m2121t2rev-sol-fs2015

# m2121t2rev-sol-fs2015 - MATH/2121 CALCULUS FOR THE LIFE...

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MATH/2121, CALCULUS FOR THE LIFE SCIENCES I TEST 2 REVIEW PROBLEMS SOLUTIONS (1) The doubling time for a certain bacteria culture is 12 hr. (a) Determine the exponential growth rate k . k = ln(2) 12 . 058 . (b) If the initial population is 200 what is the population after 10 hours? The population y = y ( t ) at time t hours is given by y = y e kt = 200 e (ln(2) / 12) t bacteria . So the population after 10 hours is y (10) = 200 e (ln(2) / 12)10 = 200 e 5 ln(2) / 6 356 bacteria . (c) How long will it take the population to triple? y = 3 × 200 = 200 e (ln(2) / 12) t = 3 × 200 = e (ln(2) / 12) t = 3 = ln(2) 12 t = ln(3) = t = 12 ln(3) ln(2) 19 hours . (2) Carbon-14 has a half-life of 5,600 years. How old is a piece of wood that has lost 70% of its carbon-14? Letting y = y ( t ) be the amount of carbon-14 after t years we have y = y e kt where y is the initial amount of carbon-14 and k = - ln(2) 5600 . 1

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Our wood has lost 70% of its carbon-14 so 30% remains, that is y = 0 . 3 y at present. Now y e kt = 0 . 3 y = e kt = 0 . 3 = kt = ln(0 . 3) = t = ln(0 . 3) k = - 5600 ln(0 . 3) ln(2) 9727 . That is our piece of wood is about 9,727 years old. (3) Plutonium-24 has a half-life of 13 years. (a) How much of a sample weighing 4 g will remain after 100 years? We have decay constant k = - ln(2) / 13 and the sample weighs y = 4 e - ln(2) t 13 grams after t years. Thus after 100 years our sample weighs y = 4 e - 100 ln(2) 13 0 . 0193 g (b) After what length of time will a 4 g sample have decayed to 3 g? y = 3 = 4 e - ln(2) t 13 = 3 = e - ln(2) t 13 = 3 4 = 0 . 75 = ⇒ - ln(2) t 13 = ln(0 . 75) = t = - 13 ln(0 . 75) ln(2) 5 . 4 years . (4) Convert the angles from degrees to radians: 30 , 45 , 60 , 90 , 180 , - 70 , - 240 , 480 30 = π 6 rad , 45 = π 4 rad , 60 = π 3 rad , 90 = π 2 rad , 180 = π rad , - 70 = - 7 π 18 rad , - 240 = - 4 π 3 rad , 480 = 8 π 3 rad . 2

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