Lesson 15: Exponential Growth and Decay (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, including radioactivity, cooling, and interest.

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Lesson 15: Exponential Growth and Decay (handout)

  1. 1. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Sec on 3.4 . Exponen al Growth and Decay V63.0121.001: Calculus I Professor Ma hew Leingang New York University March 23, 2011 . . Notes Announcements Quiz 3 next week in recita on on 2.6, 2.8, 3.1, 3.2 . . Notes Objectives Solve the ordinary differen al equa on y′ (t) = ky(t), y(0) = y0 Solve problems involving exponen al growth and decay . . . 1.
  2. 2. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Derivatives of exponential and Notes logarithmic functions y y′ ex ex ax (ln a) · ax 1 ln x x 1 1 loga x · ln a x . . Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . . 2.
  3. 3. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes What is a differential equation? Defini on A differen al equa on is an equa on for an unknown func on which includes the func on and its deriva ves. Example Newton’s Second Law F = ma is a differen al equa on, where a(t) = x′′ (t). In a spring, F(x) = −kx, where x is displacement from equilibrium and k is a constant. So k −kx(t) = mx′′ (t) =⇒ x′′ (t) + x(t) = 0. . m . Notes Showing a function is a solution Example (Con nued) Show that x(t) = A sin ωt + B cos ωt sa sfies the differen al k √ equa on x′′ + x = 0, where ω = k/m. m Solu on We have x(t) = A sin ωt + B cos ωt x′ (t) = Aω cos ωt − Bω sin ωt x′′ (t) = −Aω 2 sin ωt − Bω 2 cos ωt . . Notes The Equation y′ = 2 Example Find a solu on to y′ (t) = 2. Find the most general solu on to y′ (t) = 2. Solu on A solu on is y(t) = 2t. The general solu on is y = 2t + C. Remark . If a func on has a constant rate of growth, it’s linear. . . 3.
  4. 4. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes The Equation y′ = 2t Example Find a solu on to y′ (t) = 2t. Find the most general solu on to y′ (t) = 2t. Solu on A solu on is y(t) = t2 . The general solu on is y = t2 + C. . . Notes The Equation y′ = y Example Find a solu on to y′ (t) = y(t). Find the most general solu on to y′ (t) = y(t). Solu on A solu on is y(t) = et . The general solu on is y = Cet , not y = et + C. (check this) . . Notes Kick it up a notch: y′ = 2y Example Find a solu on to y′ = 2y. Find the general solu on to y′ = 2y. Solu on y = e2t y = Ce2t . . . 4.
  5. 5. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes In general: y′ = ky Example Find a solu on to y′ = ky. Find the general solu on to y′ = ky. Solu on Remark y=e kt What is C? Plug in t = 0: y = Cekt y(0) = Cek·0 = C · 1 = C, so y(0) = y0 , the ini al value of y. . . Constant Relative Growth =⇒ Notes Exponential Growth Theorem A func on with constant rela ve growth rate k is an exponen al func on with parameter k. Explicitly, the solu on to the equa on y′ (t) = ky(t) y(0) = y0 is y(t) = y0 ekt . . Notes Exponential Growth is everywhere Lots of situa ons have growth rates propor onal to the current value This is the same as saying the rela ve growth rate is constant. Examples: Natural popula on growth, compounded interest, social networks . . . 5.
  6. 6. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Bacteria Since you need bacteria to make bacteria, the amount of new bacteria at any moment is propor onal to the total amount of bacteria. This means bacteria popula ons grow exponen ally. . . Notes Bacteria Example Example A colony of bacteria is grown under ideal condi ons 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 ini ally? Solu on . . . 6.
  7. 7. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Bacteria Example Solution Solu on (Con nued) We have . . Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Modeling radioactive decay Radioac ve decay occurs because many large atoms spontaneously give off par cles. This means that in a sample of a bunch of atoms, we can assume a certain percentage of them will “go off” at any point. (For instance, if all atom of a certain radioac ve 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.) . . . 7.
  8. 8. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Radioactive decay as a differential equation The rela ve rate of decay is constant: y′ =k y where k is nega ve. So y′ = ky =⇒ y = y0 ekt again! It’s customary to express the rela ve rate of decay in the units of half-life: the amount of me it takes a pure sample to decay to one which is only half pure. . . Notes Computing the amount remaining Example The half-life of polonium-210 is about 138 days. How much of a 100 g sample remains a er t years? . . Notes Carbon-14 Dating The ra o of carbon-14 to carbon-12 in an organism decays exponen ally: p(t) = p0 e−kt . The half-life of carbon-14 is about 5700 years. So the equa on for p(t) is p(t) = p0 e− 5700 t = p0 2−t/5700 ln2 . . . 8.
  9. 9. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Computing age with Carbon-14 Example Suppose a fossil is found where the ra o of carbon-14 to carbon-12 is 10% of that in a living organism. How old is the fossil? Solu on . . Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Newton’s Law of Cooling Newton’s Law of Cooling states that the rate of cooling of an object is propor onal to the temperature difference between the object and its surroundings. This gives us a differen al equa on of the form dT = k(T − Ts ) dt (where k < 0 again). . . . 9.
  10. 10. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes General Solution to NLC problems ′ ′ To solve this, change the variable y(t) = T(t) − Ts . Then y = T and k(T − Ts ) = ky. The equa on now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts Plugging in t = 0, we see C = y0 = T0 − Ts . So Theorem The solu on to the equa on T′ (t) = k(T(t) − Ts ), T(0) = T0 is . T(t) = (T0 − Ts )ekt + Ts . Notes Computing cooling time with NLC Example A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. A er 5 minutes, the egg’s temperature is 38 ◦ C. Assuming the water has not warmed appreciably, how much longer will it take the egg to reach 20 ◦ C? Solu on We know that the temperature func on takes the form T(t) = (T0 − Ts )ekt + Ts = 80ekt + 18 To find k, plug in t = 5 and solve for k. . . Notes Finding k Solu on (Con nued) . . . 10.
  11. 11. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Finding t Solu on (Con nued) . . Notes Computing time of death with NLC Example A murder vic m is discovered at midnight and the temperature of the body is recorded as 31 ◦ C. One hour later, the temperature of the body is 29 ◦ C. Assume that the surrounding air temperature remains constant at 21 ◦ C. Calculate the vic m’s me of death. (The “normal” temperature of a living human being is approximately 37 ◦ C.) . . Solu on Notes . . . 11.
  12. 12. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Interest If an account has an compound interest rate of r per year compounded n mes, then an ini al deposit of A0 dollars becomes ( r )nt A0 1 + n a er t years. For different amounts of compounding, this will change. As n → ∞, we get con nously compounded interest ( r )nt A(t) = lim A0 1 + = A0 ert . n→∞ n Thus dollars are like bacteria. . . Notes 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 con nuously. If they produce the same balance a er every year, what is r? Solu on The balance for the 10% compounded quarterly account a er t years is A1 (t) = A0 (1.025)4t = P((1.025)4 )t The balance for the interest rate r compounded con nuously account a er t years is A2 (t) = A0 ert . . . 12.
  13. 13. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Solving Solu on (Con nued) A1 (t) = A0 ((1.025)4 )t A2 (t) = A0 (er )t For those to be the same, er = (1.025)4 , so r = ln((1.025)4 ) = 4 ln 1.025 ≈ 0.0988 So 10% annual interest compounded quarterly is basically equivalent to 9.88% compounded con nuously. . . Computing doubling time with Notes exponential growth Example How long does it take an ini al deposit of $100, compounded con nuously, to double? Solu on . . Notes I-banking interview tip of the day ln 2 The frac on can also r be approximated as either 70 or 72 divided by the percentage rate (as a number between 0 and 100, not a frac on between 0 and 1.) This is some mes called the rule of 70 or rule of 72. 72 has lots of factors so . it’s used more o en. . . 13.
  14. 14. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Summary When something grows or decays at a constant rela ve rate, the growth or decay is exponen al. Equa ons with unknowns in an exponent can be solved with logarithms. . . Notes . . Notes . . . 14.

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