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- 1. Section 3.4 Exponential Growth and Decay V63.0121.002.2010Su, Calculus I New York University June 2, 2010 Announcements Review in second half of class today Oﬃce Hours after class today Midterm tomorrow
- 2. Announcements Review in second half of class today Oﬃce Hours after class today Midterm tomorrow V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 2 / 37
- 3. Objectives V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 3 / 37
- 4. Outline Recall The diﬀerential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton’s Law of Cooling Continuously Compounded Interest V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 4 / 37
- 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 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 5 / 37
- 6. Outline Recall The diﬀerential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton’s Law of Cooling Continuously Compounded Interest V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 6 / 37
- 7. What is a diﬀerential equation? Deﬁnition A diﬀerential equation is an equation for an unknown function which includes the function and its derivatives. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
- 8. What is a diﬀerential equation? Deﬁnition A diﬀerential equation is an equation for an unknown function which includes the function and its derivatives. Example Newton’s Second Law F = ma is a diﬀerential equation, where a(t) = x (t). V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
- 9. What is a diﬀerential equation? Deﬁnition A diﬀerential equation is an equation for an unknown function which includes 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 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
- 10. What is a diﬀerential equation? Deﬁnition A diﬀerential equation is an equation for an unknown function which includes 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 most general solution is x(t) = A sin ωt + B cos ωt, where ω = k/m. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 7 / 37
- 11. The Equation y = 2 Example Find a solution to y (t) = 2. Find the most general solution to y (t) = 2. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
- 12. 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
- 13. 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 . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
- 14. 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’s linear. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 8 / 37
- 15. The Equation y = 2t Example Find a solution to y (t) = 2t. Find the most general solution to y (t) = 2t. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
- 16. 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) = t 2 . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
- 17. 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) = t 2 . The general solution is y = t 2 + C . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 9 / 37
- 18. The diﬀerential equation y = ky Example Find a solution to y (t) = y (t). Find the most general solution to y (t) = y (t). V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
- 19. The diﬀerential equation y = ky 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) = e t . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
- 20. The diﬀerential equation y = ky 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) = e t . The general solution is y = Ce t , not y = e t + C . (check this) V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 10 / 37
- 21. Kick it up a notch Example Find a solution to y = 2y . Find the general solution to y = 2y . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 11 / 37
- 22. Kick it up a notch Example Find a solution to y = 2y . Find the general solution to y = 2y . Solution y = e 2t y = Ce 2t V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 11 / 37
- 23. In general Example Find a solution to y = ky . Find the general solution to y = ky . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
- 24. In general Example Find a solution to y = ky . Find the general solution to y = ky . Solution y = e kt y = Ce kt V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
- 25. In general Example Find a solution to y = ky . Find the general solution to y = ky . Solution y = e kt y = Ce kt Remark What is C ? Plug in t = 0: y (0) = Ce k·0 = C · 1 = C , so y (0) = y0 , the initial value of y . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 12 / 37
- 26. Constant Relative Growth =⇒ 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) = y0 e kt V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 13 / 37
- 27. 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 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 14 / 37
- 28. Outline Recall The diﬀerential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton’s Law of Cooling Continuously Compounded Interest V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 15 / 37
- 29. 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 16 / 37
- 30. 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? V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
- 31. 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 Since y = ky for bacteria, we have y = y0 e kt . We have 10, 000 = y0 e k·3 40, 000 = y0 e k·5 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
- 32. 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 Since y = ky for bacteria, we have y = y0 e kt . We have 10, 000 = y0 e k·3 40, 000 = y0 e k·5 Dividing the ﬁrst into the second gives 4 = e 2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have 10, 000 = y0 e ln 2·3 = y0 · 8 10, 000 So y0 = = 1250. 8 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 17 / 37
- 33. Could you do that again please? We have 10, 000 = y0 e k·3 40, 000 = y0 e k·5 Dividing the ﬁrst into the second gives 40, 000 y0 e 5k = 10, 000 y0 e 3k =⇒ 4 = e 2k =⇒ ln 4 = ln(e 2k ) = 2k ln 4 ln 22 2 ln 2 =⇒ k = = = = ln 2 2 2 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 18 / 37
- 34. Outline Recall The diﬀerential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton’s Law of Cooling Continuously Compounded Interest V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 19 / 37
- 35. Modeling radioactive decay Radioactive decay occurs because many large atoms spontaneously give oﬀ particles. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 20 / 37
- 36. Modeling radioactive decay Radioactive decay occurs because many large atoms spontaneously give oﬀ particles. 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 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.) V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 20 / 37
- 37. Radioactive decay as a diﬀerential equation The relative rate of decay is constant: y =k y where k is negative. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
- 38. Radioactive decay as a diﬀerential equation The relative rate of decay is constant: y =k y where k is negative. So y = ky =⇒ y = y0 e kt again! V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
- 39. Radioactive decay as a diﬀerential equation The relative rate of decay is constant: y =k y where k is negative. So y = ky =⇒ y = y0 e kt again! It’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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 21 / 37
- 40. 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? V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
- 41. 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? Solution We have y = y0 e kt , where y0 = y (0) = 100 grams. Then 365 · ln 2 50 = 100e k·138/365 =⇒ k = − . 138 Therefore 365·ln 2 y (t) = 100e − 138 t = 100 · 2−365t/138 . V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
- 42. 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? Solution We have y = y0 e kt , where y0 = y (0) = 100 grams. Then 365 · ln 2 50 = 100e k·138/365 =⇒ k = − . 138 Therefore 365·ln 2 y (t) = 100e − 138 t = 100 · 2−365t/138 . Notice y (t) = y0 · 2−t/t1/2 , where t1/2 is the half-life. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 22 / 37
- 43. 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 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 23 / 37
- 44. 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? V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
- 45. 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 We are looking for the value of t for which p(t) = 0.1 p(0) V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
- 46. 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 We are looking for the value of t for which p(t) = 0.1 p(0) From the equation we have t ln 0.1 2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940 5700 ln 2 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
- 47. 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 We are looking for the value of t for which p(t) = 0.1 p(0) From the equation we have 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 24 / 37
- 48. Outline Recall The diﬀerential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton’s Law of Cooling Continuously Compounded Interest V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 25 / 37
- 49. Newton’s 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 26 / 37
- 50. Newton’s 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). V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 26 / 37
- 51. 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 equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
- 52. 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 equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y = ky V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
- 53. 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 equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y = ky =⇒ y = Ce kt V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
- 54. 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 equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
- 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 equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt =⇒ T = Ce kt + Ts V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
- 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 equation now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y = ky =⇒ y = Ce kt =⇒ T − Ts = Ce kt =⇒ T = Ce kt + Ts Plugging in t = 0, we see C = y0 = T0 − Ts . So Theorem The solution to the equation T (t) = k(T (t) − Ts ), T (0) = T0 is T (t) = (T0 − Ts )e kt + Ts V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 27 / 37
- 57. Computing cooling time with NLC Example A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 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? V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 28 / 37
- 58. Computing cooling time with NLC Example A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 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? Solution We know that the temperature function takes the form T (t) = (T0 − Ts )e kt + Ts = 80e kt + 18 To ﬁnd k, plug in t = 5: 38 = T (5) = 80e 5k + 18 and solve for k. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 28 / 37
- 59. Finding k 38 = T (5) = 80e 5k + 18 20 = 80e 5k 1 = e 5k 4 1 ln = 5k 4 1 =⇒ k = − ln 4. 5 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 29 / 37
- 60. Finding k 38 = T (5) = 80e 5k + 18 20 = 80e 5k 1 = e 5k 4 1 ln = 5k 4 1 =⇒ k = − ln 4. 5 Now we need to solve t 20 = T (t) = 80e − 5 ln 4 + 18 for t. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 29 / 37
- 61. Finding t 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 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 30 / 37
- 62. Computing time of death with NLC Example A murder victim 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 victim’s time of death. (The “normal” temperature of a living human being is approximately 37 ◦ C.) V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 31 / 37
- 63. 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
- 64. 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 = 10e k·1 + 21 =⇒ k = ln 0.8 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
- 65. 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 = 10e k·1 + 21 =⇒ k = ln 0.8 To ﬁnd t: 37 = 10e t·ln(0.8) + 21 1.6 = e t·ln(0.8) ln(1.6) t= ≈ −2.10 hr ln(0.8) So the time of death was just before 10:00 pm. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 32 / 37
- 66. Outline Recall The diﬀerential equation y = ky Modeling simple population growth Modeling radioactive decay Carbon-14 Dating Newton’s Law of Cooling Continuously Compounded Interest V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 33 / 37
- 67. 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
- 68. 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 e rt . n→∞ n V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
- 69. 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 e rt . n→∞ n Thus dollars are like bacteria. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 34 / 37
- 70. Computing doubling time with exponential growth Example How long does it take an initial deposit of $100, compounded continuously, to double? V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 35 / 37
- 71. Computing doubling time with exponential growth Example How long does it take an initial deposit of $100, compounded continuously, to double? Solution We need t such that A(t) = 200. In other words ln 2 200 = 100e rt =⇒ 2 = e rt =⇒ 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 V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 35 / 37
- 72. I-banking interview tip of the day ln 2 The fraction can also be r 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. V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 36 / 37
- 73. 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). V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 37 / 37

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