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# Lesson15 -exponential_growth_and_decay_021_slides

## by Matthew Leingang, Clinical Associate Professor of Mathematics at New York University on Nov 10, 2010

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## Lesson15 -exponential_growth_and_decay_021_slidesPresentation Transcript

• Section 3.4 Exponential Growth and Decay V63.0121.021, Calculus I New York University October 28, 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 2 / 40
• 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
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 4 / 40
• 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
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 6 / 40
• What is a differential equation? Definition A differential equation is an equation for an unknown function which includes the function and its derivatives. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
• 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). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
• 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 √ 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
• Showing a function is a solution Example (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
• Showing a function is a solution Example (Continued) Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation k √ x′′ + x = 0, where ω = k/m. m Solution 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 Since 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
• The Equation y′ = 2 Example 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
• 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 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’s linear. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
• The Equation y′ = 2t Example 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
• 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 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
• 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 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). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 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 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
• Kick it up a notch: y′ = 2y Example 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
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
• In general: y′ = ky Example 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
• In general: y′ = ky Example 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
• 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·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
• 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 ekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 15 / 40
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 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? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 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 Since 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
• 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 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
• 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 ekt . We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5 Dividing the first into the second gives 4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have 10, 000 = y0 eln 2·3 = y0 · 8 10, 000 So y0 = = 1250. 8 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
• Could you do that again please? We have 10, 000 = y0 ek·3 40, 000 = y0 ek·5 Dividing 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
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 20 / 40
• Modeling radioactive decay Radioactive decay occurs because many large atoms spontaneously give off particles. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
• 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.) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
• Radioactive decay as a differential equation The relative rate of decay is constant: y′ =k y where k is negative. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
• 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 ekt again! . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
• 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 ekt 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 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? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 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? Solution We have y = y0 ekt , where y0 = y(0) = 100 grams. Then 365 · ln 2 50 = 100ek·138/365 =⇒ k = − . 138 Therefore = 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
• 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 ekt , where y0 = y(0) = 100 grams. Then 365 · ln 2 50 = 100ek·138/365 =⇒ k = − . 138 Therefore = 100 · 2−365t/138 365·ln 2 y(t) = 100e− 138 t Notice 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
• 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
• 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? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 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 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
• 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 p(t) We are looking for the value of t for which = 0.1 From the p0 equation we have 2−t/5700 = 0.1 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 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 p(t) We are looking for the value of t for which = 0.1 From the p0 equation we have t 2−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
• 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 p(t) We are looking for the value of t for which = 0.1 From the p0 equation we have t ln 0.1 2−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
• 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 p(t) We are looking for the value of t for which = 0.1 From the p0 equation we have t ln 0.1 2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940 5700 ln 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 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 p(t) We are looking for the value of t for which = 0.1 From the p0 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 26 / 40
• 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
• 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). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
• 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 = Cekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
• 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 = Cekt =⇒ T − Ts = Cekt . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
• 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 = 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
• 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 = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + 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 )ekt + Ts . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
• 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? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
• 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 )ekt + Ts = 80ekt + 18 To find k, plug in t = 5: 38 = T(5) = 80e5k + 18 and solve for k. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
• Finding k Solution (Continued) 38 = T(5) = 80e5k + 18 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
• Finding k Solution (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
• Finding k Solution (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
• Finding k Solution (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
• Finding k Solution (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
• Finding k Solution (Continued) 38 = T(5) = 80e5k + 18 20 = 80e5k 1 = e5k ( )4 1 ln = 5k 4 1 =⇒ k = − ln 4. 5 Now 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
• Finding t Solution (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
• Finding t Solution (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
• Finding t Solution (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
• Finding t Solution (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
• Finding t Solution (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
• 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.) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 32 / 40
• 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
• 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
• 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
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 34 / 40
• 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
• 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
• 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
• 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? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 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 A1 (t) = A0 (1.025)4t = P((1.025)4 )t The balance for the interest rate r compounded continuously account after 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
• Solving Solution (Continued) 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 continuously. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 37 / 40
• Computing doubling time with exponential growth Example How long does it take an initial deposit of \$100, compounded continuously, to double? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
• 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 = 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
• 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
• 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