Midterm Exam 3 Solutions (2012)

Transcription

1 Mierm Exam 3 Solutions (01) November 19, 01 Directions an rules. The exam will last 70 minutes; the last five minutes of class will be use for collecting the exams. No electronic evices of any kin will be allowe, with one exception: a music player that noboy else can hear, an whose controls you o not use uring the exam (just put it on shuffle). Anything (else) with an off switch must be off. In particular, turn your cell phone off. If it says the answer must be a number, it means an integer or a ecimal number; thus 5/11 or π shoul be evaluate further. Show your work. There are 7 problems, total 100 points. Five of the problems are worth 14 points each, an the two problems with three parts are worth 15 points each (five points per part). You might get some partial creit, but only if you have one part of the problem completely correctly, not if you ve written some lines that coul perhaps be eite to be correct. 1. As a consulting scientist, you have been given a sphere of zirconium an aske to etermine its volume by measuring the iameter. The iameter is approximately centimeters. Your client wants to know the volume to the nearest cubic centimeter. How accurately must you measure the iameter to achieve the esire accuracy in volume? In other wors, you must measure the iameter accurate to within approximately how many centimeters? Use ifferentials to solve this problem. The answer must be a number. Solution: V = 4 3 πr3 = 4 3 π D3 8 V = π D D = 1 6 πd3 D = V πd = π = = cm. 1

2 . The volcano at Thera, Greece (also calle Santorini) erupte many centuries BCE (before Christian era). That eruption completely estroye the islan of Santorini an that event may be the basis for the story of the lost continent Atlantis that supposely sank into the sea. When i this eruption occur? It was iscovere in 006 that there was an olive tree burie uner the lava flow. Suppose that measurements (in 006) reveale that the carbon-14 content of the remains of the olive tree was 3.33% of what it ha been at the time of the eruption. (The true number was ifferent, so you won t get the true historical ate when you solve this problem.) Using 5730 years for the half life of carbon 14, calculate the ate of the eruption. The answer must be an integer number for the ate (not for the number of years that have elapse since that ate). To o this problem without a calculator you can make use of some of these values: Solution. log (1/0.333) = log (0.333) = 1.63 ln = ln(1/) = ln = 1.19 (1/) t/5730 = = t/5730 log (1/0.333) = t/ = t/5730 t = = 9340 So the answer is = 7334 BCE. Alternately = e ct where e 5730c = 1/; so then c = ln()/5730 = 0.693/5730 = Then ct = ln so t = ln(0.333)/c = 1.19/ = /11 = 9330 years ago, which correspons to 734 BCE, close enough to the answer we got by the first metho they iffer by only one part in a thousan, an we use some three-igit approximate numbers.

3 3. A plane flying horizontally with a constant spee of 480 km/hr passes over a groun raar station at an altitue of 6 km. At what rate is the istance from the plane to the raar station increasing one minute later? The answer must be a number an the units of that number. Solution: s = x + 6 x = 8 km/min After one minute x = 8, so s = 10 (either because you see that the triangle has the shape 3-4-5, or by calculating s = ). Now ifferentiate the main equation: s s s s s = x + 6 = x x = x x = 8xsince x/ = 8 At the particular moment of interest we have x = 8 an s = 10: 10 s s = 8 8 = 64 = (a) sketch the graph of cosh x (b) sketch the graph of sinh x (c) sketch the graph of tanh x = 6.4km/min = 384km/hr You can look these graphs up in the textbook or raw them in MathXpert. 5. Differentiate each of the following with respect to x: (a) cosh(1/x) x cosh(1/x) = sinh(1/x) x (1/x) ( = sinh(1/x) 1x ) = sinh(1/x) x 3

4 (b) tanh( x) You coul also write this as x tanh( x) = sech ( x) x ( x) = sech ( ( ) 1 x) x = sech ( x) x 1 x cosh x 6. Use the efinitions of sinh an cosh to erive the formula cosh(x) = 1 + sinh x. Show your work. cosh(x) =? 1 + sinh x e x + e x ( e x e x =? 1 + e x + e x =? + 4 (ex e x ) 4 e x + e x =? + (e x e x ) ) =? + (e x ) e x e x + (e x ) =? + e x + e x = e x + e x Since the two sies are now ientical, the ientity is verifie. Here are some hints for oing calculations with pencil an paper instea of a calculator. The sign = means approximately equal to. For small numbers x, we have ln(1 + x) = x, an we have e x = 1 + x, or even more accurately, e x = 1 + x + x /. So for example, ln = ln 1.01 = A useful number to know is ln 10 =.303. Then for example ln 11 can be foun as ln(10 1.1) = ln 10 + ln 1.1 = =.4. With these hints, you won t nee a calculator on the following problem: 7. A cup of coffee is serve at 180 F. The room temperature is 70. After one minute the coffee has coole to 170. Use Newton s Law of Cooling to answer these questions. (a) What equation oes Newton s Law of Cooling say shoul escribe the temperature of the coffee? 4

5 T 70 = (180 70)e ct = 110e ct where t is time since the coffee was serve, an c is a constant. We have to fin c. When t = 1 we have T = 170, so 100 = 110e c. Hence c = ln = ln 100 = ln 1.1 = 0.1. (b) What will the temperature of the coffee be after one more minute (that is, two minutes after it is serve)? The answer must be a number. T = e ct = e ln 1.1 since c = ln 1.1 an t = = (e ln 1.1 ) = = = = = 161 That seems reasonable: ten egrees cooler in the first minute, another nine egrees cooler in the secon minute. (c ) When will the coffee cool to 80? That is, how many minutes after being serve? The answer must be a number = 110e ct (ln 1.1)t 10 = 110e That s the answer, 4 minutes after serving. (ln 1.1)t = e ln 1 11 = t ln 1.1 ln 11 = t ln 1.1 t = ln 11 ln 1.1 = see the hints = 4 5