Math 115 First Midterm October 11, 2011

Transcription

1 Math 5 First Midterm October, 0 Name: EXAM SOLUTIONS Instructor: Section:. Do not open this exam until you are told to do so.. This exam has pages including this cover. There are 0 problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck.. Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions,so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation)for each problem, so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI-9 (or other calculator with a full alphanumeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 5 note card. 7. If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers, andremoveallheadphones. 9. You must use the methods learned in this course to solve all problems. Problem Points Score Total 00

2 Math 5 / Exam (October, 0) page. [points] Foreachpartbelow,giveanexplicitformulaforafunctionwhichsatisfiesthe given properties, if one exists. If such a function does not exist, explain why. Be sure to clearly indicate your final answer for each part. a. [ points] A continuous function, f, which is not differentiable. Solution: The function f(x) = x is continuous, but not differentiable. The function is continuous as it can be drawn without picking up one s pencil, but not differentiable because there is a corner on the graph at the point (0, 0). b. [points] Acubicpolynomial,p, withtwox-intercepts. Solution: The function p(x) =x (x ) = x x is a cubic polynomial with x-intercepts at x =0,. c. [points] Acontinuousfunction,c, satisfying lim c(x) = and lim c(x) =. x 0 + x 0 Solution: The function described here does not exist. If lim c(x) = and lim c(x) =, x 0 + x 0 then lim c(x) doesnotexistsincetherightandlefthandlimitsarenotequal. The function c(x) iscontinuousatzeromeansthelimitasx 0existsandequalsc(0). If the x 0 right hand limit and the left hand limit are not the same, limc(x) doesnotexist,andso x 0 c(x) cannotbecontinuousatx =0. d. [points] Arationalfunction,r, withaverticalasymptoteatx =andahorizontal asymptote at y =. Solution: The function r(x) = x asymptote at y =. x has a vertical asymptote at x =andahorizontal

3 Math 5 / Exam (October, 0) page. [points] TheFacebookDatateamhasdecidedtotracktheUniversity of Michigan network status updates that mention football in order to see which days to show ads for tailgating supplies. Starting at pm Saturday, they measure an aggregate Football Status Factor by calculating the percentage of status updates which mention any of a number of designated football terms every hour. They notice very quickly that the data is sinusoidal with period 68 hours (the number of hours in a week). Suppose F (t) isthispercentage,t hours after pm Saturday. a. [points] Ifthemaximumpercentageis96%atpmSaturday,and the minimum is 8% attained 84 hours later, compute the following quantities:. Midline Solution: Since the maximum value is 96 and the minimum value is 8, the midline will be the average of these two values, that is (96 + 8)/ =6.. Amplitude Solution: The amplitude of this sinusoidal function is half the distance between the maximum and minimum values, that is (96 8)/ =4. Noticethatthemidlineplus the amplitude will be the maximum 64 + = 96 and the midline minus the amplitude will be the minimum 6 4 = 8. b. [6points] Usingthevaluescomputedabove,findaformulafor F (t). Solution: From the problem, we know that this function begins at its maximum, which makes cosine the natural choice for our sinusoidal function. We need the appropriate values of A,B, and k in F (t) =A cos(bt)+k. Wearegiventhattheperiodis68hours, so the value of B =π/68. We found the amplitude A =4andthemidlinek =6in part (a). So, the formula for F (t) willbe ( ) πt F (t) =4cos c. [4points] Supposeadvertiserswanttoadvertisewhentherate at which people are talking about football is increasing the fastest. What time range would you recommend to them and why? Use a graph of F (t) tojustifyyouranswer. FSF hours

4 Math 5 / Exam (October, 0) page 4 Solution: The Football Status factor gives an hourly rate of people talking about football since it averages mentions of football in Facebook statues over the course of an hour. To find when the rate at which people are talking about football is increasing the fastest, we just need to find when the Football Status Factor or F (t) isincreasingthe fastest. The function F will tell us whether F (t) isincreasingordecreasing,buttofind maximum of F,weneedtoconsiderF,thesecondderivativeofF.Att =6hours, the graph is increasing and switches from concave up to concave down, so the derivative is at its largest when t =6. Anytimerangearoundt =6wouldbethebesttime for advertisers to advertise.

5 Math 5 / Exam (October, 0) page 5. [points] AzombieplaguehasbrokenoutinAnnArbor. Asanurse in the University of Michigan hospital, you saw the person with the first case of the plague,patientzero. a. [points] Inordertokeeptrackofthegrowingzombiepopulation in Ann Arbor, you collected the following data: Days after patient zero Number of Zombies Would a linear function or an exponential function be the bestmodel?why? Solution: In order for a function to be linear, it must have constant slope. Using the table above, we can compute the slopes between subsequent points: (9 )/6 = 4/, (7 9)/(9 6) = 6, and (8 7)/( 9) = 8. Not only are these values not constant, they are increasing, so a linear function would be a very bad model. If instead we take ratios of outputs with the same change in input we have 7/9 == 8/7. Since this value is constant, an exponential function will be a good model for this data. b. [4points] WriteafunctionZ(t) oftheappropriatetypetomodelthegrowthofthezombie population with t measured in days after patient zero. Solution: From part a), we know we are looking for an exponential function, which can take the form Z(t) =Ab t or Z(t) =Ae kt.sincez(0) =, the value of A for both types of exponential equation will be. Let s find Z(t) oftheformb t. Since Z(6) = 9 and Z(9) = 7, we can take the ratio of these two equations: 7 9 = b9 b 6 =b9 6 = b b = /.445. Therefore we can write the function exactly as Z(t) = t/,orapproximatewithz(t) =.445 t. Similarly, we can solve for Z(t) intheforme kt : Then we have Z(t) =e ln()t. 7 9 = e9k e 6k =ek ln() = k k = ln().

6 Math 5 / Exam (October, 0) page 6 c. [points] ThepopulationofNorthAmericaisapproximately50,000,000people.Using your model, how long will it take until all but one person are infected? Solution: To solve for the time it takes for all but one of 50,000,000 people to get infected, we need to set Z(t) = andsolvefort. t/ = t ln() = ln( ) t = ln( ) days ln() d. [points] Usingyourtable,approximatetheinstantaneousrateofchangeofthezombie population on the ninth day. Solution: To approximate the instantaneous rate of change from the table, we need to compute the average rate of change either between the sixth day and the ninth day, the ninth day and the twelfth day, or the sixth day and the twelfth day. Average rate of change between 6th and 9th day = =6zombiesperday Average rate of change between 9th and th day = =8zombiesperday Average rate of change between 6th and th day = =zombiesperday

7 Math 5 / Exam (October, 0) page 7 4. [ points] The Twitter Celebrity Index(TCI) measures the celebrity of Twitter users; the function T (x) takesthenumberoffollowers(inmillions)ofagivenuserand returns a TCI value from 0 to 0. Below is a graph of this function Twitter Celebrity Index Followers in Millions Use the graph above to help you answer the following questions. a. [points] ExplaininpracticaltermswhatT (.7) = 8.67 means. Solution: is When a Twitter user has.7 million followers, their Twitter Celebrity index b. [points] ExplaininpracticaltermswhatT (4.5) = 4.88 means. Solution: followers. When a user has a Twitter Celebrity index of 4.5, they have 4.88 million c. [points] ExplaininpracticaltermswhatT (0) = 0.78 means. Solution: When a Twitter user has 0 million followers, adding 00, 000 followers will increase their celebrity index by roughly.078. d. [points]explaininpracticaltermswhat(t ) (7.8) = 0.7 means. Solution: When a Twitter user has celebrity index 7.8, and increase of. totheir index corresponds to gaining approximately.07 million (7,000) followers.

8 Math 5 / Exam (October, 0) page 8 5. [6points] Findanumberk so that the following function is continuous on any interval. { (t +4) t< j(t) = kt t Using your value of k, explainwhythisfunctioniscontinuousonanyinterval. Solution: On intervals not containing t =, this function is continuous since the functions (t +4) and kt are both polynomials, and thus continuous, regardless of the valueofk. So we must find the value of k which makes j(t) continuousatt =. So we set lim j(t) t =( +4) = k = lim j(t). t + Solving this, we get k = 4. Now for any interval containing t = wehavethatj(t) is continuous. 6. [5 points] Using the limit definition of the derivative, write an explicit expression for the derivative of the function E(x) =x cos x at x =. Donottrytocalculatethisderivative. Solution: E ( + h) cos(+h) cos () = lim. h 0 h

9 Math 5 / Exam (October, 0) page 9 7. [0points] Ontheaxesbelowsketchawell-labeledgraphofacontinuousfunction,g, which satisfies all of the following properties. a. g (x) =for<x< b. g (x) = for<x< c. g(0) = d. g() = 0 e. g is decreasing for x> f. g (x) < 0forx> g. g is concave down for x< 4 x

10 Math 5 / Exam (October, 0) page 0 8. [9points] Thegraphbelowshowsarunner sdistance,p, inmilesfromherstartingpointt minutes after she began to run..0.5 p (miles) Using the graph, estimate the following. t (minutes) a. [points] Alltimesduringherrunwherehervelocitywaszero. t = 0, 0, 60, 75, 90 minutes b. [ points] Her average velocity over the first 45 minutes of her run. velocity= /45 miles per minute c. [points] Heraveragespeedoverthefirst45minutesofherrun. speed= /5 miles per minute d. [points] Hervelocity80minutesaftershebeganrunning. velocity= /0 miles per minute

11 Math 5 / Exam (October, 0) page 9. [points] Considerthegraphbelowofg(x): y x The four graphs below are shifts or stretches of g(x). Write each function below in terms of g(x) x y x y 5 h(x) = g(x)+4 k(x) = g(x +) x y x y 5 l(x) = g(x ) + m(x) = g(x)

12 Math 5 / Exam (October, 0) page 0. [0points] Facebooktrackstheaveragenumberofcharacters used by its users to write their status updates. Below is the graph for a random (talkative) user from the beginning of 006 to the beginning of 0. Use the graph to answer the following questions. 400 average characters per update a. [points] Whenwerethisuser sstatusupdatesthelongest?howlongwerethey? year Solution: From the graph, the status updates are the longest when Facebook began tracking the length of status updates. The status updates are longestatthebeginning of 006 when they were an average of 0 characters. b. [points] Whenwasthelengthoftheuser sstatusupdatesdecreasing? Increasing? Solution: The status updates are decreasing from the beginning of 006 until the beginning of 00. The length of status updates is increasing fromthebeginningof00 to the beginning of 0. c. [points] Whenwasthelengthofstatusupdatesshrinkingthe fastest? Solution: The graph appears to have steepest negative slope around the beginning of 008, so the length of status updates will be shrinking the fastest around then. d. [points] Isthisfunctioncontinuous?Isitinvertible?Justify your answer. Solution: The function is continuous since there are no gaps in the function over the domain given. However, it is not invertible because it attains the same value at the beginning of 0 and the beginning of 009, so it fails the horizontal line test and is not invertible.