First Solution. Let t be the number of years from the eruption until Then t/5730 is the number of half-lives, so. 1 = 0.

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1 SOLUTIONS TO MIDTERM EXAM 3 Directions and rules. The exam will last 70 minutes; the last five minutes of class will be used for collecting the exams. No electronic devices of any kind will be allowed, with one exception: a music player that nobody else can hear, and whose controls you do not use during the exam (just put it on shuffle). Anything (else) with an off switch must be off. In particular, turn your cell phone off. Show your work. Scoring: 6 problems, each worth 6 points, plus 4 free points = 00.. The volcano at Thera, Greece (also called Santorini) erupted many centuries BCE (before Christian era). That eruption completely destroyed the island of Santorini and that event may be the basis for the story of the lost continent Atlantis that supposedly sank into the sea. When did this eruption occur? It was discovered in 2006 that there was an olive tree buried under the lava flow. Measurements (in 2006) revealed that the carbon-4 content of the remains of the olive tree was 64.65% of what it had been at the time of the eruption. Using years for the half life of carbon 4, calculate the date of the eruption. To do this problem without a calculator you can make use of some of these values: log 2 (/0.6465) = log 2 (0.6465) = ln 2 = ln(/2) = ln = First Solution. Let t be the number of years from the eruption until Then t/ is the number of half-lives, so (/2) t = = t 0 = log 2 (0.6465) + t t = log 2 (0.6465) t = t = 3604 years So the date we want is = 598 BCE. Second Solution. Let t be the number of years from the eruption until Then for some positive constant c we have = e ct Date: November 22, 200.

2 2 SOLUTIONS TO MIDTERM EXAM 3 To find c we use the half life: 2 = e c ln(/2) = ln 2 = c c = ln 2 Putting this value for c into the first equation, we have t(ln 2)/ = e Taking natural log of both sides we have ln = t ln 2 ln t = ln 2 These numbers are all given in the problem so you can plug them in and do the arithmetic; or to shorten the arithmetic proceed as follows. In a quotient of logarithms we can change the base, as the base-changing factor will cancel out. t = log log 2 2 = log = = 3604 years So the date we want is = 598 BCE. 2. As a consulting scientist, you have been given a cube of zirconium and asked to determine its volume by measuring the length of a side. The volume is approximately 8 cubic centimeters. Your client wants to know the volume to the nearest 0.00 cubic centimeter. How accurately must you measure the side of the cube to achieve the desired accuracy in volume? In other words, you must measure the side of the cube accurate to within approximately how many centimeters? Use differentials to solve this problem.

3 SOLUTIONS TO MIDTERM EXAM 3 3 Solution. V = x 3 dv = 3x 2 V = 3x 2 x x = V 3x 2 = = = = cm.

4 4 SOLUTIONS TO MIDTERM EXAM 3 3. An FBI agent with a powerful spyglass is located in a boat anchored 400 meters offshore. A gangster under surveillance is walking along the shore. Assuming the shoreline is straight and that the gangster is walking at the rate of 2 km/hr, how fast must the FBI agent rotate the spyglass to track the gangster when the gangster is km from the point on the shore nearest to the boat? Convert your answer to degrees/minute. (a) Introduce two variables (in addition to t for time) and draw a picture showing what the variables mean. Let x be the distance between the gangster G and the point P on shore opposite the boat B. Let θ be the angle between P B and P G. What we are trying to find is dθ/dt. It s difficult to get a picture to appear in this electronic document, so I m not posting that part of the solution. (b) Write an equation connecting your variables that is good all the time the gangster is under observation. and also tan θ = x 400 dt = 2000 = km/min 60 (c) Write more equations that are only good at one special time. x = 000 (d) Finish solving the problem, to get the answer in radians per hour. Differentiating the equation in part (b) we have Solving for dθ/dt we have sec 2 θ dθ dt = 400 dt = = 5 dθ dt = 5 sec 2 θ = 5 + tan 2 θ = = = 0.69 Alternately dθ dt = 5 sec 2 θ = 5 cos2 θ and cos 2 θ = 6/6 since the hypotenuse of the triangle in the picture is , so dθ dt = 5 cos2 θ = = 80 6 = 0.69 (e) Convert your answer, as the problem asks, to degrees per minute. It may help to know that (80/π)/60 =

5 SOLUTIONS TO MIDTERM EXAM radians/hr (80/π) degrees/radian /60 minutes/hr = = 0.66 degrees per minute. 4. Most people got the graphs of cosh, sinh, and tanh right, and if you didn t, you can ask MathXpert to graph them for you; it s not very easy to get those graphs into this document. 5. Differentiate each of the following with respect to x: (a) sinh(/x) (b) tanh( x) d sinh(/x) = d cosh(/x) (/x) = cosh(/x)( /x 2 ) d tanh( x) = sech 2 ( x) d x = sech 2 ( x) 2 x 6. (a) Use the definitions of sinh and cosh to derive the formula cosh 2 x sinh 2 x =. Show your work. sinh x = 2 (ex e x ) cosh x = 2 (ex + e x ) cosh 2 x sin 2 x = ( (e x + e x ) 2 (e x e x ) 2) 4 = ( (e x ) 2 + 2e x e x + (e x ) 2 ((e x ) 2 2xe x e x + (e x ) 2) 4 = 4 (4ex e x ) = 4 4 = (b) Use implicit differentiation to find the formula for the derivative of arcsinh x with respect to x. In the textbook arcsinh x is written sinh x. You may use the formula from part (a) in your work.

6 6 SOLUTIONS TO MIDTERM EXAM 3 y = arcsinh x x = sinh y = d sinh y = cosh y dy dy = cosh y = + sinh 2 y = + x 2 since x = sinh y