Solutions. .5 = e k k = ln(.5) Now that we know k we find t for which the exponential function is = e kt

Transcription

1 MATH Exponential Growth and Decay Spring 08 Solutions 1. (#15 from 6.5.) Cesium 137 and strontium 90 were two radioactive chemicals released at the Chernobyl nuclear reactor in April The half life of cesium 137 is years, and that of strontium 90 is 28.8 years. In what year will the amount of cesium 137 be equal to 1% of what was released? Answer this question for strontium 90. We assume that the cesium decays exponentially, so it satisfies the equation m(t) = m 0 e kt for some constant k. We find k by using the fact that the half life is years..5 = e k k = ln(.5) Now that we know k we find t for which the exponential function is.01. t = ln(.01) k.01 = e kt = ln(.01) ln(.5) So the amount of cesium 137 will be 1% of the amount released in A similar calculation for the strontium gives the year A batch of brownies cooks in a 350 oven. The temperature in the room is 70 and the temperature inside the refrigerator is 38. The brownies are taken out of the oven and placed on the counter. After 30 minutes you try a brownie but burn yourself, they are still 210. How much longer must you wait until the brownies reach 110 and are safe to eat? Newton s law of cooling gives us the formula T (t) = T A + (T 0 T A )e kt 1

2 with T A = 70 and T 0 = 350. We also know that T (30) = 210 = 70 + (350 70)e 30k = e 30k k = 1 30 ln( ) = ln(2) 30 Knowing k we can now solve for the time when the brownies will be 110. T (t) = 110 = 70 + (350 70)e kt = e kt t = 1 k ln(1 7 ) = ln(7) k So the brownies will be cool in another 54 minutes. = 30 ln(7) ln(2) Determine from the following table of values which function is most likely linear, quadratic, and exponential. x a(x) b(x) c(x) Hint: Estimate the derivatives and consider how they are changing. Estimate the derivative of each function by calculating the slope of line joining consecutive points in the table. a (0) 2 ( 17) 50 0 = a (50) =

3 It looks like a(x) has constant slope, so it is most likely linear. b (0) = 2651 = b (50) = 7649 = b (100) = = The slope is increasing, so b(x) is not linear. The b (x) appears to be increasing by 100 each time x increases by 50. If b (x) is linear then b(x) is quadratic. c (0) =.5 50 =.01 c (50) = =.015 c (100) = = Since c (100) c (50) is greater than c (50) c (0) we guess that c(x) is exponential. 4. Lines and exponential functions are both determined by two points. Suppose f(x) is linear and g(x) is exponential. Find formulas for f and g and fill in the missing values in the table. Then sketch graphs of f and g on the same set of coordinate axes. x f(x) 2 32 g(x) 2 32 If f(x) is linear we use the two points to find the slope, m = = 15 Then the equation for the line is or, solving for y, y 2 x 1 = 15 f(x) = y = 15x 13 3

4 The function g(x) is exponential, so it has the form g(x) = g 0 e kt. g(1) = 2 = g 0 e k g(3) = 32 = g 0 e 3k Divide one equation by the other to get rid of g 0, then solve for k = g 0e 3k g 0 e k = e2k k = 1 ln(16) = 2 ln(2) 2 Now use this value of k in the first equation to find g 0. 2 = g 0 e 2 ln(2) = 4g 0 g 0 =.5 So g(t) =.5e 2 ln(2)t =.5 4 t x f(x) g(x) x Figure 1: Plots of f(x) and g(x). 4

5 5. Lines and exponential functions are also determined by one point and the derivative at that point. Suppose f(x) is linear and g(x) is exponential and: f(4) = g(4) = 10 f (4) = g (4) = 5 Find formulas for f and g. Other than the point (4, 10), do the graphs of f(x) and g(x) intersect? Find these intersections or explain why none exist. If y = f(x) then we have so y 10 x 4 = 5 f(x) = y = 5x 10 g(x) = g 0 e kt k = g (x) g(x) In particular, when x = 4 k = 5 10 =.5 g(4) = 10 = g 0 e.5 4 = g 0 e 2 g 0 = 10 e 2 g(x) = 10 e 2 e.5t The graphs of f(x) and g(x) intersect at (4, 10), but nowhere else. Since f(x) and g(x) have the same slope at (4, 10), f(x) is the tangent line to g(x) at x = 4. g(x) is concave up so it bends up and away from its tangent lines. 6. A line is determined by one point and the derivative at any point, since the slope is constant. This is not true for exponential functions. Demonstrate this by considering 2 x and 3 x. Show that there is some a such that 2 a = 3 a. Then show that there is some b such that the derivative of 2 x at b is equal to the derivative of 3 x at b. Since 2 x and 3 x are different 5

6 functions (they differ at x = 1, for example) conclude that one value and one derivative are not enough to determine an exponential function. Let a = 0. Then 2 a = 2 0 = 1 = 3 0 = 3 a. d dx 2x = ln(2)2 x d dx 3x = ln(3)3 x ln(2)2 b = ln(3)3 b ( ) b ln(2) 3 ln(3) = 2 ( ) ( ) ln(2) 3 ln = b ln ln(3) 2 ( ) ln ln(2) ln(3) b = ln ( ) (See #45 in 6.5.) In 2004 the population of the world was people and was increasing at a rate of = people per year. Suppose it takes.5 acres of land to support one person, there are mi 2 of arable land in the world, and one square mile is 640 acres. Calculate the maximum population that the world can support. Use each of the three models of population growth to estimate the year in which world population will surpass 90% of the world s carrying capacity. The three models of population growth are: exponential, saturation, and logistic. L = mi acres mi 2 2 people acre = Exponential Growth P (t) = P 0 e kt 6

7 where (t) = kp (t). Now (0) = kp dt dt 0, so k = (0) dt = P = L = P 0 e kt.9 L P 0 = e kt t = ln(.9 L P 0 ) k In this model world population reaches.9l in Saturation Growth P (t) = L (L P 0 )e jt where (t) = j(l P (t)). (0) = j(l P dt dt 0), so j = (0) dt L P 0.9 L = L (L P 0 )e jt e jt =.1L L P ( 0 ).1L jt = ln L P 0 t = 1 ( ).1L j ln L P 0 In this model world population reaches.9l in Logistic P (t) = LP 0 P 0 + (L P 0 )e Lht where (t) = hp (t)(l P (t)). Now (0) = hp dt dt 0(L P 0 ), so h = (0) dt P 0 (L P 0 ) 7

8 .9 L = LP 0 P 0 + (L P 0 )e Lht P 0 e Lht.9 = P 0 P 0 = L P 0 9(L P 0 ) ( ) P 0 Lht = ln 9(L P 0 ) t = 1 Lh ln ( P 0 9(L P 0 ) ) 130 In this model world population reaches.9l in # # # # # # # # x Figure 2: Plots of the three models of world population. 8