# Section 4.6

:::{prf:definition}
:label: expModelDef
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

::::{prf:example}
:label: expGrowthProb1
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.

:::{dropdown} 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?

:::{dropdown} 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?

:::{dropdown} 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

::::{prf:example}
:label: decayModelProb
Suppose 800 grams of radioactive substance reduces to 400 grams of substance 2.5 years later. Find the exponential model for this situation.

:::{dropdown} 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?

:::{dropdown} 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

:::{prf:definition}
:label: newLawCoolDef

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.
:::

::::{prf:example}
:label: newCoolProblem

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.

:::{dropdown}Solution:

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?

:::{dropdown} 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.
:::
::::