You may have a simple scientific calculator to assist with arithmetic, but no graphing calculators are allowed on this exam.

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1 Math 131 Exam 3 Solutions You may have a simple scientific calculator to assist ith arithmetic, but no graphing calculators are alloed on this exam. Part I consists of 14 multiple choice questions (orth 5 points each) and 5 true/false question (orth 1 point each), for a total of 75 points. Mark the correct ansers for Q1-19 on the anser card. On Part I, only the ansers on the card ill be graded. 1. Find the tangent line at œ! to the graph of C œ 0ÐÑ œ ln Ðsec tan ÑÞ A) Cœ! ) Cœ " C) Cœ D) Cœ$ E) Cœ F) Cœln G) CœÐ ln Ñ H) Cœ I) Cœ ln J) CœsecÐ"Ñ The tangent line goes through Ð!ß 0Ð!ÑÑ œ Ð!ß lnð sec! tan!ññ œ Ð!ß ln "Ñ œ Ð!ß!ÑÞ " " 0 ÐÑ œ sec tan Ð sec tan sec Ñ œ sec tan sec Ð tan sec Ñ œ sec ß so 0 Ð!Ñ œ sec! œ "Î cos! œ "Þ So the tangent line has equation ÐC!Ñ œ "Ð!Ñ or C œ Þ 2 sin 2. Find 0 ÐÑ if 0ÐÑ œ ln Ð ( Ñ ( "$ ( Ð ln Ñ cos ÐlnÐÑÑ cos A) ) C) D) 2 sin 2 sin ( sin ( sin E) ln cot ( " F) Ð ln Ñ cot cos G) (' "$ ' ( sin H) 2 sin ( I) % J) ( Ñln Ð Ñ ( sin 2 0ÐÑ œ ln Ð sin ( ( Ñ œ lnð Ñ lnðsin Ñ lnð Ñ œ ln lnðsin Ñ ( ln cos ( ( sin œ ln œ ln cot

2 3. A container holds 10g of a radioactive substance that has a half-life of 2 days, so the amount E >Î of the substance remaining after > days is given by E œ "!. Ho fast is E changing hen >œ'? & $ ) % A) g/day ) g/day C) "Þ& g/day D) Þ'& g/day E) & ln $ ln g/day F) g/day G) "Þ& ln g/day H) 'Þ' ln g/day ) & ln "! "! ln & ln ) I) g/day J) g/day %.E >Î ".> The derivative gives the rate of change: œ"! Ð ln ÑÐ Ñœ & ln Þ.E $ &.> >œ' ) When >œ'ß e get l œ & ln Ð Ñœ ln g/day. >Î 4. Suppose 0Ð%Ñ œ "Þ The graph of 0 ÐÑ is given belo. Using either a linear approximation or a differential, estimate the value of 0Ð$Þ)ÑÞ Is this an overestimate or an underestimate? A) 1., overestimate ) 1., underestimate C) 0.85, overestimate D) 0.85,underestimate E) 0.8, overestimate F) 0.8, underestimate G) 1.1, overestimate H) 1.1, underestimate I) 0.9, overestimate J) 0.9, underestimate Near + œ % À 0ÐÑ PÐÑ œ 0Ð%Ñ 0 Ð%ÑÐ %ÑÞ J Ï From the graph, 0 Ð%Ñ œ "ß so near %, 0ÐÑ PÐÑœ" Ð "ÑÐ %Ñœ& Þ So 0Ð$Þ)Ñ & $Þ) œ "ÞÞ Since 0 is decreasing near %ß 0 is concave don near % and the graph of PÐÑ œ the tangent line at 4 lies above the graph of 0ÐÑÞ So PÐ$Þ)Ñoverestimates 0Ð$Þ)ÑÞ

3 5. A tank has the shape of a rectangular box hose base is a 10m 10m square. Water is $ floing into the tank at a rate of 10 m /minute. Ho fast is the depth of the ater in the tank increasing hen the ater is 6m deep? A) 2 m/min ) 1 m/min C) 3 m/min D) 0.7 m/min E) 60 m/min F) 5 m/min G) 0.6 m/min H) 0.3 m/min I) 0.1 m/min J) 0.06 m/min When the ater is m deep in the tank, the volume of the ater is Z œ Ðarea of baseñ œ "! ß so.z œ "!!. Þ.>.>.Z $.. "!.>.>.> "!! Since œ "! m Î minute e get "! œ "!! ß so œ œ!þ" m/min.. (Note: it turns out that.> is constant (for this tank, it doesn't depend on œthe depth of the ater.) For the function Cœ0ÐÑœsin cos on the interval Ò ß Ó. The absolute maximum value is: " " $ A) ) C) D) E) F) G) " $ H) " I) J) $ is continuous on Ò ß Óso the Extreme Value Theorem guarantees that 0 has an absolute maximum (and minimum). The absolute maximum must occur at an endpoint, or at one of the 1 1 critical points Ð ß ÑÞ Setting 0ÐÑœcos Ð sin Ñœcos sin œ! gives cos œ sin sin 1 1 or cos œ tan œ " and, since e ant critical points in Ð ß Ñß this means that 1 œ Þ % So the absolute maximum occurs at one of the three points: % œ ßœ or œ Þ 1 1 Since 0Ð Ñ œ "! œ "ß 0Ð % Ñ œ Ð Ñ œ ß and 1 0Ð Ñ œ "! œ ", e see that the absolute maximum value is " (occurring at the right endpoint, œ 1 ).

4 7. A spherical snoball has radius 8 cm. After a fe minutes, melting has reduced the radius to 7.9 cm. Use a differential to estimate the change in the volume of the snoball. $ $ $ $ A) &Þ'1 cm ) &Þ& 1cm C) &Þ% 1cm D) &Þ$ 1cm $ E) &Þ 1cm $ $ F) &Þ" 1cm G) &Þ! 1cm $ H) %Þ* 1cm I) %Þ) 1cm J) %Þ( 1cm % $.Z The volume of the snoball is Z œ $ 1< ß so.< œ % 1<, and therefore, in differential notation,.z œ % 1 <.<Þ When < changes from ) to (Þ*ß e have.< œ!þ"ß so the change in Z is $ approximately.z œ % 1Ð) ÑÐ!Þ"Ñ œ &' 1Ð!Þ"Ñ œ &Þ' 1cm. 8. The function Cœ0ÐÑœ/ " has a local minimum at hich of the folloing points? A) % ) $ C) D) " E)! F) " G) H) $ I) % J) & " " " " 0 ÐÑ œ Ð/ Ñ Ð/ ÑÐÑ œ / Ð Ñ œ / Ð Ñ so the critical points are œ! and œ ÞFrom the formula for 0ÐÑÀ so for À 0ÐÑ! so 0is increasing for!à 0ÐÑ! so 0is decreasing for! À 0ÐÑ! so 0is increasing 0 has a local minimum at œ!þ $ sin 9. Suppose that Ÿ0ÐÑŸ ln for! "Þ What is 0ÐÑ? A) ) $ C) " D)! E) 1 " F) 2 G) H) 3 I) J) not enough information to decide Ä! $ sin $ sin Ä! Ä! œ œ$ "œßand, using L'Hopital's Rule, " Ä! ln Ä! ln Ä! "Î Ä! œ œ œ œþ Therefore 0ÐÑ œ! Ä! also (by the Squeeze Theorem).

5 10. A cattle rancher ants to put up a fence to create a rectangular enclosure and then subdivide it into 3 smaller pens using fences parallel to one side of the rectangle. He has 530 feet of fencing available to do the job. What is the largest possible total area for the 3 pens? (Round your ansers to the nearest integer.) A) 5280 ft ) 5971 ft C) 6354 ft D) 7254 ft E) 7778 ft F) 8213 ft G) 8517 ft H) 8778 ft I) 8934 ft J) 9251 ft Let 6 be the length of the rectangular enclosure and let be the length of the fences that subdivide it into smaller pens. The total amount of fencing required for this is 6 %ß and there is only &$! feet of fencing available, so 6 % œ &$!ß or 6 œ '&. The total area is E œ 6 œ Ð'& Ñ œ '&, and E ÐÑ œ '& %Þ '& Then EÐÑœ! gives the only critical point, œ Þ The graph of A is a concave don parabola so e kno that E has a maximum value: the vertex of the parabola must be at the critical point. So e don't need any further tests to see hether the critical point actually produces a maximum. '& '& '& '& % % % ) œ gives the maximum area: E œ Ð'& ÑÐ Ñ œ )(() ft. %

6 11. The function Cœ0ÐÑœ cos ( is concave up henever A) sin Ð(Ñ! ) cosð(ñ! C) sin Ð(Ñ! ( D) cos Ð(Ñ! E) sin Ð Ñ! F) " cosð(ñ! ( G) " sin Ð(Ñ! H) cos Ð Ñ! I) sin Ð(Ñ! J) cos Ð(Ñ! 0 ÐÑ œ " ( sin (ß so 0 ÐÑ œ %* cos(þ The function C œ 0ÐÑ ill be concave up herever 0ÐÑ!ß and that's herever cos (!Þ 12. The function Cœ0ÐÑis defined on Ò"ß&Ó. The derivative 0ÐÑis pictured belo. Does 0 have a local maximum? If so, at hat point? A) œ" ) œ"þ& C) œ D) œþ& E) œ$ F) œ$þ& G) œ% H) œ%þ& I) œ& J) No local maximum There are to critical points for 0ÐÑ À 0ÐÑ œ! for œ and œ $Þ Since 0ÐÑchanges from positive to negative at œß0changes from increasing to decreasing at œþso œ gives a local maximum. Since 0ÐÑchanges from negative to positive at œ$, 0changes from decreasing to increasing at œ$à so œ$ is a local minimum, not a local maximum.

7 13. Find lnð-./ Ñ (here +ß,ß -ß. are positive constants). Ä +, + + A) - ). C) +. D). E) +, + F)! G),. H) I) / J). Ä +, lnð-./ Ñ has the form ß so e can use L'Hopital's Rule: +, + -./ Ä lnð-./ Ñ œ " + Ä Ð Ñ./ Ä./ -./ œ Þ so e can use L'Hopital's Rule again: + Ä This it also has form, -./././ Ä./ œ+ œ+ "œ+þ Ä

8 14. A hot air balloon is rising vertically at a rate of 60m/min. An observer is lying on the ground 30m from the point here the balloon left the ground and keeps his eye on the balloon. Ho fast is the angle beteen the ground and his line of sight groing hen the balloon is 50m from the observer? ") & $ " ( % "& & A) 2 rad/min ) rad/min C) rad/min D) rad/min E) rad/min " $ "$ "" * F) rad/min G) & rad/min H) "' rad/min I) "' rad/min J) "' rad/min.c.>.>.) We ant to find at the instant hen D œ &!Þ We kno that œ '! m/min..c.d.c.d.>.>.>.> We kno that $! C œ D ß so that C œ D, so C œ D.D %! '! When D œ &!ß C œ %!ß so, at that instant œ œ %) m/min..> &! $!.) $!.D %! % D.> D.> &! & Since cos ) œ, e have sin ) œ Þ When D œ &!ß sin ) œ œ Þ %.) $!.) $! & ") &.> &!.> &! % & So at that instant: œ Ð%)Ñ, so œ Ð%)Ñ œ rad/min.

9 Q15-19 are true/false questions. Don't forget to mark these ansers on the anser card. 15. If 0ÐÑ has a local maximum, but no absolute maximum, in the interval Ò 11 ß Ó, then 0ÐÑ cannot be continuous on the interval Ò 11 ß Ó. A) True ) False True: If 0 is continuous on Ò 11 ß Ó, then the Extreme Value Theorem says 0 must have an absolute maximum on this interval. If 0 does not, then 0 cannot be continuous on Ò 11 ß ÓÞ 16. Ä "!!!!!! œ! / A) True ) False False: The it has form so e can use L'Hopital's Rule: œ / hich again is. So / Ä Ä / Þ Ä Ð******Ñ ) / Ä G Ä "!!!!!! œ We can repeat this process over and over and after steps, e get that the it is the same as, here G is some constant. So the it does not exist ( œ ). ÐThe constant G œ "!!!!!!x œ Ð"!!!!!!ÑÐ******ÑÐ*****)Ñ ÞÞÞ " Ñ / 17. Suppose Cœ0ÐÑœsin ÐÑÞ According to the Mean Value Theorem, there must be a point - beteen! and 1 here 0 Ð-Ñ œ!þ A) True ) False True: 0 is differentiable on Ò!ß 1 Óso the Mean Value Theorem says that there is a - beteen 0Ð 1Ñ 0Ð!Ñ sinð 1Ñ sinð!ñ 1 1! 1! and here 0 Ð-Ñ œ - œ!þ 18. If 0 Ð"Ñexists and 0 Ð"Ñ Á!, then 0 cannot have a local minimum at ". A) True ) False True: If 0Ð"Ñexists and 0Ð"ÑÁ!, then " is not a critical point of 0. ut Fermat's Theorem says that if 0 has a local maximum or minimum at 1, then 1 must be one of the critical points of The function 0ÐÑ œ ll has exactly one critical point. A) True ) False 0Ð!Ñ does not exist, so! is a critical point. 0ÐÑ is never! so œ! is the only critical point.

10 Name WU ID Disc Sec Letter Part II: (25%) In each problem, clearly sho your solution in the space provided. 20. a) In each part, find the derivative. Once you have completed a formula that gives C œ ÞÞÞ then no further simplification is necessary. i) Cœ log ( $Ñ Cœlog Ð$Ñœ log $ log ßso C œ " ln ii) C œ sin sin ln C œ lnð Ñ œ Ðsin ÑÐln Ñ, so C C " œ Ðsin ÑÐ Ñ Ðcos ÑÐln ÑÞ Therefore " sin sin C œ CÐÐsin ÑÐ Ñ Ðcos ÑÐln ÑÑ œ Ð Ðcos ÑÐln ÑÑ

11 20. b) For one of the folloing its (but not for the other) it is appropriate to use L'Hopital's Rule. Pick the one here you can apply L'Hopital's Rule and use it to find the it. You do not need to find the other it. Ä! " " % " " % or Ä! Ä! " " % œ! ÞThis is not an indeterminate form, so L'Hopital's Rule doesn't apply. In fact, this it doesn't existþ Ä! " " %!! œ so L'Hopital's Rule applies and e get: Ä! " " % œ Ä! " " " " % Ð %Ñ " " Ä! " " % œ œ" œ$þ

12 "Þ Here is a function C œ 0ÐÑ defined on Ò "ß %ÓÞ To save time, the formulas for 0 ÐÑ and 0 ÐÑ are also given: $ 0ÐÑ œ Ð %Ñ 0ÐÑœ%Ð $Ñ 0 ÐÑ œ Ð Ñ a) Find the intervals on hich 0 is increasing or decreasing and determine hich of the critical points is a local maximum or a local minimum. Setting 0ÐÑœ%Ð $Ñœ! gives the critical points: œ! and œ$þ if! À 0 ÐÑ! so 0ÐÑ is decreasing if! $À 0ÐÑ! so 0ÐÑis decreasing if $ À 0 ÐÑ! so 0ÐÑ is increasing This means that there is no local maximum or minimum at œ!ß local minimum at œ$þ but there is a b) Find the intervals on hich 0 is concave up or concave don and find any inflection points. 0 ÐќРќ! hen œ! or œþ if! 0 ÐÑ! so 0ÐÑ is concave up if! 0 ÐÑ! so 0ÐÑ is concave don if 0 ÐÑ! so 0ÐÑ is concave up There 0 changes concavity at œ! and œ : these are the inflection points. c) Find the absolute maximum and minimum values of 0 on the interval Ò "ß%Ó ecause 0 is continuous on this closed interval, the Extreme Value Theorem guarantees that there are absolute maximum and minimum values. They must occur at either an endpoint or at one of the critical points. So the absolute maximum and absolute minimum occur at one of: œ "ßœ!ßœ$ßœ%Þ Since 0Ð "Ñœ&ß 0Ð!Ñœ!ß 0Ð$Ñœ ( and 0Ð%Ñœ!, e conclude that the absolute maximum value is & and the absolute minimum value is (Þ Note: these points are useful to plot for sketching the graph in part d)

13 d) On the grid belo, dra a reasonably accurate graph for CœÐ 4Ñ on the interval Ò "ß %Ó We can plot the points computed in c): 0Ð "Ñœ&ß 0Ð!Ñœ!ß 0Ð$Ñœ ( and 0Ð%Ñœ!. $ Also, e can plot the other inflection point: 0ÐÑ œ "'Þ from a) and b), e get something like: With those points and the information