Section 4.6

Definition 52

Let \(y_0\) be the amount or number present at time \(t=0\). Then, under certain conditions, the amount \(y\) at any time \(t\) is modeled by

\[y=y_0e^{kt}\]

where \(k\) is some constant.

• If \(k>0\), then the model is called exponential growth.

• If \(k<0\), then the model is called exponential decay.

Exponential Growth Model

Example 88

In the year 1990 the amount of atmospheric carbon dioxide was \(353\) parts per million and then in the year 2000 the amount was \(375\) parts per million. Since the model is exponential find the function.

Solution:

We want to find the function \(f(x)=y_0e^{kx}\) where \(x\) is the number of years after 1990 and \(f(x)\) is the amount of atmospheric carbon dioxide in units of parts per million.

First, we will use the fact that \(f(0)=353\) to find \(y_0\).

\[\begin{align*} f(0) & = y_0e^{k(0)}\\ 353 & = y_0e^{0}\\ y_0 & = 353 \end{align*}\]

This means the function is \(f(x)=353e^{kx}\). The year 2000 means \(x=10\). We will then use \(f(10)=375\) to solve for \(k\).

\[\begin{align*} f(10) & = 353e^{k(10)}\\ 375 & = 353e^{10k}\\ e^{10k} & = \frac{375}{353} \end{align*}\]

After composing both sides by the natural log function we have

\[\begin{align*} \ln(e^{10k}) & = \ln\left(\frac{375}{353}\right)\\ 10k\ln(e) & = \ln\left(\frac{375}{353}\right)\\ k & = \frac{1}{10}\cdot\ln\left(\frac{375}{353}\right) \end{align*}\]

This gives the function:

\[f(x)=353e^{\frac{1}{10}\cdot\ln\left(\frac{375}{353}\right)x}\]

How much atmospheric carbon dioxide will there be in the year 2020?

Solution:

The year 2020, \(x=30\). That means we want to evaluate \(f(30)\).

\[f(30)=353e^{\frac{1}{10}\cdot\ln\left(\frac{375}{353}\right)(30)}\approx 423\]

That is, in the year 2020 there will be 423 parts per million atomspheric carbon dioxide.

In what year will the amount be 560 parts per million?

Solution:

We want to find \(x\) such that \(f(x)=560\).

\[\begin{align*} f(x) & = 560\\ 353e^{\frac{1}{10}\cdot\ln\left(\frac{375}{353}\right)x} & = 560\\ e^{\frac{1}{10}\cdot\ln\left(\frac{375}{353}\right)x} & = \frac{560}{353}\\ \ln\left(e^{\frac{1}{10}\cdot\ln\left(\frac{375}{353}\right)x}\right) & =\ln\left(\frac{560}{353}\right)\\ \frac{1}{10}\cdot\ln\left(\frac{375}{353}\right)x\ln(e) & = \ln\left(\frac{560}{353}\right)\\ x & = 10\cdot \frac{\ln\left(\frac{560}{353}\right)}{\ln\left(\frac{375}{353}\right)}\\ & \approx 76 \end{align*}\]

That is, it will take about 76 years for the atomospheric carbon dioxide to double from the year 2000 amount.

Exponential Decay Model

Example 89

Suppose 800 grams of radioactive substance reduces to 400 grams of substance 2.5 years later. Find the exponential model for this situation.

Solution:

We are trying to find the function:

\[f(x)=y_0e^{kx}\]

Since, the substance starts with 800 grams we have:

\[\begin{align*} f(0) & = 800\\ y_0e^{k(0)} & = 800\\ y_0(1) & = 800 \end{align*}\]

Which then gives: \(y_0=800\).

Next, we want to use the fact that \(f(2.5)=400\) to find \(k\).

\[\begin{align*} f(2.5) & = 400\\ 800e^{k(2.5)} & = 400\\ e^{2.5k} & = \frac{400}{800} \text{ or } \frac12 \end{align*}\]

After composing both sides by the natural log we have

\[\begin{align*} \ln(e^{2.5k}) & \ln(\frac12)\\ 2.5k\ln(e) & = -\ln(2)\\ 2.5k & = -\ln(2)\\ k & = -\frac{\ln(2)}{2.5} \end{align*}\]

Therefore, the equation of the model is:

\[f(x)=800e^{-\frac{\ln(2)}{2.5}x}\]

How much of the substance will be present after 4 years?

Solution:

Here we want to evaluate \(f(4)\). We will then get

\[f(4)=800e^{-\frac{\ln(2)}{2.5}(4))}\approx 264\]

That is, there will be 264 grams of the substance.

Exponential Find \(k\)

{prf:example}

label:

findKExam1

The model for Carbon-14 is

\[y=y_0e^{-0.0001216t}\]

Find the half life of Carbon-14.

Solution:

We don’t know what the initial amount is; therefore, we will have the following equation to solve.

\[\frac{1}{2}y_0 = y_0e^{-0.0001216t}\]

Dividing by \(y_0\) to both sides we will be able to solve for \(t\).

\[\begin{align*} \frac{1}{2}y_0 & = y_0e^{-0.0001216t}\\ \frac{1}{2} = e^{-0.0001216t}\\ \ln(\frac{1}{2}) & = \ln(e^{-0.0001216t})\\ \ln(\frac{1}{2}) & = (-0.0001216t)\ln(e)\\ \frac{\ln(\frac12)}{-0.0001216} & = t \end{align*}\]

That is, it will take \(\frac{\ln(\frac12)}{-0.0001216} \approx 5700\) years.

Newton’s Law of Cooling

Definition 53

The temperature \(f(t)\) of the body at time \(t\) in appropriate units after being introduced into an environment having constant temperature \(T_0\) is

\[f(t) = T_0 + Ce^{-kt}\]

where \(C\) and \(k\) are constants.

Example 90

A New York Strip is pulled from the refrigerator with a temperature of \(34^{\circ}\text{F}\). Then the strip is moved to an oven preheated to \(350^{\circ}\text{F}\). After one hour the internal temperature of the meat is \(70^{\circ}\). Find the model that represents the internal temperature.

We start with \(f(t)=T_0+Ce^{-kt}\). Want to find \(T_0\), \(C\), and \(-k\). First, by definition the meat is being introduced into an environment with constant temperature \(350^{\circ}\). This means \(T_0=350\).

Next, we will use the fact \(T_0=350\) and \(f(0)=34\) to solve for \(C\).

\[\begin{align*} f(0) & = 34\\ 350 + Ce^{-k(0)} & = 34\\ C & = -316 \end{align*}\]

Next, we will solve for \(k\) using the fact that \(f(1)=70\).

\[\begin{align*} f(1) & = 70\\ 350 - 316e^{-k(1)} & = 70\\ -316e^{-k} & = -280\\ e^{-k} & = -\frac{280}{316}\text{ or }\frac{70}{79}\\ -k\ln(e) & = \ln(\frac{70}{79}) \end{align*}\]

That is, \(-k=\ln(\frac{70}{79})\)

Therefore, the model is defined by

\[f(t)=350 - 316e^{\ln(\frac{70}{79})t}\]

What will be the internal temp after 90 minutes?

Solution:

Here we will evaluate \(f\) when \(t=1.5\).

\[\begin{align*} f(1.5) & = 350-316e^{\ln(\frac{70}{79})(1.5)}\\ & \approx 86 \end{align*}\]

The internal tempt of the meat after 90 minutes or 1.5 hours is approximately \(86^{\circ}\).

How long will it take the meat to reach an internal temp of \(145^{\circ}\)?

{dropdown} Solution:

Want to solve \(f(t)=145\).

\[\begin{align*} 350-316e^{\ln(\frac{70}{79})t} & = 145\\ -316e^{\ln(\frac{70}{79})t} & = -205\\ e^{\ln(\frac{70}{79})t} & = \frac{205}{316}\\ \ln(\frac{70}{79})t\cdot \ln(e) & = \ln(\frac{205}{316})\\ t & = \frac{\ln(\frac{205}{316})}{\ln(\frac{70}{79})} \approx 3.5 \end{align*}\]

It will take about 3.5 hours to cook the meat to 145 degree temp.