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Modeling with Exponential and Logarithmic Functions

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The mathematical model for exponential growth or decay is given by f (t) = A 0 e kt or A = A 0 e kt. If k > 0, the function models the amount or size of a growing entity. A 0 is the original amount or size of the growing entity at time t = 0. A is the amount at time t, and k is a constant representing the growth rate. If k < 0, the function models the amount or size of a decaying entity. A 0 is the original amount or size of the decaying entity at time t = 0. A is the amount at time t, and k is a constant representing the decay rate. The mathematical model for exponential growth or decay is given by f (t) = A 0 e kt or A = A 0 e kt. If k > 0, the function models the amount or size of a growing entity. A 0 is the original amount or size of the growing entity at time t = 0. A is the amount at time t, and k is a constant representing the growth rate. If k < 0, the function models the amount or size of a decaying entity. A 0 is the original amount or size of the decaying entity at time t = 0. A is the amount at time t, and k is a constant representing the decay rate. decreasing A0A0 x y increasing y = A 0 e kt k > 0 x y y = A 0 e kt k < 0 A0A0 Exponential Growth and Decay Models

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The graph below shows the growth of the Mexico City metropolitan area from 1970 through 2000. In 1970, the population of Mexico City was 9.4 million. By 1990, it had grown to 20.2 million. Find the exponential growth function that models the data. By what year will the population reach 40 million? 20 15 10 5 25 30 1970198019902000 Population (millions) Year Example

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Solution a. We use the exponential growth model A = A 0 e kt in which t is the number of years since 1970. This means that 1970 corresponds to t = 0. At that time there were 9.4 million inhabitants, so we substitute 9.4 for A 0 in the growth model. A = 9.4 e kt We are given that there were 20.2 million inhabitants in 1990. Because 1990 is 20 years after 1970, when t = 20 the value of A is 20.2. Substituting these numbers into the growth model will enable us to find k, the growth rate. We know that k > 0 because the problem involves growth. A = 9.4 e kt Use the growth model with A 0 = 9.4. 20.2 = 9.4 e k20 When t = 20, A = 20.2. Substitute these values. Example cont.

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Solution We substitute 0.038 for k in the growth model to obtain the exponential growth function for Mexico City. It is A = 9.4 e 0.038t where t is measured in years since 1970. Example cont. 20.2/ 9.4 = e k20 Isolate the exponential factor by dividing both sides by 9.4. ln(20.2/ 9.4) = lne k20 Take the natural logarithm on both sides. 0.038 = k Divide both sides by 20 and solve for k. 20.2/ 9.4 = 20k Simplify the right side by using ln e x = x.

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Solution b. To find the year in which the population will grow to 40 million, we substitute 40 in for A in the model from part (a) and solve for t. Because 38 is the number of years after 1970, the model indicates that the population of Mexico City will reach 40 million by 2008 (1970 + 38). A = 9.4 e 0.038t This is the model from part (a). 40 = 9.4 e 0.038t Substitute 40 for A. Example cont. ln(40/9.4) = lne 0.038t Take the natural logarithm on both sides. ln(40/9.4)/0.038 =t Solve for t by dividing both sides by 0.038 ln(40/9.4) =0.038t Simplify the right side by using ln e x = x. 40/9.4 = e 0.038t Divide both sides by 9.4.

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Use the fact that after 5715 years a given amount of carbon-14 will have decayed to half the original amount to find the exponential decay model for carbon-14. In 1947, earthenware jars containing what are known as the Dead Sea Scrolls were found by an Arab Bedouin herdsman. Analysis indicated that the scroll wrappings contained 76% of their original carbon-14. Estimate the age of the Dead Sea Scrolls. Solution We begin with the exponential decay model A = A 0 e kt. We know that k < 0 because the problem involves the decay of carbon-14. After 5715 years (t = 5715), the amount of carbon-14 present, A, is half of the original amount A 0. Thus we can substitute A 0 /2 for A in the exponential decay model. This will enable us to find k, the decay rate. Text Example

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Substituting for k in the decay model, the model for carbon-14 is A = A 0 e –0.000121t. k = ln(1/2)/5715=-0.000121 Solve for k. 1/2= e kt5715 Divide both sides of the equation by A 0. Solution A 0 /2= A 0 e k5715 After 5715 years, A = A 0 /2 ln(1/2) = ln e k5715 Take the natural logarithm on both sides. ln(1/2) = 5715k ln e x = x. Text Example cont.

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Solution The Dead Sea Scrolls are approximately 2268 years old plus the number of years between 1947 and the current year. A = A 0 e -0.000121t This is the decay model for carbon-14. 0.76A 0 = A 0 e -0.000121t A =.76A 0 since 76% of the initial amount remains. 0.76 = e -0.000121t Divide both sides of the equation by A 0. ln 0.76 = ln e -0.000121t Take the natural logarithm on both sides. ln 0.76 = -0.000121t ln e x = x. Text Example cont. t=ln(0.76)/(-0.000121) Solver for t.

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Welcome to Interactive Chalkboard Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION.

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