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3.8 Exponential Growth and Decay

DIFFERENTIATION RULES. 3.8 Exponential Growth and Decay. In this section, we will: Use differentiation to solve real-life problems involving exponentially growing quantities. EXPONENTIAL GROWTH & DECAY. In many natural phenomena, quantities grow or decay at a rate

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3.8 Exponential Growth and Decay

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  1. DIFFERENTIATION RULES 3.8 Exponential Growth and Decay In this section, we will: Use differentiation to solve real-life problems involving exponentially growing quantities.

  2. EXPONENTIAL GROWTH & DECAY • In many natural phenomena, • quantities grow or decay at a rate • proportional to their size.

  3. EXAMPLE • For instance, suppose y = f(t) is • the number of individuals in a population • of animals or bacteria at time t. • Then,it seems reasonable to expect that the rate of growth f’(t) is proportional to the population f(t). • That is, f’(t)= kf(t) for some constant k.

  4. EXPONENTIAL GROWTH & DECAY • Indeed, under ideal conditions—unlimited • environment, adequate nutrition, and • immunity to disease—the mathematical • model given by the equation f’(t) = kf(t) • predicts what actually happens fairly • accurately.

  5. EXAMPLE • Another example occurs in nuclear • physics where the mass of a radioactive • substance decays at a rate proportional • to the mass.

  6. EXAMPLE • In finance, the value of a savings • account with continuously compounded • interest increases at a rate proportional • to that value.

  7. EXPONENTIAL GROWTH & DECAY Equation 1 • In general, if y(t) is the value of a quantity y • at time t and if the rate of change of y with • respect to t is proportional to its size y(t) • at any time, then • where k is a constant.

  8. EXPONENTIAL GROWTH & DECAY • Equation 1 is sometimes called the law of • natural growth (if k > 0) or the law of natural • decay (if k < 0).

  9. EXPONENTIAL GROWTH & DECAY • It’s not hard to think of a solution of • Equation 1. • The equation asks us to find a function whose derivative is a constant multiple of itself. • Any exponential function of the form y(t) = Cekt, where C is a constant, satisfies

  10. EXPONENTIAL GROWTH & DECAY • To see the significance of the constant C, we observe that • Therefore, C is the initial value of the function.

  11. POPULATION GROWTH • What is the significance of • the proportionality constant k? • It is called the relative growth rate.

  12. RELATIVE GROWTH RATE • Notice that the relative growth rate k • appears as the coefficient of t in the • exponential function Cekt.

  13. RELATIVE GROWTH RATE • For instance, if • and t is measured in years, then the relative • growth rate is k = 0.02 and the population • grows at a relative rate of 2% per year. • If the population at time 0 is P0, then the expression for the population is: P(t) = P0e0.02t

  14. POPULATION GROWTH Example 1 • Use the fact that the world population was • 2,560 million in 1950 and 3,040 million in • 1960 to model the population in the second • half of the 20th century. (Assume the growth • rate is proportional to the population size.) • What is the relative growth rate? • Use the model to estimate the population in 1993 and to predict the population in 2020.

  15. POPULATION GROWTH Example 1 • We measure the time t in years and let • t = 0 in 1950. • We measure the population P(t) in millions • of people. • Then, P(0) =2560 and P(10) =3040

  16. POPULATION GROWTH Example 1 • Since we are assuming dP/dt = kP, • Theorem 2 gives:

  17. POPULATION GROWTH Example 1 • The relative growth rate is about 1.7% • per year and the model is: • We estimate that the world population in 1993 was: • (Actual population was 5100 million) • The model predicts that the population in 2020 will be:

  18. POPULATION GROWTH Example 1 • The graph shows that the model is fairly • accurate to the end of the 20th century. • The dots represent the actual population.

  19. RADIOACTIVE DECAY • Radioactive substances decay by • spontaneously emitting radiation. • It follows that where k is a • negative constant. • In other words, radioactive substances decay at a rate proportional to the remaining mass. • This means we can use Theorem 2 to show that the mass decays exponentially:

  20. HALF-LIFE • Physicists express the rate of decay • in terms of half-life. • This is the time required for half of any given quantity to decay.

  21. RADIOACTIVE DECAY Example 2 • The half-life of radium-226 is 1590 years. • A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample that remains after t years. • Find the mass after 1,000 years correct to the nearest milligram. • When will the mass be reduced to 30 mg?

  22. RADIOACTIVE DECAY Example 2 a • Let m(t) be the mass of radium-226 • (in milligrams) that remains after t years. • Then, dm/dt = km and y(0) =100. • So, Theorem 2 gives: m(t) = m(0)ekt = 100ekt

  23. RADIOACTIVE DECAY Example 2 a • To determine the value of k , we use the fact • that y(1590) = ½(100). • Thus, 100e1590k = 50. So, e1590k = ½. • Also, 1590k = ln½ = ln 2-1 = -ln 2 • So, m(t) = 100e-(ln 2)t/1590 = 100e(ln 2)(-t/1590) = 100(2)–t/1590

  24. RADIOACTIVE DECAY Example 2 b • The mass after 1,000 years is: • m(1000) = 100e-(ln 2)1000/1590 • ≈ 65 mg

  25. RADIOACTIVE DECAY Example 2 c • We want to find the value of t such that • m(t) =30, that is, • 100e-(ln 2)t/1590 = 30 or e-(ln 2)t/1590 = 0.3 • We solve this equation for t by taking the natural logarithm of both sides: • Thus,

  26. RADIOACTIVE DECAY • As a check on our work in the example, we • use a graphing device to draw the graph of • m(t) together with the horizontal line m =30. • These curves intersect when t≈ 2800. • This agrees with the answer to (c).

  27. CONTINUOUSLY COMPD. INT. Example 3 • If $1000 is invested at 6% interest, • compounded annually, then: • After 1 year, the investment is worth $1000(1.06) = $1060 • After 2 years, it’s worth $[1000(1.06)] 1.06 = $1123.60 • After t years, it’s worth $1000(1.06)t = $1000(1 + .06)t

  28. CONTINUOUSLY COMPD. INT. Example 3 • In general, if an amount A0 is invested • at an interest rate r (r = 0.06 in this • example), then after t years it’s worth • A0(1+ r)t.

  29. CONTINUOUSLY COMPD. INT. Example 3 • Usually, however, interest is compounded • more frequently—say, n times a year. • Then, in each compounding period, the interest rate is r/n and there are nt compounding periods in t years. • So, the value of the investment is:

  30. CONTINUOUSLY COMPD. INT. Example 3 • For instance, after 3 years at 6% interest, • a $1000 investment will be worth:

  31. CONTINUOUSLY COMPD. INT. Example 3 • You can see that the interest paid • increases as the number of • compounding periods (n) increases.

  32. CONTINUOUSLY COMPD. INT. Example 3 • If we let n→ ∞, then we will be compounding • the interest continuously and the value of • the investment will be:

  33. CONTINUOUSLY COMPD. INT. Example 3 • However, the limit in this expression is • equal to the number e. (See Equation 6 • in Section 3.6) • So, with continuous compounding of interest at interest rate r, the amount after t years is: A(t) = A0ert

  34. CONTINUOUSLY COMPD. INT. Example 4 • If we differentiate this function, • we get: • This states that, with continuous compounding of interest, the rate of increase of an investment is proportional to its size.

  35. CONTINUOUSLY COMPD. INT. Example 4 • Returning to the example of $1000 invested • for 3 years at 6% interest, we see that, • with continuous compounding of interest, • the value of the investment will be: • Notice how close this is to the amount we calculated for daily compounding, $1197.20

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