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# Lesson 15: Exponential Growth and Decay (Section 041 handout)

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Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, …

Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.

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• 1. Section 3.4 Exponential Growth and Decay V63.0121.041, Calculus I New York University October 27, 2010 Announcements Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2 Announcements Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2 V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 2 / 40 Objectives Solve the ordinary di&#xFB00;erential equation y (t) = ky(t), y(0) = y0 Solve problems involving exponential growth and decay V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 3 / 40 Notes Notes Notes 1 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 2. Outline Recall The di&#xFB00;erential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton&#x2019;s Law of Cooling Continuously Compounded Interest V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 4 / 40 Derivatives of exponential and logarithmic functions y y ex ex ax (ln a) &#xB7; ax ln x 1 x loga x 1 ln a &#xB7; 1 x V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 5 / 40 Outline Recall The di&#xFB00;erential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton&#x2019;s Law of Cooling Continuously Compounded Interest V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 6 / 40 Notes Notes Notes 2 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 3. What is a di&#xFB00;erential equation? De&#xFB01;nition A di&#xFB00;erential equation is an equation for an unknown function which includes the function and its derivatives. Example Newton&#x2019;s Second Law F = ma is a di&#xFB00;erential equation, where a(t) = x (t). In a spring, F(x) = &#x2212;kx, where x is displacement from equilibrium and k is a constant. So &#x2212;kx(t) = mx (t) =&#x21D2; x (t) + k m x(t) = 0. The most general solution is x(t) = A sin &#x3C9;t + B cos &#x3C9;t, where &#x3C9; = k/m. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 7 / 40 Showing a function is a solution Example (Continued) Show that x(t) = A sin &#x3C9;t + B cos &#x3C9;t satis&#xFB01;es the di&#xFB00;erential equation x + k m x = 0, where &#x3C9; = k/m. Solution We have x(t) = A sin &#x3C9;t + B cos &#x3C9;t x (t) = A&#x3C9; cos &#x3C9;t &#x2212; B&#x3C9; sin &#x3C9;t x (t) = &#x2212;A&#x3C9;2 sin &#x3C9;t &#x2212; B&#x3C9;2 sin &#x3C9;t Since &#x3C9;2 = k/m, the last line plus k/m times the &#xFB01;rst line result in zero. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 8 / 40 The Equation y = 2 Example Find a solution to y (t) = 2. Find the most general solution to y (t) = 2. Solution A solution is y(t) = 2t. The general solution is y = 2t + C. Remark If a function has a constant rate of growth, it&#x2019;s linear. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 9 / 40 Notes Notes Notes 3 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 4. The Equation y = 2t Example Find a solution to y (t) = 2t. Find the most general solution to y (t) = 2t. Solution A solution is y(t) = t2 . The general solution is y = t2 + C. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 10 / 40 The Equation y = y Example Find a solution to y (t) = y(t). Find the most general solution to y (t) = y(t). Solution A solution is y(t) = et . The general solution is y = Cet , not y = et + C. (check this) V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 11 / 40 Kick it up a notch: y = 2y Example Find a solution to y = 2y. Find the general solution to y = 2y. Solution y = e2t y = Ce2t V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 12 / 40 Notes Notes Notes 4 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 5. In general: y = ky Example Find a solution to y = ky. Find the general solution to y = ky. Solution y = ekt y = Cekt Remark What is C? Plug in t = 0: y(0) = Cek&#xB7;0 = C &#xB7; 1 = C, so y(0) = y0, the initial value of y. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 13 / 40 Constant Relative Growth =&#x21D2; Exponential Growth Theorem A function with constant relative growth rate k is an exponential function with parameter k. Explicitly, the solution to the equation y (t) = ky(t) y(0) = y0 is y(t) = y0ekt V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 14 / 40 Exponential Growth is everywhere Lots of situations have growth rates proportional to the current value This is the same as saying the relative growth rate is constant. Examples: Natural population growth, compounded interest, social networks V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 15 / 40 Notes Notes Notes 5 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 6. Outline Recall The di&#xFB00;erential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton&#x2019;s Law of Cooling Continuously Compounded Interest V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 16 / 40 Bacteria Since you need bacteria to make bacteria, the amount of new bacteria at any moment is proportional to the total amount of bacteria. This means bacteria populations grow exponentially. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 17 / 40 Bacteria Example Example A colony of bacteria is grown under ideal conditions in a laboratory. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are 40,000. How many bacteria were present initially? Solution V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 18 / 40 Notes Notes Notes 6 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 7. Outline Recall The di&#xFB00;erential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton&#x2019;s Law of Cooling Continuously Compounded Interest V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 20 / 40 Modeling radioactive decay Radioactive decay occurs because many large atoms spontaneously give o&#xFB00; particles. This means that in a sample of a bunch of atoms, we can assume a certain percentage of them will &#x201C;go o&#xFB00;&#x201D; at any point. (For instance, if all atom of a certain radioactive element have a 20% chance of decaying at any point, then we can expect in a sample of 100 that 20 of them will be decaying.) V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 21 / 40 Radioactive decay as a di&#xFB00;erential equation The relative rate of decay is constant: y y = k where k is negative. So y = ky =&#x21D2; y = y0ekt again! It&#x2019;s customary to express the relative rate of decay in the units of half-life: the amount of time it takes a pure sample to decay to one which is only half pure. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 22 / 40 Notes Notes Notes 7 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 8. Computing the amount remaining of a decaying sample Example The half-life of polonium-210 is about 138 days. How much of a 100 g sample remains after t years? V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 23 / 40 Carbon-14 Dating The ratio of carbon-14 to carbon-12 in an organism decays exponentially: p(t) = p0e&#x2212;kt . The half-life of carbon-14 is about 5700 years. So the equation for p(t) is p(t) = p0e&#x2212; ln2 5700 t Another way to write this would be p(t) = p02&#x2212;t/5700 V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 24 / 40 Computing age with Carbon-14 content Example Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is 10% of that in a living organism. How old is the fossil? Solution V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 25 / 40 Notes Notes Notes 8 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 9. Outline Recall The di&#xFB00;erential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton&#x2019;s Law of Cooling Continuously Compounded Interest V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 26 / 40 Newton&#x2019;s Law of Cooling Newton&#x2019;s Law of Cooling states that the rate of cooling of an object is proportional to the temperature di&#xFB00;erence between the object and its surroundings. This gives us a di&#xFB00;erential equation of the form dT dt = k(T &#x2212; Ts) (where k &lt; 0 again). V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 27 / 40 General Solution to NLC problems To solve this, change the variable y(t) = T(t) &#x2212; Ts. Then y = T and k(T &#x2212; Ts) = ky. The equation now looks like dT dt = k(T &#x2212; Ts) &#x21D0;&#x21D2; dy dt = ky Now we can solve! y = ky =&#x21D2; y = Cekt =&#x21D2; T &#x2212; Ts = Cekt =&#x21D2; T = Cekt + Ts Plugging in t = 0, we see C = y0 = T0 &#x2212; Ts. So Theorem The solution to the equation T (t) = k(T(t) &#x2212; Ts), T(0) = T0 is T(t) = (T0 &#x2212; Ts)ekt + Ts V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 28 / 40 Notes Notes Notes 9 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 10. Computing cooling time with NLC Example A hard-boiled egg at 98 &#x25E6; C is put in a sink of 18 &#x25E6; C water. After 5 minutes, the egg&#x2019;s temperature is 38 &#x25E6; C. Assuming the water has not warmed appreciably, how much longer will it take the egg to reach 20 &#x25E6; C? Solution We know that the temperature function takes the form T(t) = (T0 &#x2212; Ts)ekt + Ts = 80ekt + 18 To &#xFB01;nd k, plug in t = 5: 38 = T(5) = 80e5k + 18 and solve for k. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 29 / 40 Finding k Solution (Continued) V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 30 / 40 Finding t Solution (Continued) V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 31 / 40 Notes Notes Notes 10 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 11. Computing time of death with NLC Example A murder victim is discovered at midnight and the temperature of the body is recorded as 31 &#x25E6; C. One hour later, the temperature of the body is 29 &#x25E6; C. Assume that the surrounding air temperature remains constant at 21 &#x25E6; C. Calculate the victim&#x2019;s time of death. (The &#x201C;normal&#x201D; temperature of a living human being is approximately 37 &#x25E6; C.) V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 32 / 40 Solution V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 33 / 40 Outline Recall The di&#xFB00;erential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton&#x2019;s Law of Cooling Continuously Compounded Interest V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 34 / 40 Notes Notes Notes 11 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 12. Interest If an account has an compound interest rate of r per year compounded n times, then an initial deposit of A0 dollars becomes A0 1 + r n nt after t years. For di&#xFB00;erent amounts of compounding, this will change. As n &#x2192; &#x221E;, we get continously compounded interest A(t) = lim n&#x2192;&#x221E; A0 1 + r n nt = A0ert . Thus dollars are like bacteria. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 35 / 40 Continuous vs. Discrete Compounding of interest Example Consider two bank accounts: one with 10% annual interested compounded quarterly and one with annual interest rate r compunded continuously. If they produce the same balance after every year, what is r? Solution The balance for the 10% compounded quarterly account after t years is B1(t) = P(1.025)4t = P((1.025)4 )t The balance for the interest rate r compounded continuously account after t years is B2(t) = Pert V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 36 / 40 Solving Solution (Continued) B1(t) = P((1.025)4 )t B2(t) = P(er )t For those to be the same, er = (1.025)4 , so r = ln((1.025)4 ) = 4 ln 1.025 &#x2248; 0.0988 So 10% annual interest compounded quarterly is basically equivalent to 9.88% compounded continuously. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 37 / 40 Notes Notes Notes 12 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010
• 13. Computing doubling time with exponential growth Example How long does it take an initial deposit of \$100, compounded continuously, to double? Solution V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 38 / 40 I-banking interview tip of the day The fraction ln 2 r can also be approximated as either 70 or 72 divided by the percentage rate (as a number between 0 and 100, not a fraction between 0 and 1.) This is sometimes called the rule of 70 or rule of 72. 72 has lots of factors so it&#x2019;s used more often. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 39 / 40 Summary When something grows or decays at a constant relative rate, the growth or decay is exponential. Equations with unknowns in an exponent can be solved with logarithms. V63.0121.041, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 27, 2010 40 / 40 Notes Notes Notes 13 Section 3.4 : Exponential Growth and DecayV63.0121.041, Calculus I October 27, 2010