ams151finalsolution

ams151finalsolution - AMS 151.01 Applied Calculus I Spring...

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Unformatted text preview: AMS 151.01 Applied Calculus I Spring 2008, Pratice Final Exam Solution Sketch 1. A population of animals oscillates sinusoidally between a low of 300 on January 1st, 2006 and a high of 1000 on July 1st, 2006. (a) Graph the population against time. (b) Use sine as your base function, find a formula for the population P as a function of time, t , measured in months since the start of the year. Solution: (b) Let P = A sin( Bt + C ) + D . The amplitude A can be computed by A = max- min 2 = 1000- 300 2 = 350 . B is determined by the period of the function according to the following equation B = 2 period Since period equal to 12 in this example, B = 2 / 12 = / 6. Then vertical shift D obtained by D = max- A = 1000- 350 = 650 . In order to find C , the phase shift, we use the fact that function reaches its minimum on January 1st, i.e. when t = 0. Thus sin ( C ) =- 1. We can pick any value of C as long as it satisfies this equation, for example, C =- / 2 ,C = 3 / 2. 2. In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones of people alive at the time. If the half-life of strontium-90 is 40 years, (a) Write a formula for the quantity of strontium-90, Q , left after t years, if the initial quantity is Q . (b) If the initial quantity of strontium-90 was measured in 1960s, what percentage of the original amount will be left in 2006? Solution: (a) Let Q ( t ) = Q a t . a is the only unknown parameter. Because the half-life is 40 years, we have a 40 = 1 2 which yields a = 1 2 1 / 40 ....
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ams151finalsolution - AMS 151.01 Applied Calculus I Spring...

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