Em December, Pge 3 of 5 3. ( points) Consider the function f() () Is f() continuous t? Justify your nswer. { ( + ) + e if < + if ( points) By definition f() is continuous t if f() f(). We compute the left nd right its t, nd f(). Indeed, using direct substitution we get tht f() nd [ f() ( + ) + e ] +. nd f() Hence, we hve f() f(), nd so the function is continuous. (b) Is f() differentible t? Justify your nswer. (6 points) Since this is piecewise function, we need to use the definition of the derivtive to find the derivtive t point where two brnches meet. By definition, f() is differentible t if the following it eists f() f(). We compute the left nd the right hnd side its: this it hs the form f() f() + + +, nd so we cn use L Hospitl s rule Now, the it from the left hnd side, lso of the form f() f() ( + ) + e e 6. so we cn use L Hospitl s rule, is Since the left hnd it nd the right side it re not the sme the function f() is not differentible t.

4 Mth, Autumn Solutions to Finl Em December, Pge of 5. (8 points) Two people leve the sme city to two different destintions t the sme time. Person A leves due Est t constnt speed of 5 mi/hr nd Person B leves due South Est t constnt speed 6 mi/hr. How fst is the distnce between them incresing hours lter? Your finl nswer must be in terms of numbers lone, but you do not need to simplify it. Your finl nswer must hve units too. Let be the distnce from the city to Person A, y the distnce from the city to Person B, nd z the distnce between Person A nd B. Then it is given tht d dy dz 5, 6 nd we re looking for dt dt dt. The tringle formed by the city, Person A nd Person B is not necessrily right tringle, but we know tht the ngle opposite the side joining Person A nd B hs π/ rdins, so we cn use the lw of cosines: ( π ) z + y y cos + y y Tking derivtive with respect to t: z dz dt d dy + y dt dt d dt y dy dt Now, hours lter, nd y, so z And therefore: dz dt d dy dt + y dt d dt y dy dt z mi/hr. So the distnce between them is incresing t rte of mi/hr.

5 Mth, Autumn Solutions to Finl Em December, Pge 5 of 5 5. ( points) Consider the eqution () Find n intervl of length one where solution of the eqution is contined. Justify your nswer nd mke sure you mention which theorems you re using. (7 points) Let f() We seek n intervl of length one contining solution to f(). Note tht f is continuous on ll of R (, ) becuse it is polynomil. We compute f( ) 3( ) 7 +35( ) < nd f() 3() 7 +35()+5 5 >. By the intermedite vlue theorem, there eists number c in the intervl (, ) such tht f(c), tht is, the intervl (, ) [which hs length ] contins solution to f() s desired. (b) Show tht this eqution hs unique solution. (Hint: Study the grph) (5 points) We compute, using f() s bove, tht f () Since 6 for ll, it follows tht f () 35 > for ll. In prticulr, f() is incresing on R (, ). It follows tht f cn hve t most one root in R, since if c, d re two roots then f(c) f(d), but if c < d then f(c) < f(d), which doesn t mke sense, nd similrly if c > d then f(c) > f(d), which lso doesn t mke sense. If you think bout it, wht we re using is tht if function is incresing, then it is one-to-one. In terms of the grph, ll we re sying is tht n incresing function cnnot cross the -is more thn once! So f() hs t most one solution, nd by prt () it hs t lest one solution, so therefore it hs unique solution. Alternte solution: Suppose nd b re two different solutions to f(), so b. Then f(b) f() b. Since f is polynomil, it is everywhere differentible, so the Men Vlue Theorem pplies, nd thus there is number c between nd b with f (c) f(b) f() b. But this is impossible becuse f (c) c >. b (Note tht the lternte method is relly just unwinding the proof tht function with positive derivtive is incresing!)

6 Mth, Autumn Solutions to Finl Em December, Pge 6 of 5 6. ( points) The rnge in miles of projectile lunched over flt ground from the origin with positive velocity v in mi/hr t n ngle θ given in rdins is given by where g is positive constnt. R v sin(θ), g () Given fied velocity, v, wht ngle will mimize the rnge of the projectile? nswer. Justify your (8 points) The domin of R, s function in θ, is [, π/] becuse θ is the ngle between the ground nd the forwrd direction of the projectile. Then since v nd g re constnts, dr dθ v cos θ g Then in our domin, cos θ if nd only if θ π/. The derivtive is defined in the whole [, π/], so θ π/ is the only criticl number. R reches globl mimum nd globl minimum, becuse R is continuous function defined on closed intervl, nd it my hppen only t the criticl points or t the endpoints. We check: R() v sin /g R(π/) v sin π/g R(π/) v sin(π/)/g v /g > Therefore the mimum occurs t θ π/ rdins. Note: The domin for this function could hve lso been interpreted s being (, π/), [, π/), (, π/], (, π), [, π), (, π] or [, π]. (b) For this prt, ssume v, g nd θ ws mesured to be π/6 rdins. If there ws possible error in the mesurement of θ of. rdins, estimte the corresponding error in the computtion of the rnge using differentils or liner pproimtion. Your finl nswer must be in terms of numbers lone, but you do not need to simplify it. Give the units of your finl nswer. ( points) By Liner Approimtion, when θ is ner π/6, ( π ) R(θ) R 6 Then if we hve the rnge of θ s θ π/6 <., ( + R π ) ( θ π ) ( π ) R + ( θ π ) R(θ) R(π/6) θ π < (.) Therefore the possible error in the rnge is pproimtely. miles.

7 Mth, Autumn Solutions to Finl Em December, Pge 7 of 5 7. (9 points) () Give precise sttement of both prts of the Fundmentl Theorem of Clculus. (3 points) Prt : Let f be continuous function on [, b]. Then, the function g() f(t)dt is differentible for < < b nd d d f(t)dt f() Prt : Let F be ny ntiderivtive of n integrble function f defined on [, b]. Then b f(t)dt F (b) F () (b) Find the derivtive with respect to of the function g() sin () cos () ll your steps. Your finl nswer must not contin ny symbols. ln(t) dt. Show nd justify (6 points) By the Fundmentl Theorem of Clculus, we know there eists n ntiderivtive F (t) of ln(t), tht is, F (t) ln t. Then by the Evlution Theorem, g() Tking derivtive on both sides, sin () cos () ln tdt F (sin ()) F (cos ()) g () F (sin ())( sin() cos()) F (cos ())( cos() ( sin())) by the chin rule. Thus substituting F (t) ln(t), we hve: g () ln( sin ()) sin() cos() + ln( cos ()) cos() sin().

8 Mth, Autumn Solutions to Finl Em December, Pge 8 of 5 8. ( points) Instruments bord submrine which is trveling on stright line record the velocity in mi/hr once every hour. During the -hour period from t to t, they recorded: t v(t) () Without clculting it, wht does the quntity v(t)dt represent? Epress your nswer in terms relevnt to this sitution, nd mke it understndble to someone who does not know ny clculus; be sure to use ny units tht re pproprite, nd lso eplin wht the sign of this quntity would signify. ( points) The quntity v(t) dt represents the net displcement or net chnge in position of the submrine during the -hour period, mesured in miles. A positive number signifies tht the submrine moved forwrds by tht mount, while negtive number signifies tht it moved bckwrds. (On the other hnd, the distnce trveled is represented by the quntity v(t) dt.) (b) Use the Left Endpoint Rule with n 3 to estimte epression in terms of numbers lone, but you do not hve to simplify it. 9 v(t)dt; give your finl nswer s n ( points) We subdivide the intervl t 9 into 3 subintervls of length t 3: [, 3], [3, 6], nd [6, 9]. In ech, we use the left endpoint s our smple point, t i : 9 v(t) dt 3 v(t i ) t 3(v() + v(3) + v(6)) 3(3 + + ) 5. i (c) Use the Midpoint Rule with n 5 to estimte of numbers lone. v(t) dt; gin give your finl nswer in terms ( points) Now we subdivide [, ] into 5 subintervls of length t : [, ], [, ], [, 6], [6, 8], [8, ]. In ech, we use the midpoint s our smple point, t i : v(t) dt 5 v(t i ) t ( v() + v(3) + v(5) + v(7) + v(9) ) (++6++9) 7. i

9 Mth, Autumn Solutions to Finl Em December, Pge 9 of 5 9. ( points) Mrk ech sttement below s true or flse by circling T or F. No justifiction is necessry. T F If f is odd nd continuous everywhere, ll of its ntiderivtives must be even. True. If f is odd, we hve f(t)dt. But lso F. So F () F ( ), tht is, F is even. f(t)dt F () F ( ) for ny ntiderivtive T F If two differentible functions stisfy f() g() for ll, then f () g () for ll. Flse. Pick f() sin() nd g(). Then f () cos() nd g (). But cos() is not lwys negtive, for emple cos() >. T F The most generl ntiderivtive of g() tn() is H() ln( cos() ) + C, where C is constnt. Flse. The domin of tn() hs more thn one intervl, so the most generl ntiderivtive is piecewise function tht equls ln( cos() ) plus different constnt in ech of those intervls. T F b f(t) dt b f() dt for ny function f integrble on [, b]. Flse. We hve b f() dt f()(b ), so it is function of, while emple, if f(), nd b, then T F If f is continuous on [, b], then b dt, while f(t) dt b b t dt /. f(t) dt. f(t) dt is number. For True. If f is continuous on [, b], by the Fundmentl Theorem of Clculus g() f(t) dt is differentible on [, b], nd therefore continuous on [, b]. So g(b) g(), which is precisely wht b the sttement sys. T F If we pply Newton s method to the eqution f() nd get 3, then f( ). True. Since 3 f( ) f ( ), then f( ) f ( ) nd therefore f( ). T F e 3 d e3 + C, where C is constnt. 3 Flse. The derivtive of e3 e e 3 + C is 3 (3 ) nd this is not equl to e 3. T F If function is not integrble on [, b], then it is not differentible on [, b].

10 Mth, Autumn Solutions to Finl Em December, Pge of 5 True. If function is not integrble on [, b], then it is not continuous on [, b], nd so it is not differentible on [, b]. T F The function g() e t dt is the only ntiderivtive of f() e with g() True. The function g() is n ntiderivtive of f() by the Fundmentl Theorem of Clculus nd g(). Any other ntiderivtive would be of the form g() + C, so the only wy this is zero when is when C. T F 3 3 ( + 7 cos() + 5) d 3 3 ( + 5) d True. The function 7 cos() is odd, so 3 3 ( + 5) d cos() d cos() d, nd therefore ( + 5) d 3 3 ( + 7 cos()+5) d

11 Mth, Autumn Solutions to Finl Em December, Pge of 5. (5 points) () Suppose f() ( ). Let R be the region in the y-plne bounded by the curve y f() nd the lines y,, nd. Find the re of R by evluting the it of Riemnn sum tht uses the Right Endpoint Rule; show ll resoning. Give your finl nswer s n epression in terms of numbers lone, but you do not hve to simplify it. (9 points) Let ( ) n n (note tht this should lwys be positive s it is the width of the rectngles we will use to pproimte the re). For the Right Endpoint Rule we define i + i + i n. For given n the Riemnn sum is given by Now, by definition R n n f( i ) i n i n i n i ( ( + i ) n ) n ( i ) n n ( i n + i n ) n n n n n i + n n i i i i n n ( n ) + n n + ( n 3 + 3n ) + n n 6 ( + ) ( + n n + ) n. ( ) d n R n n n n f( i ) i [ ( + ) ( + n n + )] n

12 Mth, Autumn Solutions to Finl Em December, Pge of 5 (b) Epress the it n n n ( ) + i n i s definite integrl, nd then compute its vlue using the Evlution Theorem. Give your finl nswer s n epression in terms of numbers lone, but you do not hve to simplify it. (6 points) It is cler tht n b n. Hence, b + nd i i n, which implies tht. We hve then tht f( i ). Hence, the definite integrl is + i + d rctn rctn() rctn() rctn().

13 Mth, Autumn Solutions to Finl Em December, Pge 3 of 5. (8 points) Show ll resoning when solving ech of the problems below. Your finl nswer must not contin ny symbols. () 3 + d (6 points) We use substitution with u +, so du d. Tht mens u nd the its chnge to nd d + d 5 (u ) u du 5 (u 3/ u / ) du (b) [ u 5/ 5/ ] 5 [ ] 5 u 3/ u 5/ 3/ 5 u3/ 3 [ ] 55/ 5 53/ 3 5/ 5 3/ 3 t t dt (6 points) We use substitution with u t, so tht du dt. Then t u +. t (u + ) t dt du u u + u + du u (u + + u ) du u + u + ln( u ) + C (t ) + (t ) + ln( t ) + C

14 Mth, Autumn Solutions to Finl Em December, Pge of 5 (c) rctn ( ) d (6 points) First we use integrtion by prts with u rctn du + (/) d d nd v. + ( ) nd dv d, nd therefore rctn ( ) ( ) d rctn d rctn + ( ) + + d Now we use substitution with w + so tht dw d. rctn rctn ( ) + ( ) dw rctn + w ln( w ) + C ( ) + ln( + ) + C

15 Mth, Autumn Solutions to Finl Em December, Pge 5 of 5. ( points) () The function f stisfies f( + ) f() for ll. Use subtitution to show tht 8 f() d 6 f() d. (6 points) We strt with the fct tht 8 f()d 6 f()d show tht 6 f()d 8 6 f()d. f()d. So we just need to Net note tht 6 f()d 6 f( + )d 8 6 f(u)du where the first equlity is becuse f() f(+) nd the second one comes from u-substitution with u +, du d. Since 8 6 f(u)du nd 8 6 f()d men the sme thing, we hve shown tht 6 f()d 8 6 f()d. Thus, nd so we re done. 8 f()d f()d + f()d + f()d f()d f()d (b) The function f() stisfies f(π) nd f (). Find n ntiderivtive G of g() f() sin() sin() tht stisfies G(π). Your finl nswer must not contin ny symbols. (Hint: Use integrtion by prts) (6 points) The ntiderivtive of f() sin() is the indefinite integrl G() f() sin()d. To find G(), we use integrtion by prts twice: f() sin()d cos()f() cos()f ()d where u f(), du f ()d, dv sin()d, v cos() cos()f() + f () sin() d sin() sin() where u f (), du f ()d d/ sin(), dv cos()d, v sin() cos()f() + f () sin() d so, G() cos()f() + f () sin() + C We know tht G(π) nd f(π). Since G(π) cos(π)f(π) + f (π) sin(π) π + C π + C, we must hve C π. Thus, G() cos()f() + f () sin() + π