Exponential Growth Using Calculus

Introduction to Exponential Growth and Decay	Exponential Growth Word Problems
Solving Exponential Growth Problems Using Differential Equations

We can use Calculus to measure Exponential Growth and Decay by using Differential Equations and Separation of Variables. Note that we studied Exponential Functions here and Differential Equations here in earlier sections.

Introduction to Exponential Growth and Decay

Remember that Exponential Growth or Decay means something is increasing or decreasing at an exponential rate (faster than if it were linear). We usually see Exponential Growth and Decay problems relating to populations, bacteria, temperature, and so on, usually as a function of time.

Solving Exponential Growth Problems using Differential Equations

It turns out that if a function is exponential, as many applications are, the rate of change of a variable is proportional to the value of that variable. Thus, $ \displaystyle \frac{{dy}}{{dt}}=ky$ or $ {y}’=ky$. This is where the Calculus comes in: we can use a differential equation to get the following:

Exponential Growth and Decay Formula

For a function $ y>0$ that is differentiable function of $ t$, and $ {y}’=ky$:

$ y=C{{e}^{{kt}}}$

$ C$ is the initial value of $ y$, $ k$ is the proportionality constant. For $ k>0$, we have exponential growth, and for $ k<0$, we have exponential decay.

Here’s how we got to this equation (using a Differential Equation), which is good to know for future problems. Note that since $ {{e}^{C}}$ is a constant, we can just turn this into another constant “$ C$”. (Note that $ y>0$.)

$ \displaystyle \frac{{dy}}{{dt}}=ky;\,\,\,dy=ky\cdot dt$

$ \displaystyle \frac{{dy}}{y}=k\,dt;\,\,\,\int{{\frac{1}{y}\,dy}}=\int{{kdt}}$

$ \displaystyle \begin{array}{c}\ln \left( y \right)=kt+{{C}_{1}}\\{{e}^{{\ln \left( y \right)}}}={{e}^{{kt+{{C}_{1}}}}}\\y={{e}^{{kt}}}\cdot {{e}^{{{{C}_{1}}}}}={{e}^{{kt}}}\cdot C=C{{e}^{{kt}}}\end{array}$

Before we get into the Exponential Growth problems, let’s do a few practice differential equation problems using Separation of Variables. Remember that we can cross-multiply to get started:

Differential Equation Problem	Solution
Solve the differential equation: $ \displaystyle {y}’=\frac{{\sqrt{x}}}{{2y}}$	$ \displaystyle \frac{{dy}}{{dx}}=\frac{{\sqrt{x}}}{{2y}};\,\,\,\,\,\,\,2y\,dy={{x}^{{\frac{1}{2}}}}\,dx;\,\,\,\,\,\int{{2y\,dy}}=\int{{{{x}^{{\frac{1}{2}}}}\,dx}}$ $ \displaystyle {{y}^{2}}=\frac{2}{3}{{x}^{{\frac{3}{2}}}}+C;\,\,\,\,\,\,y=\pm \sqrt{{\frac{2}{3}{{x}^{{\frac{3}{2}}}}+C}}$
The rate of change of $ Q$ with respect to $ t$ is inversely proportional to the cube of $ t$.	$ \begin{align}\frac{{dQ}}{{dt}}&=\frac{k}{{{{t}^{3}}}};\,\,dQ=k{{t}^{{-3}}}\,dt\\\int{{dQ}}&=\int{{k{{t}^{{-3}}}\,dt}}\\Q&=-\frac{k}{{2{{t}^{2}}}}+C\end{align}$

Differential Equation Problem

Solution

Solve the differential equation:

$ \displaystyle {y}’=\frac{{\sqrt{x}}}{{2y}}$

$ \displaystyle \frac{{dy}}{{dx}}=\frac{{\sqrt{x}}}{{2y}};\,\,\,\,\,\,\,2y\,dy={{x}^{{\frac{1}{2}}}}\,dx;\,\,\,\,\,\int{{2y\,dy}}=\int{{{{x}^{{\frac{1}{2}}}}\,dx}}$

$ \displaystyle {{y}^{2}}=\frac{2}{3}{{x}^{{\frac{3}{2}}}}+C;\,\,\,\,\,\,y=\pm \sqrt{{\frac{2}{3}{{x}^{{\frac{3}{2}}}}+C}}$

The rate of change of $ Q$ with respect to $ t$ is inversely proportional to the cube of $ t$.

$ \begin{align}\frac{{dQ}}{{dt}}&=\frac{k}{{{{t}^{3}}}};\,\,dQ=k{{t}^{{-3}}}\,dt\\\int{{dQ}}&=\int{{k{{t}^{{-3}}}\,dt}}\\Q&=-\frac{k}{{2{{t}^{2}}}}+C\end{align}$

Use this same technique to solve an Exponential Growth problem:

Differential Equation Problem	Solution
The rate of change of $ R$ with respect to $ t$ is proportional to the value of $ R$: when $ t=0,\,\,R=300$ and when $ t=1,\,\,R=500$. Write and solve the differential equation that models this situation.	$ \displaystyle \begin{align}\frac{{dR}}{{dt}}=kR;\,\,\,dR&=kR\,dt;\,\,\,\frac{{dR}}{R}=k\,dt\\\int{{\frac{{dR}}{R}}}&=\int{{k\,dt}}\\\ln \left( R \right)&=kt+{{C}_{1}}\\R={{e}^{{kt+{{C}_{1}}}}}&={{e}^{{kt}}}\cdot {{e}^{{{{C}_{1}}}}}=C{{e}^{{kt}}}\end{align}$ $ \begin{array}{c}\text{For point}\left( {0,300} \right)\text{for}\left( {t,R} \right):\\300=C{{e}^{{k\cdot 0}}};\,\,\,\,\,C\cdot 1=300;\,\,\,\,\,C=300\\\text{For point }\left( {1,500} \right):\\R=300{{e}^{{kt}}};\,\,\,\,\,500=300{{e}^{{k\cdot 1}}}\\{{e}^{k}}=\frac{{500}}{{300}};\,\,\,\,\,\,k=\ln \left( {\frac{{500}}{{300}}} \right)\approx .511\\\text{Equation: }R=300{{e}^{{.511t}}}\end{array}$
Find a solution to the differential equation $ \displaystyle \frac{{dy}}{{dx}}+6{{x}^{2}}y=0$ that contains the point $ \displaystyle \left( {0,e} \right)$.	$ \displaystyle \frac{{dy}}{{dx}}=-6{{x}^{2}}y;\,\,\,\,\,\,\,\frac{{dy}}{y}=-6{{x}^{2}}\,dx;\,\,\,\,\,\int{{\frac{1}{y}\,}}dy=\int{{-6{{x}^{2}}\,dx}}$ $ \displaystyle \begin{array}{c}\ln \left( y \right)=-2{{x}^{3}}+C;\,\,\,{{e}^{{\ln \left( y \right)}}}={{e}^{{-2{{x}^{3}}+C}}}\\y={{e}^{{-2{{x}^{3}}+C}}}\,\,\,\,\text{or}\,\,\,\,y=C{{e}^{{-2{{x}^{3}}}}}\end{array}$ You can see that you get the same equation two different ways: $ \displaystyle \begin{array}{l}y={{e}^{{-2{{x}^{3}}+C}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=C{{e}^{{-2{{x}^{3}}}}}\\e={{e}^{{-2{{{\left( 0 \right)}}^{3}}+C}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e=C{{e}^{{-2{{{\left( 0 \right)}}^{3}}}}}\\e={{e}^{c}};\,\,\,C=1\,\,\,\,\,\,\,\,\,\,\,\,e=C\cdot 1;\,\,\,\,C=e\\y={{e}^{{-2{{x}^{3}}+1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=e\cdot {{e}^{{-2{{x}^{3}}}}}={{e}^{{-2{{x}^{3}}+1}}}\end{array}$

Differential Equation Problem

Solution

The rate of change of $ R$ with respect to $ t$ is proportional to the value of $ R$: when $ t=0,\,\,R=300$ and when $ t=1,\,\,R=500$.

Write and solve the differential equation that models this situation.

$ \displaystyle \begin{align}\frac{{dR}}{{dt}}=kR;\,\,\,dR&=kR\,dt;\,\,\,\frac{{dR}}{R}=k\,dt\\\int{{\frac{{dR}}{R}}}&=\int{{k\,dt}}\\\ln \left( R \right)&=kt+{{C}_{1}}\\R={{e}^{{kt+{{C}_{1}}}}}&={{e}^{{kt}}}\cdot {{e}^{{{{C}_{1}}}}}=C{{e}^{{kt}}}\end{align}$ $ \begin{array}{c}\text{For point}\left( {0,300} \right)\text{for}\left( {t,R} \right):\\300=C{{e}^{{k\cdot 0}}};\,\,\,\,\,C\cdot 1=300;\,\,\,\,\,C=300\\\text{For point }\left( {1,500} \right):\\R=300{{e}^{{kt}}};\,\,\,\,\,500=300{{e}^{{k\cdot 1}}}\\{{e}^{k}}=\frac{{500}}{{300}};\,\,\,\,\,\,k=\ln \left( {\frac{{500}}{{300}}} \right)\approx .511\\\text{Equation: }R=300{{e}^{{.511t}}}\end{array}$

Find a solution to the differential equation $ \displaystyle \frac{{dy}}{{dx}}+6{{x}^{2}}y=0$ that contains the point $ \displaystyle \left( {0,e} \right)$.

$ \displaystyle \frac{{dy}}{{dx}}=-6{{x}^{2}}y;\,\,\,\,\,\,\,\frac{{dy}}{y}=-6{{x}^{2}}\,dx;\,\,\,\,\,\int{{\frac{1}{y}\,}}dy=\int{{-6{{x}^{2}}\,dx}}$

$ \displaystyle \begin{array}{c}\ln \left( y \right)=-2{{x}^{3}}+C;\,\,\,{{e}^{{\ln \left( y \right)}}}={{e}^{{-2{{x}^{3}}+C}}}\\y={{e}^{{-2{{x}^{3}}+C}}}\,\,\,\,\text{or}\,\,\,\,y=C{{e}^{{-2{{x}^{3}}}}}\end{array}$

You can see that you get the same equation two different ways:

$ \displaystyle \begin{array}{l}y={{e}^{{-2{{x}^{3}}+C}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=C{{e}^{{-2{{x}^{3}}}}}\\e={{e}^{{-2{{{\left( 0 \right)}}^{3}}+C}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e=C{{e}^{{-2{{{\left( 0 \right)}}^{3}}}}}\\e={{e}^{c}};\,\,\,C=1\,\,\,\,\,\,\,\,\,\,\,\,e=C\cdot 1;\,\,\,\,C=e\\y={{e}^{{-2{{x}^{3}}+1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=e\cdot {{e}^{{-2{{x}^{3}}}}}={{e}^{{-2{{x}^{3}}+1}}}\end{array}$

Exponential Growth Word Problems

Now let’s do some Exponential Growth and Decay Calculus problems:

Exponential Growth Problem	Solution
In a 60°F room, a cup of liquid cools from 100°F to 80°F in 10 minutes. How much longer will it take to cool to 70°F? Use the differential equation $ \displaystyle \frac{{dT}}{{dt}}=k\left( {T-60} \right)$, where $ t$ is in minutes, and first solve this differential equation, including finding the value for $ k$. (Note that this problem is an example of Newton’s Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference in its temperature and the temperature of its surrounding.)	$ \displaystyle \frac{{dT}}{{dt}}=k\left( {T-60} \right);\,\,\,\,\,\frac{{dT}}{{T-60}}=k\,dt;\,\,\,\,\,\,\int{{\frac{{dT}}{{T-60}}}}=\int{{k\,dt}}$ $ \begin{array}{c}\ln \left\| {T-60} \right\|=kt+C;\,\,\,\,{{e}^{{\ln \left\| {T-60} \right\|}}}={{e}^{{kt+C}}};\,\,\,\,\,\,\left\| {T-60} \right\|=C{{e}^{{kt}}}\\T=C{{e}^{{kt}}}+60\,\,\,\,\,\,\text{(Can remove }\left\| {\,\,} \right\|\text{ since 70}\le \text{T}\le 100)\end{array}$ $ \text{Initial Condition }\left( {0,100} \right):\text{ }100=C{{e}^{{k\cdot 0}}}+60\,;\,\,\,\,\,C=40;\,\,\,\,T=40{{e}^{{kt}}}+60$ Find $ k$, if after 10 minutes, the temperature is 80: $ \displaystyle 80=40{{e}^{{k\left( {10} \right)}}}+60;\,\,\,\,\,{{e}^{{10k}}}=.5;\,\,\,\,\,10k=\ln \left( {.5} \right);\,\,\,\,\,k=\frac{{\ln \left( {.5} \right)}}{{10}}\approx -.0693147$ Now find the time it takes to cool to 70: $ \displaystyle 70=40{{e}^{{-.0693147t}}}+60;\,\,\,{{e}^{{-.0693147t}}}=\frac{1}{4};\,\,\,-.0693147t=\ln \left( {.25} \right)$ $ \displaystyle t=\frac{{\ln \left( {.25} \right)}}{{-.0693147}}\approx 20 \,\,\text{minutes}$
Exponential Growth is used to model the number of bacteria in an experiment. If there are 200 bacteria after 2 hours and 800 bacteria after 5 hours, how many bacteria were present initially? After how many hours will the bacteria be 50,000?	Use the equation $ y=C{{e}^{{kt}}}$, where $ C$ is the initial amount, and $ k$ is the proportionality constant. We have two unknowns ($ C$ and $ k$) and two $ \left( {t,y} \right)$ data points: $ \left( {2,200} \right)$and $ \left( {5,800} \right)$. Plug in the points and solve for $ C$: $ \displaystyle y=C{{e}^{{kt}}};\,\,\,\,\,\,200=C{{e}^{{2k}}};\,\,C=\frac{{200}}{{{{e}^{{2k}}}}};\,\,\,\,\,\,800=C{{e}^{{5k}}};\,\,C=\frac{{800}}{{{{e}^{{5k}}}}}$ Set the $ C$’s together (substitution) and cross-multiply to solve for $ k$: $ \displaystyle \frac{{200}}{{{{e}^{{2k}}}}}=\frac{{800}}{{{{e}^{{5k}}}}};\,\,\,\,200{{e}^{{5k}}}=800{{e}^{{2k}}};\,\,\,\,{{e}^{{5k}}}=4{{e}^{{2k}}};\,\,\,\ln \left( {{{e}^{{5k}}}} \right)=\ln \left( {4{{e}^{{2k}}}} \right)$ $ \require {cancel} \displaystyle \,\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{5k}}}} \right)=\ln 4+\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{2k}}}} \right);\,\,5k=\ln 4+2k;\,\,3k=\ln 4\,$ $ \displaystyle k=\frac{{\ln 4}}{3}\approx .462$ Now solve for $ C$ using either equation above: $ \displaystyle C=\frac{{200}}{{{{e}^{{2k}}}}}=\frac{{200}}{{{{e}^{{2\left( {\frac{{\ln 4}}{3}} \right)}}}}}\approx 79.37$ (bacteria initially). Thus, $ y=79.37{{e}^{{.462t}}}$. (Note that you can also put the equations $ 200=C{{e}^{{2k}}}$ and $ 800=C{{e}^{{5k}}}$ in the graphing calculator for $ \displaystyle {{Y}_{1}}\,\,(=\frac{{200}}{{{{e}^{{2x}}}}})$ and $ \displaystyle {{Y}_{2}}\,\,(=\frac{{800}}{{{{e}^{{5x}}}}})$, and use Calc Intercept to solve the system for $ C$ and $ t$. ) To get how many hours it takes for the bacteria to be 50,000: $ \displaystyle 50000=79.37{{e}^{{.462t}}};\,\,\,\,\frac{{50000}}{{79.37}}={{e}^{{.462t}}};\,\,\,\,\ln \left( {\frac{{50000}}{{79.37}}} \right)=.462t$ $ \displaystyle t\approx 13.951\,\,\text{hours}$

Exponential Growth Problem

Solution

In a 60°F room, a cup of liquid cools from 100°F to 80°F in 10 minutes. How much longer will it take to cool to 70°F?

Use the differential equation $ \displaystyle \frac{{dT}}{{dt}}=k\left( {T-60} \right)$, where $ t$ is in minutes, and first solve this differential equation, including finding the value for $ k$.

(Note that this problem is an example of Newton’s Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference in its temperature and the temperature of its surrounding.)

$ \displaystyle \frac{{dT}}{{dt}}=k\left( {T-60} \right);\,\,\,\,\,\frac{{dT}}{{T-60}}=k\,dt;\,\,\,\,\,\,\int{{\frac{{dT}}{{T-60}}}}=\int{{k\,dt}}$

$ \begin{array}{c}\ln \left| {T-60} \right|=kt+C;\,\,\,\,{{e}^{{\ln \left| {T-60} \right|}}}={{e}^{{kt+C}}};\,\,\,\,\,\,\left| {T-60} \right|=C{{e}^{{kt}}}\\T=C{{e}^{{kt}}}+60\,\,\,\,\,\,\text{(Can remove }\left| {\,\,} \right|\text{ since 70}\le \text{T}\le 100)\end{array}$

$ \text{Initial Condition }\left( {0,100} \right):\text{ }100=C{{e}^{{k\cdot 0}}}+60\,;\,\,\,\,\,C=40;\,\,\,\,T=40{{e}^{{kt}}}+60$

Find $ k$, if after 10 minutes, the temperature is 80:

$ \displaystyle 80=40{{e}^{{k\left( {10} \right)}}}+60;\,\,\,\,\,{{e}^{{10k}}}=.5;\,\,\,\,\,10k=\ln \left( {.5} \right);\,\,\,\,\,k=\frac{{\ln \left( {.5} \right)}}{{10}}\approx -.0693147$

Now find the time it takes to cool to 70:

$ \displaystyle 70=40{{e}^{{-.0693147t}}}+60;\,\,\,{{e}^{{-.0693147t}}}=\frac{1}{4};\,\,\,-.0693147t=\ln \left( {.25} \right)$

$ \displaystyle t=\frac{{\ln \left( {.25} \right)}}{{-.0693147}}\approx 20 \,\,\text{minutes}$

Exponential Growth is used to model the number of bacteria in an experiment.

If there are 200 bacteria after 2 hours and 800 bacteria after 5 hours, how many bacteria were present initially?

After how many hours will the bacteria be 50,000?

Use the equation $ y=C{{e}^{{kt}}}$, where $ C$ is the initial amount, and $ k$ is the proportionality constant. We have two unknowns ($ C$ and $ k$) and two $ \left( {t,y} \right)$ data points: $ \left( {2,200} \right)$and $ \left( {5,800} \right)$. Plug in the points and solve for $ C$:

$ \displaystyle y=C{{e}^{{kt}}};\,\,\,\,\,\,200=C{{e}^{{2k}}};\,\,C=\frac{{200}}{{{{e}^{{2k}}}}};\,\,\,\,\,\,800=C{{e}^{{5k}}};\,\,C=\frac{{800}}{{{{e}^{{5k}}}}}$

Set the $ C$’s together (substitution) and cross-multiply to solve for $ k$:

$ \displaystyle \frac{{200}}{{{{e}^{{2k}}}}}=\frac{{800}}{{{{e}^{{5k}}}}};\,\,\,\,200{{e}^{{5k}}}=800{{e}^{{2k}}};\,\,\,\,{{e}^{{5k}}}=4{{e}^{{2k}}};\,\,\,\ln \left( {{{e}^{{5k}}}} \right)=\ln \left( {4{{e}^{{2k}}}} \right)$

$ \require {cancel} \displaystyle \,\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{5k}}}} \right)=\ln 4+\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{2k}}}} \right);\,\,5k=\ln 4+2k;\,\,3k=\ln 4\,$

$ \displaystyle k=\frac{{\ln 4}}{3}\approx .462$

Now solve for $ C$ using either equation above: $ \displaystyle C=\frac{{200}}{{{{e}^{{2k}}}}}=\frac{{200}}{{{{e}^{{2\left( {\frac{{\ln 4}}{3}} \right)}}}}}\approx 79.37$ (bacteria initially). Thus, $ y=79.37{{e}^{{.462t}}}$. (Note that you can also put the equations $ 200=C{{e}^{{2k}}}$ and $ 800=C{{e}^{{5k}}}$ in the graphing calculator for $ \displaystyle {{Y}_{1}}\,\,(=\frac{{200}}{{{{e}^{{2x}}}}})$ and $ \displaystyle {{Y}_{2}}\,\,(=\frac{{800}}{{{{e}^{{5x}}}}})$, and use Calc Intercept to solve the system for $ C$ and $ t$. )

To get how many hours it takes for the bacteria to be 50,000:

$ \displaystyle 50000=79.37{{e}^{{.462t}}};\,\,\,\,\frac{{50000}}{{79.37}}={{e}^{{.462t}}};\,\,\,\,\ln \left( {\frac{{50000}}{{79.37}}} \right)=.462t$

$ \displaystyle t\approx 13.951\,\,\text{hours}$

Here are a few more Exponential Growth problems:

Exponential Growth Problem	Solution
A city had a population of 2 million people in 2000, and a population of 2.5 million in 2010. Find the exponential growth model $ y=C{{e}^{{kt}}}$ for the population growth of this city, and use this model to predict its population in the year 2030.	For the equation $ y=C{{e}^{{kt}}}$, we already have $ y=2{{e}^{{kt}}}$ (in millions), since we can begin counting at year 2000 (make $ t=0$); this $ \left( {t,y} \right)$ data point is $ \left( {0,2} \right)$. (Here is the work: $ y=C{{e}^{{kt}}};\,\,2=C{{e}^{{k\left( 0 \right)}}};\,\,2=C\left( 1 \right);\,\,C=2$). Use the other $ \left( {t,y} \right)$ data point $ \left( {10,2.5} \right)$ to solve for $ k$ (the population growth rate, or proportionally constant): $ \displaystyle y=2{{e}^{{kt}}};\,\,\,\,\,2.5=2{{e}^{{10k}}};\,\,\,\,\,\,1.25={{e}^{{10k}}};\,\,\,\,\,k=\frac{{\ln \left( {1.25} \right)}}{{10}}\approx .0223$ Thus, the exponential growth model for the population is $ y=2{{e}^{{.0223t}}}$. Find the population in year 2030 (when $ t=30$): $ y=2{{e}^{{.0223\cdot 30}}}\approx 3.9 \,\text{million}$.
The rate of decay of a certain substance is directly proportional to the amount remaining. The half-life of this substance is 150 days. After how many days will the sample have disintegrated 90%? If the sample initially weighs 30 grams, what is the decay rate of change of this new sample on its 100^th day?	Use the equation $ y=C{{e}^{{kt}}}$, and remember how we can get continuous exponential decay using this formula: $ \text{ending}=\text{beginning}\times {{e}^{{kt}}}$. Since we end up with half of the substance after 150 days, we have: $ \displaystyle .5=1{{e}^{{k\cdot 150}}};\,\,\,\,\,\ln \left( {.5} \right)=150k;\,\,\,\,\,k=\frac{{\ln \left( {.5} \right)}}{{150}}\approx -.00462$ Thus, $ y=C{{e}^{{-.00462t}}}$. Since we want the number of days until the sample has disintegrated 90%, find the number of days it has 10% left: $ \displaystyle .1=1{{e}^{{-.00462\cdot t}}};\,\,\,\,\,\ln \left( {.1} \right)=-.00462t;\,\,\,\,\,t=\frac{{\ln \left( {.1} \right)}}{{-.00462}}\approx 498\,\,\text{days}$ The last question is tricky; since we want a decay rate of change, we take the derivative of the decay function (using initial condition $ \left( {0,30} \right)$), and then use $ t=100$ after taking this derivative: $ \displaystyle y=30{{e}^{{-.00462t}}};\,\,\,\,\,{y}’=30\cdot -.00462\cdot {{e}^{{-.00462t}}};\,\,\,\,\,{y}’=-.1386{{e}^{{-.00462\cdot 100}}}\approx -.08732$

Exponential Growth Problem

Solution

A city had a population of 2 million people in 2000, and a population of 2.5 million in 2010.

Find the exponential growth model $ y=C{{e}^{{kt}}}$ for the population growth of this city, and use this model to predict its population in the year 2030.

For the equation $ y=C{{e}^{{kt}}}$, we already have $ y=2{{e}^{{kt}}}$ (in millions), since we can begin counting at year 2000 (make $ t=0$); this $ \left( {t,y} \right)$ data point is $ \left( {0,2} \right)$. (Here is the work: $ y=C{{e}^{{kt}}};\,\,2=C{{e}^{{k\left( 0 \right)}}};\,\,2=C\left( 1 \right);\,\,C=2$).

Use the other $ \left( {t,y} \right)$ data point $ \left( {10,2.5} \right)$ to solve for $ k$ (the population growth rate, or proportionally constant):

$ \displaystyle y=2{{e}^{{kt}}};\,\,\,\,\,2.5=2{{e}^{{10k}}};\,\,\,\,\,\,1.25={{e}^{{10k}}};\,\,\,\,\,k=\frac{{\ln \left( {1.25} \right)}}{{10}}\approx .0223$

Thus, the exponential growth model for the population is $ y=2{{e}^{{.0223t}}}$. Find the population in year 2030 (when $ t=30$): $ y=2{{e}^{{.0223\cdot 30}}}\approx 3.9 \,\text{million}$.

The rate of decay of a certain substance is directly proportional to the amount remaining. The half-life of this substance is 150 days.

After how many days will the sample have disintegrated 90%?

If the sample initially weighs 30 grams, what is the decay rate of change of this new sample on its 100^th day?

Use the equation $ y=C{{e}^{{kt}}}$, and remember how we can get continuous exponential decay using this formula: $ \text{ending}=\text{beginning}\times {{e}^{{kt}}}$. Since we end up with half of the substance after 150 days, we have:

$ \displaystyle .5=1{{e}^{{k\cdot 150}}};\,\,\,\,\,\ln \left( {.5} \right)=150k;\,\,\,\,\,k=\frac{{\ln \left( {.5} \right)}}{{150}}\approx -.00462$

Thus, $ y=C{{e}^{{-.00462t}}}$. Since we want the number of days until the sample has disintegrated 90%, find the number of days it has 10% left:

$ \displaystyle .1=1{{e}^{{-.00462\cdot t}}};\,\,\,\,\,\ln \left( {.1} \right)=-.00462t;\,\,\,\,\,t=\frac{{\ln \left( {.1} \right)}}{{-.00462}}\approx 498\,\,\text{days}$

The last question is tricky; since we want a decay rate of change, we take the derivative of the decay function (using initial condition $ \left( {0,30} \right)$), and then use $ t=100$ after taking this derivative:

$ \displaystyle y=30{{e}^{{-.00462t}}};\,\,\,\,\,{y}’=30\cdot -.00462\cdot {{e}^{{-.00462t}}};\,\,\,\,\,{y}’=-.1386{{e}^{{-.00462\cdot 100}}}\approx -.08732$

Learn these rules and practice, practice, practice!

On to Derivatives and Integrals of Inverse Trig Functions — you are ready!