Lesson 15: Exponential Growth and Decay (Section 021 slides)

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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 V63.0121.021, Calculus I New York University October 28, 2010Announcements Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2 . . . . . .
  • 2. Announcements Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 2 / 40
  • 3. Objectives Solve the ordinary differential equation y′ (t) = ky(t), y(0) = y0 Solve problems involving exponential growth and decay . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 3 / 40
  • 4. OutlineRecallThe differential equation y′ = kyModeling simple population growthModeling radioactive decay Carbon-14 DatingNewton’s Law of CoolingContinuously Compounded Interest . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 4 / 40
  • 5. Derivatives of exponential and logarithmic functions y y′ ex ex ax (ln a) · ax 1 ln x x 1 1 loga x · ln a x . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 5 / 40
  • 6. OutlineRecallThe differential equation y′ = kyModeling simple population growthModeling radioactive decay Carbon-14 DatingNewton’s Law of CoolingContinuously Compounded Interest . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 6 / 40
  • 7. What is a differential equation?DefinitionA differential equation is an equation for an unknown function whichincludes the function and its derivatives. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
  • 8. What is a differential equation?DefinitionA differential equation is an equation for an unknown function whichincludes the function and its derivatives.Example Newton’s Second Law F = ma is a differential equation, where a(t) = x′′ (t). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
  • 9. What is a differential equation?DefinitionA differential equation is an equation for an unknown function whichincludes the function and its derivatives.Example Newton’s Second Law F = ma is a differential equation, 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
  • 10. What is a differential equation?DefinitionA differential equation is an equation for an unknown function whichincludes the function and its derivatives.Example Newton’s Second Law F = ma is a differential equation, 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 The √ general solution is x(t) = A sin ωt + B cos ωt, where most ω = k/m. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
  • 11. Showing a function is a solutionExample (Continued)Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation k √x′′ + x = 0, where ω = k/m. m . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
  • 12. Showing a function is a solutionExample (Continued)Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation k √x′′ + x = 0, where ω = k/m. mSolutionWe 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 ωtSince ω 2 = k/m, the last line plus k/m times the first line result in zero. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
  • 13. The Equation y′ = 2Example Find a solution to y′ (t) = 2. Find the most general solution to y′ (t) = 2. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
  • 14. The Equation y′ = 2Example Find a solution to y′ (t) = 2. Find the most general solution to y′ (t) = 2.Solution A solution is y(t) = 2t. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
  • 15. The Equation y′ = 2Example 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
  • 16. The Equation y′ = 2Example 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.RemarkIf a function has a constant rate of growth, it’s linear. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
  • 17. The Equation y′ = 2tExample Find a solution to y′ (t) = 2t. Find the most general solution to y′ (t) = 2t. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
  • 18. The Equation y′ = 2tExample Find a solution to y′ (t) = 2t. Find the most general solution to y′ (t) = 2t.Solution A solution is y(t) = t2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
  • 19. The Equation y′ = 2tExample 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
  • 20. The Equation y′ = yExample Find a solution to y′ (t) = y(t). Find the most general solution to y′ (t) = y(t). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
  • 21. The Equation y′ = yExample 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 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
  • 22. The Equation y′ = yExample 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
  • 23. Kick it up a notch: y′ = 2yExample Find a solution to y′ = 2y. Find the general solution to y′ = 2y. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
  • 24. Kick it up a notch: y′ = 2yExample Find a solution to y′ = 2y. Find the general solution to y′ = 2y.Solution y = e2t y = Ce2t . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
  • 25. In general: y′ = kyExample Find a solution to y′ = ky. Find the general solution to y′ = ky. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
  • 26. In general: y′ = kyExample Find a solution to y′ = ky. Find the general solution to y′ = ky.Solution y = ekt y = Cekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
  • 27. In general: y′ = kyExample Find a solution to y′ = ky. Find the general solution to y′ = ky.Solution y = ekt y = CektRemarkWhat is C? Plug in t = 0: y(0) = Cek·0 = C · 1 = C,so y(0) = y0 , the initial value of y. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
  • 28. Constant Relative Growth =⇒ Exponential GrowthTheoremA function with constant relative growth rate k is an exponentialfunction with parameter k. Explicitly, the solution to the equation y′ (t) = ky(t) y(0) = y0is y(t) = y0 ekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 14 / 40
  • 29. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 15 / 40
  • 30. OutlineRecallThe differential equation y′ = kyModeling simple population growthModeling radioactive decay Carbon-14 DatingNewton’s Law of CoolingContinuously Compounded Interest . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 16 / 40
  • 31. 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 17 / 40
  • 32. Bacteria ExampleExampleA colony of bacteria is grown under ideal conditions in a laboratory. Atthe end of 3 hours there are 10,000 bacteria. At the end of 5 hoursthere are 40,000. How many bacteria were present initially? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
  • 33. Bacteria ExampleExampleA colony of bacteria is grown under ideal conditions in a laboratory. Atthe end of 3 hours there are 10,000 bacteria. At the end of 5 hoursthere are 40,000. How many bacteria were present initially?SolutionSince y′ = ky for bacteria, we have y = y0 ekt . We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
  • 34. Bacteria ExampleExampleA colony of bacteria is grown under ideal conditions in a laboratory. Atthe end of 3 hours there are 10,000 bacteria. At the end of 5 hoursthere are 40,000. How many bacteria were present initially?SolutionSince y′ = ky for bacteria, we have y = y0 ekt . We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
  • 35. Bacteria ExampleExampleA colony of bacteria is grown under ideal conditions in a laboratory. Atthe end of 3 hours there are 10,000 bacteria. At the end of 5 hoursthere are 40,000. How many bacteria were present initially?SolutionSince y′ = ky for bacteria, we have y = y0 ekt . We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5Dividing the first into the second gives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have 10, 000 = y0 eln 2·3 = y0 · 8 10, 000So y0 = = 1250. 8 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
  • 36. Could you do that again please?We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5Dividing the first into the second gives 40, 000 y e5k = 0 3k 10, 000 y0 e =⇒ 4 = e2k =⇒ ln 4 = ln(e2k ) = 2k ln 4 ln 22 2 ln 2 =⇒ k = = = = ln 2 2 2 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 19 / 40
  • 37. OutlineRecallThe differential equation y′ = kyModeling simple population growthModeling radioactive decay Carbon-14 DatingNewton’s Law of CoolingContinuously Compounded Interest . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 20 / 40
  • 38. Modeling radioactive decayRadioactive decay occurs because many large atoms spontaneouslygive off particles. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
  • 39. Modeling radioactive decayRadioactive decay occurs because many large atoms spontaneouslygive off particles.This means that in a sample ofa bunch of atoms, we canassume a certain percentage ofthem will “go off” at any point.(For instance, if all atom of acertain radioactive elementhave a 20% chance of decayingat any point, then we canexpect in a sample of 100 that20 of them will be decaying.) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
  • 40. Radioactive decay as a differential equationThe relative rate of decay is constant: y′ =k ywhere k is negative. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
  • 41. Radioactive decay as a differential equationThe relative rate of decay is constant: y′ =k ywhere k is negative. So y′ = ky =⇒ y = y0 ektagain! . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
  • 42. Radioactive decay as a differential equationThe relative rate of decay is constant: y′ =k ywhere k is negative. So y′ = ky =⇒ y = y0 ektagain!It’s customary to express the relative rate of decay in the units ofhalf-life: the amount of time it takes a pure sample to decay to onewhich is only half pure. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
  • 43. Computing the amount remaining of a decayingsampleExampleThe half-life of polonium-210 is about 138 days. How much of a 100 gsample remains after t years? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
  • 44. Computing the amount remaining of a decayingsampleExampleThe half-life of polonium-210 is about 138 days. How much of a 100 gsample remains after t years?SolutionWe have y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 · ln 2 50 = 100ek·138/365 =⇒ k = − . 138Therefore = 100 · 2−365t/138 365·ln 2 y(t) = 100e− 138 t . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
  • 45. Computing the amount remaining of a decayingsampleExampleThe half-life of polonium-210 is about 138 days. How much of a 100 gsample remains after t years?SolutionWe have y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 · ln 2 50 = 100ek·138/365 =⇒ k = − . 138Therefore = 100 · 2−365t/138 365·ln 2 y(t) = 100e− 138 tNotice y(t) = y0 · 2−t/t1/2 , where t1/2 is the half-life. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
  • 46. Carbon-14 Dating The ratio of carbon-14 to carbon-12 in an organism decays exponentially: p(t) = p0 e−kt . The half-life of carbon-14 is about 5700 years. So the equation for p(t) is ln2 p(t) = p0 e− 5700 t Another way to write this would be p(t) = p0 2−t/5700 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 24 / 40
  • 47. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 48. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?Solution p(t)We are looking for the value of t for which = 0.1 p0 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 49. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?Solution p(t)We are looking for the value of t for which = 0.1 From the p0equation we have2−t/5700 = 0.1 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 50. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?Solution p(t)We are looking for the value of t for which = 0.1 From the p0equation we have t2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 5700 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 51. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?Solution p(t)We are looking for the value of t for which = 0.1 From the p0equation we have t ln 0.12−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 5700 ln 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 52. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?Solution p(t)We are looking for the value of t for which = 0.1 From the p0equation we have t ln 0.12−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940 5700 ln 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 53. Computing age with Carbon-14 contentExampleSuppose a fossil is found where the ratio of carbon-14 to carbon-12 is10% of that in a living organism. How old is the fossil?Solution p(t)We are looking for the value of t for which = 0.1 From the p0equation we have t ln 0.12−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940 5700 ln 2So the fossil is almost 19,000 years old. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
  • 54. OutlineRecallThe differential equation y′ = kyModeling simple population growthModeling radioactive decay Carbon-14 DatingNewton’s Law of CoolingContinuously Compounded Interest . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 26 / 40
  • 55. Newtons Law of Cooling Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
  • 56. Newtons Law of Cooling Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. This gives us a differential equation of the form dT = k(T − Ts ) dt (where k < 0 again). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
  • 57. General Solution to NLC problemsTo solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ andk(T − Ts ) = ky. The equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
  • 58. General Solution to NLC problemsTo solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ andk(T − Ts ) = ky. The equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dtNow we can solve! y′ = ky . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
  • 59. General Solution to NLC problemsTo solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ andk(T − Ts ) = ky. The equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dtNow we can solve! y′ = ky =⇒ y = Cekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
  • 60. General Solution to NLC problemsTo solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ andk(T − Ts ) = ky. The equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dtNow we can solve! y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
  • 61. General Solution to NLC problemsTo solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ andk(T − Ts ) = ky. The equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dtNow we can solve! y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
  • 62. General Solution to NLC problemsTo solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ andk(T − Ts ) = ky. The equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dtNow we can solve! y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + TsPlugging in t = 0, we see C = y0 = T0 − Ts . SoTheoremThe solution to the equation T′ (t) = k(T(t) − Ts ), T(0) = T0 is T(t) = (T0 − Ts )ekt + Ts . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
  • 63. Computing cooling time with NLCExampleA hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 5minutes, the egg’s temperature is 38 ◦ C. Assuming the water has notwarmed appreciably, how much longer will it take the egg to reach20 ◦ C? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
  • 64. Computing cooling time with NLCExampleA hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 5minutes, the egg’s temperature is 38 ◦ C. Assuming the water has notwarmed appreciably, how much longer will it take the egg to reach20 ◦ C?SolutionWe know that the temperature function takes the form T(t) = (T0 − Ts )ekt + Ts = 80ekt + 18To find k, plug in t = 5: 38 = T(5) = 80e5k + 18and solve for k. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
  • 65. Finding kSolution (Continued) 38 = T(5) = 80e5k + 18 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
  • 66. Finding kSolution (Continued) 38 = T(5) = 80e5k + 18 20 = 80e5k . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
  • 67. Finding kSolution (Continued) 38 = T(5) = 80e5k + 18 20 = 80e5k 1 = e5k 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
  • 68. Finding kSolution (Continued) 38 = T(5) = 80e5k + 18 20 = 80e5k 1 = e5k ( )4 1 ln = 5k 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
  • 69. Finding kSolution (Continued) 38 = T(5) = 80e5k + 18 20 = 80e5k 1 = e5k ( )4 1 ln = 5k 4 1 =⇒ k = − ln 4. 5 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
  • 70. Finding kSolution (Continued) 38 = T(5) = 80e5k + 18 20 = 80e5k 1 = e5k ( )4 1 ln = 5k 4 1 =⇒ k = − ln 4. 5Now we need to solve for t: t 20 = T(t) = 80e− 5 ln 4 + 18 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
  • 71. Finding tSolution (Continued) t 20 = 80e− 5 ln 4 + 18 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
  • 72. Finding tSolution (Continued) t 20 = 80e− 5 ln 4 + 18 t 2 = 80e− 5 ln 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
  • 73. Finding tSolution (Continued) t 20 = 80e− 5 ln 4 + 18 t 2 = 80e− 5 ln 4 1 t = e− 5 ln 4 40 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
  • 74. Finding tSolution (Continued) t 20 = 80e− 5 ln 4 + 18 t 2 = 80e− 5 ln 4 1 t = e− 5 ln 4 40 t − ln 40 = − ln 4 5 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
  • 75. Finding tSolution (Continued) t 20 = 80e− 5 ln 4 + 18 t 2 = 80e− 5 ln 4 1 t = e− 5 ln 4 40 t − ln 40 = − ln 4 5 ln 40 5 ln 40 =⇒ t = 1 = ≈ 13 min 5 ln 4 ln 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
  • 76. Computing time of death with NLCExampleA murder victim is discovered atmidnight and the temperature ofthe body is recorded as 31 ◦ C.One hour later, the temperatureof the body is 29 ◦ C. Assumethat the surrounding airtemperature remains constantat 21 ◦ C. Calculate the victim’stime of death. (The “normal”temperature of a living humanbeing is approximately 37 ◦ C.) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 32 / 40
  • 77. Solution Let time 0 be midnight. We know T0 = 31, Ts = 21, and T(1) = 29. We want to know the t for which T(t) = 37. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
  • 78. Solution Let time 0 be midnight. We know T0 = 31, Ts = 21, and T(1) = 29. We want to know the t for which T(t) = 37. To find k: 29 = 10ek·1 + 21 =⇒ k = ln 0.8 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
  • 79. Solution Let time 0 be midnight. We know T0 = 31, Ts = 21, and T(1) = 29. We want to know the t for which T(t) = 37. To find k: 29 = 10ek·1 + 21 =⇒ k = ln 0.8 To find t: 37 = 10et·ln(0.8) + 21 1.6 = et·ln(0.8) ln(1.6) t= ≈ −2.10 hr ln(0.8) So the time of death was just before 10:00pm. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
  • 80. OutlineRecallThe differential equation y′ = kyModeling simple population growthModeling radioactive decay Carbon-14 DatingNewton’s Law of CoolingContinuously Compounded Interest . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 34 / 40
  • 81. Interest If an account has an compound interest rate of r per year compounded n times, then an initial deposit of A0 dollars becomes ( r )nt A0 1 + n after t years. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
  • 82. Interest If an account has an compound interest rate of r per year compounded n times, then an initial deposit of A0 dollars becomes ( r )nt A0 1 + n after t years. For different amounts of compounding, this will change. As n → ∞, we get continously compounded interest ( r )nt A(t) = lim A0 1 + = A0 ert . n→∞ n . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
  • 83. Interest If an account has an compound interest rate of r per year compounded n times, then an initial deposit of A0 dollars becomes ( r )nt A0 1 + n after t years. For different amounts of compounding, this will change. As n → ∞, we get continously compounded interest ( r )nt A(t) = lim A0 1 + = A0 ert . n→∞ n Thus dollars are like bacteria. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
  • 84. Continuous vs. Discrete Compounding of interestExampleConsider two bank accounts: one with 10% annual interestedcompounded quarterly and one with annual interest rate r compundedcontinuously. If they produce the same balance after every year, whatis r? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
  • 85. Continuous vs. Discrete Compounding of interestExampleConsider two bank accounts: one with 10% annual interestedcompounded quarterly and one with annual interest rate r compundedcontinuously. If they produce the same balance after every year, whatis r?SolutionThe balance for the 10% compounded quarterly account after t years is A1 (t) = A0 (1.025)4t = P((1.025)4 )tThe balance for the interest rate r compounded continuously accountafter t years is A2 (t) = A0 ert . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
  • 86. SolvingSolution (Continued) A1 (t) = A0 ((1.025)4 )t A2 (t) = A0 (er )tFor those to be the same, er = (1.025)4 , so r = ln((1.025)4 ) = 4 ln 1.025 ≈ 0.0988So 10% annual interest compounded quarterly is basically equivalentto 9.88% compounded continuously. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 37 / 40
  • 87. Computing doubling time with exponential growthExampleHow long does it take an initial deposit of $100, compoundedcontinuously, to double? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
  • 88. Computing doubling time with exponential growthExampleHow long does it take an initial deposit of $100, compoundedcontinuously, to double?SolutionWe need t such that A(t) = 200. In other words ln 2 200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t = . rFor instance, if r = 6% = 0.06, we have ln 2 0.69 69 t= ≈ = = 11.5 years. 0.06 0.06 6 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
  • 89. I-banking interview tip of the day ln 2 The fraction can also r 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’s used more often. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 39 / 40
  • 90. 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. Your friend list is like culture of bacteria (no offense). . . . . . .V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 40 / 40