# Bounded Growth and Decay

### Learning Objectives

- Use Newton's law of cooling
- Use a logistic growth model
- Choose an appropriate model for data

## Use Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a**vertical shift**of the generic

**exponential decay function**. This translation leads to

**Newton’s Law of Cooling**, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

[latex]T\left(t\right)=a{e}^{kt}+{T}_{s}[/latex]

This formula is derived as follows:[latex]\begin{array}{l}T\left(t\right)=A{b}^{ct}+{T}_{s}\hfill & \hfill \\ T\left(t\right)=A{e}^{\mathrm{ln}\left({b}^{ct}\right)}+{T}_{s}\hfill & \text{Laws of logarithms}.\hfill \\ T\left(t\right)=A{e}^{ct\mathrm{ln}b}+{T}_{s}\hfill & \text{Laws of logarithms}.\hfill \\ T\left(t\right)=A{e}^{kt}+{T}_{s}\hfill & \text{Rename the constant }c \mathrm{ln} b,\text{ calling it }k.\hfill \end{array}[/latex]

### A General Note: Newton’s Law of Cooling

The temperature of an object,*T*, in surrounding air with temperature [latex]{T}_{s}[/latex] will behave according to the formula [latex-display]T\left(t\right)=A{e}^{kt}+{T}_{s}[/latex-display] where

*t*is time*A*is the difference between the initial temperature of the object and the surroundings*k*is a constant, the continuous rate of cooling of the object

### How To: Given a set of conditions, apply Newton’s Law of Cooling.

- Set [latex]{T}_{s}[/latex] equal to the
*y*-coordinate of the horizontal asymptote (usually the ambient temperature). - Substitute the given values into the continuous growth formula [latex]T\left(t\right)=A{e}^{k}{}^{t}+{T}_{s}[/latex] to find the parameters
*A*and*k*. - Substitute in the desired time to find the temperature or the desired temperature to find the time.

### Example: Using Newton’s Law of Cooling

A cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\circ\text{F}[/latex], and is placed into a [latex]35^\circ\text{F}[/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\circ\text{F}[/latex]. If we must wait until the cheesecake has cooled to [latex]70^\circ\text{F}[/latex] before we eat it, how long will we have to wait?Answer: Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation

[latex]T\left(t\right)=A{e}^{kt}+35[/latex]

We know the initial temperature was 165, so [latex]T\left(0\right)=165[/latex].[latex]\begin{array}{l}165=A{e}^{k0}+35\hfill & \text{Substitute }\left(0,165\right).\hfill \\ A=130\hfill & \text{Solve for }A.\hfill \end{array}[/latex]

We were given another data point, [latex]T\left(10\right)=150[/latex], which we can use to solve for*k*.

[latex]\begin{array}{l}\text{ }150=130{e}^{k10}+35\hfill & \text{Substitute (10, 150)}.\hfill \\ \text{ }115=130{e}^{k10}\hfill & \text{Subtract 35}.\hfill \\ \text{ }\frac{115}{130}={e}^{10k}\hfill & \text{Divide by 130}.\hfill \\ \text{ }\mathrm{ln}\left(\frac{115}{130}\right)=10k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{ }k=\frac{\mathrm{ln}\left(\frac{115}{130}\right)}{10}=-0.0123\hfill & \text{Divide by the coefficient of }k.\hfill \end{array}[/latex]

This gives us the equation for the cooling of the cheesecake: [latex]T\left(t\right)=130{e}^{-0.0123t}+35[/latex]. Now we can solve for the time it will take for the temperature to cool to 70 degrees.[latex]\begin{array}{l}70=130{e}^{-0.0123t}+35\hfill & \text{Substitute in 70 for }T\left(t\right).\hfill \\ 35=130{e}^{-0.0123t}\hfill & \text{Subtract 35}.\hfill \\ \frac{35}{130}={e}^{-0.0123t}\hfill & \text{Divide by 130}.\hfill \\ \mathrm{ln}\left(\frac{35}{130}\right)=-0.0123t\hfill & \text{Take the natural log of both sides}\hfill \\ t=\frac{\mathrm{ln}\left(\frac{35}{130}\right)}{-0.0123}\approx 106.68\hfill & \text{Divide by the coefficient of }t.\hfill \end{array}[/latex]

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\circ\text{F}[/latex].### Try It

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?Answer: 6.026 hours

## Choose an Appropriate Model

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015. Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered. In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down. A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection. After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.### Example: Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed below? Find the model, and use a graph to check your choice.x |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

y |
0 | 1.386 | 2.197 | 2.773 | 3.219 | 3.584 | 3.892 | 4.159 | 4.394 |

Answer:
First, plot the data on a graph as in the graph below. For the purpose of graphing, round the data to two significant digits.
Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\mathrm{ln}\left(bx\right)[/latex]. Plugging in the first point, [latex]\left(\text{1,0}\right)[/latex], gives [latex]0=a\mathrm{ln}b[/latex]. We reject the case that *a *= 0 (if it were, all outputs would be 0), so we know

[latex]\mathrm{ln}\left(b\right)=0[/latex]. Thus *b *= 1 and [latex]y=a\mathrm{ln}\left(\text{x}\right)[/latex]. Next we can use the point [latex]\left(\text{9,4}\text{.394}\right)[/latex] to solve for *a*:
[latex]\begin{array}{l}y=a\mathrm{ln}\left(x\right)\hfill \\ 4.394=a\mathrm{ln}\left(9\right)\hfill \\ a=\frac{4.394}{\mathrm{ln}\left(9\right)}\hfill \end{array}[/latex]

*x*> 0. A quick check confirms this conclusion: [latex]y=\mathrm{ln}\left({x}^{2}\right)=2\mathrm{ln}\left(x\right)[/latex] for

*x*> 0.However, if

*x*< 0, the graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] includes a "extra" branch, as shown below. This occurs because, while [latex]y=2\mathrm{ln}\left(x\right)[/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] can have negative domain values.

### Try It

Does a linear, exponential, or logarithmic model best fit the data in the table below? Find the model.x |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

y |
3.297 | 5.437 | 8.963 | 14.778 | 24.365 | 40.172 | 66.231 | 109.196 | 180.034 |

Answer: Exponential. [latex]y=2{e}^{0.5x}[/latex].

## Expressing an Exponential Model in Base *e*

While powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[/latex] and [latex]e[/latex]. In science and mathematics, the base *e*is often preferred. We can use laws of exponents and laws of logarithms to change any base to base

*e*.

### How To: Given a model with the form [latex]y=a{b}^{x}[/latex], change it to the form [latex]y={A}_{0}{e}^{kx}[/latex].

- Rewrite [latex]y=a{b}^{x}[/latex] as [latex]y=a{e}^{\mathrm{ln}\left({b}^{x}\right)}[/latex].
- Use the power rule of logarithms to rewrite y as [latex]y=a{e}^{x\mathrm{ln}\left(b\right)}=a{e}^{\mathrm{ln}\left(b\right)x}[/latex].
- Note that [latex]a={A}_{0}[/latex] and [latex]k=\mathrm{ln}\left(b\right)[/latex] in the equation [latex]y={A}_{0}{e}^{kx}[/latex].

### Example: Changing to base *e*

Change the function [latex]y=2.5{\left(3.1\right)}^{x}[/latex] so that this same function is written in the form [latex]y={A}_{0}{e}^{kx}[/latex].
Answer: The formula is derived as follows

[latex]\begin{array}{l}y=2.5{\left(3.1\right)}^{x}\hfill & \hfill \\ =2.5{e}^{\mathrm{ln}\left({3.1}^{x}\right)}\hfill & \text{Insert exponential and its inverse}\text{.}\hfill \\ =2.5{e}^{x\mathrm{ln}3.1}\hfill & \text{Laws of logs}\text{.}\hfill \\ =2.5{e}^{\left(\mathrm{ln}3.1\right)}{}^{x}\hfill & \text{Commutative law of multiplication}\hfill \end{array}[/latex]

### Try It

Change the function [latex]y=3{\left(0.5\right)}^{x}[/latex] to one having*e*as the base.

Answer: [latex]y=3{e}^{\left(\mathrm{ln}0.5\right)x}[/latex]

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