#### Transcript Section 8B - Gordon State College

8 Exponential Astonishment Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 1 Unit 8B Doubling Time and Half-Life Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 2 Doubling and Halving Times The time required for each doubling in exponential growth is called doubling time. The time required for each halving in exponential decay is called halving time. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 3 Doubling Time After a time t, an exponentially growing quantity with a doubling time of Tdouble increases in size by a factor t Tdouble of 2 . The new value of the growing quantity is related to its initial value (at t = 0) by New value = initial value × 2t/Tdouble Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 4 Example Compound interest (Unit 4B) produces exponential growth because an interest-bearing account grows by the same percentage each year. Suppose your bank account has a doubling time of 13 years. By what factor does your balance increase in 50 years? Solution The doubling time is Tdouble = 13 years, so after t = 50 years your balance increases by a factor of 2t/Tdouble = 250 yr/13 yr = 23.8462 ≈ 14.382 For example, if you start with a balance of $1000, in 50 years it will grow to $1000 × 14.382 = $14,382. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 5 Approximate Double Time Formula (The Rule of 70) For a quantity growing exponentially at a rate of P% per time period, the doubling time is approximately 70 Tdouble P This approximation works best for small growth rates and breaks down for growth rates over about 15%. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 6 Example World population doubled in the 40 years from 1960 to 2000. What was the average percentage growth rate during this period? Contrast this growth rate with the 2012 growth rate of 1.1% per year. Solution We answer the question by solving the approximate doubling time formula for P. Multiplying both sides of the formula by P and dividing both sides by Tdouble, we have P 70 Tdouble Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 7 Example (cont) Substituting Tdouble = 40 years, we find P 70 Tdouble 70 1.75 / yr 40 yr The average population growth rate between 1960 and 2000 was about P% = 1.75% per year. This is significantly higher (by 0.65 percentage point) than the 2012 growth rate of 1.1% per year. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 8 Exponential Decay and Half-Life After a time t, an exponentially decaying quantity with a half-life time of Thalf decreases in size by a t Thalf factor of (1 2) . The new value of the decaying quantity is related to its initial value (at t = 0) by New value = initial value x (1/2)t/Thalf Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 9 Example Radioactive carbon-14 has a half-life of about 5700 years. It collects in organisms only while they are alive. Once they are dead, it only decays. What fraction of the carbon-14 in an animal bone still remains 1000 years after the animal has died? Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 10 Example (cont) Solution The half-life is Thalf = 5700 years, so the fraction of the initial amount remaining after t = 1000 years is 1 2 t /Thalf 1000 yr/5700 yr 1 2 0.885 For example, if the bone originally contained 1 kilogram of carbon-14, the amount remaining after 1000 years is approximately 0.885 kilogram. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 11 Example Suppose that 100 pounds of Pu-239 is deposited at a nuclear waste site. How much of it will still be present in 100,000 years? Solution The half-life of Pu-239 is Thalf = 24,000 years. Given an initial amount of 100 pounds, the amount remaining after t = 100,000 years is 1 new value = initial value 2 t /Thalf 100,000 yr/24,000 yr 1 100 lb 2 5.6 lb About 5.6 pounds of the original 100 pounds of Pu-239 will still be present in 100,000 years. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 12 The Approximate Half-Life Formula For a quantity decaying exponentially at a rate of P% per time period, the half-life is approximately Thalf 70 P This approximation works best for small decay rates and breaks down for decay rates over about 15%. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 13 Example Suppose that inflation causes the value of the Russian ruble to fall at a rate of 12% per year (relative to the dollar). At this rate, approximately how long does it take for the ruble to lose half its value? Solution We can use the approximate half-life formula because the decay rate is less than 15%. The 12% decay rate means we set P = 12/yr. Thalf 70 70 5.8 yr P 12/ yr The ruble loses half its value (against the dollar) in 6 yr. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 14 Exact Doubling Time and Half-Life Formulas For more precise work, use the exact formulas. These use the fractional growth rate, r = P/100. For an exponentially growing quantity with a fractional grow rate r, the doubling time is Tdouble log10 2 log10(1 + r) For a exponentially decaying quantity, in which the fractional decay rate r is negative (r < 0), the half-life is Thalf log10 2 log10(1 + r) Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 15 Example A population of rats is growing at a rate of 80% per month. Find the exact doubling time for this growth rate and compare it to the doubling time found with the approximate doubling time formula. Solution The growth rate of 80% per month means P = 80/mo or r = 0.8/mo. The doubling time is Tdouble log10 2 0.301030 0.301030 1.18 mo log10 1 0.8 log10 1.8 0.255273 or about 1.2 mo. Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 16