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

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### Lesson 15: Exponential Growth and Decay (slides)

1. 1. Sec on 3.4 Exponen al Growth and Decay V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 23, 2011.
2. 2. Announcements Quiz 3 next week in recita on on 2.6, 2.8, 3.1, 3.2
3. 3. Objectives Solve the ordinary diﬀeren al equa on y′ (t) = ky(t), y(0) = y0 Solve problems involving exponen al growth and decay
4. 4. 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
5. 5. Derivatives of exponential andlogarithmic functions y y′ ex ex ax (ln a) · ax 1 ln x x 1 1 loga x · ln a x
6. 6. 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
7. 7. 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.
8. 8. 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).
9. 9. 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
10. 10. 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
11. 11. 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
12. 12. The Equation y′ = 2 Example Find a solu on to y′ (t) = 2. Find the most general solu on to y′ (t) = 2.
13. 13. 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.
14. 14. 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.
15. 15. 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.
16. 16. The Equation y′ = 2t Example Find a solu on to y′ (t) = 2t. Find the most general solu on to y′ (t) = 2t.
17. 17. 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 .
18. 18. 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.
19. 19. 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).
20. 20. 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 .
21. 21. 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)
22. 22. Kick it up a notch: y′ = 2y Example Find a solu on to y′ = 2y. Find the general solu on to y′ = 2y.
23. 23. 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
24. 24. In general: y′ = ky Example Find a solu on to y′ = ky. Find the general solu on to y′ = ky.
25. 25. In general: y′ = ky Example Find a solu on to y′ = ky. Find the general solu on to y′ = ky. Solu on y = ekt y = Cekt
26. 26. In general: y′ = ky Example Find a solu on to y′ = ky. Find the general solu on to y′ = ky. Solu on Remark y = ekt 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.
27. 27. Constant Relative Growth =⇒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
28. 28. 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
29. 29. 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
30. 30. 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.
31. 31. 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?
32. 32. 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 Since y′ = ky for bacteria, we have y = y0 ekt . We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5
33. 33. Bacteria Example Solution Solu on (Con nued) We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5 Dividing the ﬁrst into the second gives 40, 000 y0 e5k = 3k =⇒ 4 = e2k =⇒ ln 4 = ln(e2k ) = 2k 10, 000 y0 e ln 4 ln 22 2 ln 2 =⇒ k = = = = ln 2 2 2 2
34. 34. Solu on (Con nued)Since y = y0 et ln 2 , at me t = 3 we have 10, 000 10, 000 = y0 e3 ln 2 = y0 · 8 =⇒ y0 = = 1250 8
35. 35. 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
36. 36. Modeling radioactive decay Radioac ve decay occurs because many large atoms spontaneously give oﬀ par cles.
37. 37. 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.)
38. 38. Radioactive decay as a diﬀerential equation The rela ve rate of decay is constant: y′ =k y where k is nega ve.
39. 39. 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!
40. 40. 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.
41. 41. 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?
42. 42. 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? Solu on We have y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 · ln 2 50 = 100ek·138/365 =⇒ k = − . 138 Therefore y(t) = 100e− = 100 · 2−365t/138 365·ln 2 138 t
43. 43. 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? Solu on We have y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 · ln 2 50 = 100ek·138/365 =⇒ k = − . 138 Therefore y(t) = 100e− = 100 · 2−365t/138 365·ln 2 138 t
44. 44. 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
45. 45. 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?
46. 46. 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 p(t) We are looking for the value of t for which = 0.1. p0
47. 47. 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 p(t) We are looking for the value of t for which = 0.1. From the equa on we have p0 2−t/5700 = 0.1
48. 48. 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 p(t) We are looking for the value of t for which = 0.1. From the equa on we have p0 t 2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 5700
49. 49. 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 p(t) We are looking for the value of t for which = 0.1. From the equa on we have p0 t ln 0.1 2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 5700 ln 2
50. 50. 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 p(t) We are looking for the value of t for which = 0.1. From the equa on we have p0 t ln 0.1 2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940 5700 ln 2
51. 51. 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 p(t) We are looking for the value of t for which = 0.1. From the equa on we have p0 t ln 0.1 2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940 5700 ln 2 So the fossil is almost 19,000 years old.
52. 52. 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
53. 53. 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.
54. 54. 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).
55. 55. 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
56. 56. 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
57. 57. 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
58. 58. 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
59. 59. 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
60. 60. 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
61. 61. 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?
62. 62. 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.
63. 63. Finding k Solu on (Con nued) 38 = T(5) = 80e5k + 18
64. 64. Finding k Solu on (Con nued) 38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k
65. 65. Finding k Solu on (Con nued) 38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k 1 = e5k 4
66. 66. Finding k Solu on (Con nued) 38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k ( ) 1 1 = e5k =⇒ ln = 5k 4 4
67. 67. Finding k Solu on (Con nued) 38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k ( ) 1 1 1 = e5k =⇒ ln = 5k =⇒ k = − ln 4. 4 4 5
68. 68. Finding k Solu on (Con nued) 38 = T(5) = 80e5k + 18 =⇒ 20 = 80e5k ( ) 1 1 1 = e5k =⇒ ln = 5k =⇒ k = − ln 4. 4 4 5 Now we need to solve for t: 20 = T(t) = 80e− 5 ln 4 + 18 t
69. 69. Finding t Solu on (Con nued) 20 = 80e− 5 ln 4 + 18 t
70. 70. Finding t Solu on (Con nued) 20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 t t
71. 71. Finding t Solu on (Con nued) 1 20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 =⇒ = e− 5 ln 4 t t t 40
72. 72. Finding t Solu on (Con nued) 1 20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 =⇒ = e− 5 ln 4 t t t 40 t − ln 40 = − ln 4 5
73. 73. Finding t Solu on (Con nued) 1 20 = 80e− 5 ln 4 + 18 =⇒ 2 = 80e− 5 ln 4 =⇒ = e− 5 ln 4 t t t 40 t ln 40 5 ln 40 − ln 40 = − ln 4 =⇒ t = 1 = ≈ 13 min 5 5 ln 4 ln 4
74. 74. Computing time of death with NLCExampleA murder vic m is discovered atmidnight and the temperature of thebody is recorded as 31 ◦ C. One hourlater, the temperature of the body is29 ◦ C. Assume that the surroundingair temperature remains constant at21 ◦ C. Calculate the vic m’s me ofdeath. (The “normal” temperature ofa living human being is approximately37 ◦ C.)
75. 75. Solu on Let me 0 be midnight. We know T0 = 31, Ts = 21, and T(1) = 29. We want to know the t for which T(t) = 37.
76. 76. Solu on Let me 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
77. 77. Solu on Let me 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 me of death was just before 10:00 .
78. 78. 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
79. 79. 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.
80. 80. 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
81. 81. 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.
82. 82. 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?
83. 83. 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 = A0 ((1.025)4 )t The balance for the interest rate r compounded con nuously account a er t years is A2 (t) = A0 ert
84. 84. 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.
85. 85. Computing doubling time Example How long does it take an ini al deposit of \$100, compounded con nuously, to double?
86. 86. Computing doubling time Example How long does it take an ini al deposit of \$100, compounded con nuously, to double? Solu on We need t such that A(t) = 200. In other words ln 2 200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t = . r For instance, if r = 6% = 0.06, we have ln 2 0.69 69 t= ≈ = = 11.5 years. 0.06 0.06 6
87. 87. I-banking interview tip of the day ln 2 The frac on can also be r 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.
88. 88. 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. Your friend list is like culture of bacteria (no oﬀense).