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Bounded Growth and Decay

Learning Outcomes

• Use Newton's Law of Cooling.
• Use a logistic growth model.
• Choose an appropriate model for data.

tip for success

The applications in this section rely upon all the knowledge we've built up from our first study of the language and notation of functions to the most recent work you've done in exponential and logarithmic functions. If you need a refresher, return to Algebra Basics, the module on function basics,or any section you've covered with regard to modeling using functions.

Using 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. The formula is derived as follows:

$\begin{array}{l}T\left(t\right)=A{b}^{ct}+{T}_{s}\hfill & \hfill \\ T\left(t\right)=A{e}^{\mathrm{ln}\left({b}^{ct}\right)}+{T}_{s}\hfill & \text{Properties of logarithms}.\hfill \\ T\left(t\right)=A{e}^{ct\mathrm{ln}b}+{T}_{s}\hfill & \text{Properties of logarithms}.\hfill \\ T\left(t\right)=A{e}^{kt}+{T}_{s}\hfill & \text{Rename the constant }c \mathrm{ln} b,\text{ calling it }k.\hfill \end{array}$

A General Note: Newton’s Law of Cooling

The temperature of an object, T, in surrounding air with temperature ${T}_{s}$ 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

1. Set ${T}_{s}$ equal to the y-coordinate of the horizontal asymptote (usually the ambient temperature).
2. Substitute the given values into the continuous growth formula $T\left(t\right)=A{e}^{k}{}^{t}+{T}_{s}$ to find the parameters A and k.
3. 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 $165^\circ\text{F}$ and is placed into a $35^\circ\text{F}$ refrigerator. After 10 minutes, the cheesecake has cooled to $150^\circ\text{F}$. If we must wait until the cheesecake has cooled to $70^\circ\text{F}$ 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

$T\left(t\right)=A{e}^{kt}+35$

We know the initial temperature was 165, so $T\left(0\right)=165$.

$\begin{array}{l}165=A{e}^{k0}+35\hfill & \text{Substitute }\left(0,165\right).\hfill \\ A=130\hfill & \text{Solve for }A.\hfill \end{array}$

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

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

This gives us the equation for the cooling of the cheesecake: $T\left(t\right)=130{e}^{-0.0123t}+35$. Now we can solve for the time it will take for the temperature to cool to 70 degrees.

$\begin{array}{l}70=130{e}^{-0.0123t}+35\hfill & \text{Substitute in 70 for }T\left(t\right).\hfill \\ 35=130{e}^{-0.0123t}\hfill & \text{Subtract 35 from both sides}.\hfill \\ \frac{35}{130}={e}^{-0.0123t}\hfill & \text{Divide both sides by 130}.\hfill \\ \mathrm{ln}\left(\frac{35}{130}\right)=-0.0123t\hfill & \text{Take the natural log of both sides}.\hfill \\ t=\frac{\mathrm{ln}\left(\frac{35}{130}\right)}{-0.0123}\approx 106.68\hfill & \text{Divide both sides by the coefficient of }t.\hfill \end{array}$

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to $70^\circ\text{F}$.

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

[ohm_question]114392[/ohm_question]
Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually an exponential model must begin to approach some limiting value and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model although the exponential growth model is still useful over a short term before approaching the limiting value. The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound called the carrying capacity. For constants a, b, and c, the logistic growth of a population over time x is represented by the model

$f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$

The graph below shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate at which the rate of increase decreases.

A General Note: Logistic Growth

The logistic growth model is

$f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$

where
• $\frac{c}{1+a}$ is the initial value
• c is the carrying capacity or limiting value
• b is a constant determined by the rate of growth.

Example: Using the Logistic-Growth Model

An influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population. For example, at time = 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is = 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

Answer: We substitute the given data into the logistic growth model

$f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is = 1000. To find a, we use the formula that the number of cases at time = 0 is $\frac{c}{1+a}=1$, from which it follows that = 999. This model predicts that, after ten days, the number of people who have had the flu is $f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}\approx 293.8$. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, = 1000.

Analysis of the Solution

Remember that because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values. The graph below gives a good picture of how this model fits the data.
The graph of $f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}$.

Try It

Using the model in the previous example, estimate the number of cases of flu on day 15.

Answer: 895 cases on day 15

[ohm_question]5801[/ohm_question]

Choosing 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 although 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 down 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 $y=a\mathrm{ln}\left(bx\right)$. Plugging in the first point, $\left(\text{1,0}\right)$, gives $0=a\mathrm{ln}b$. We reject the case that = 0 (if it were, all outputs would be 0), so we know

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

Because $a=\frac{4.394}{\mathrm{ln}\left(9\right)}\approx 2$, an appropriate model for the data is $y=2\mathrm{ln}\left(x\right)$. To check the accuracy of the model, we graph the function together with the given points.
The graph of $y=2\mathrm{ln}x$.
We can conclude that the model is a good fit to the data. Compare the figure above to the graph of $y=\mathrm{ln}\left({x}^{2}\right)$ shown below.
The graph of $y=\mathrm{ln}\left({x}^{2}\right)$
The graphs appear to be identical when > 0. A quick check confirms this conclusion: $y=\mathrm{ln}\left({x}^{2}\right)=2\mathrm{ln}\left(x\right)$ for > 0. However, if < 0, the graph of $y=\mathrm{ln}\left({x}^{2}\right)$ includes an "extra" branch as shown below. This occurs because while $y=2\mathrm{ln}\left(x\right)$ cannot have negative values in the domain (as such values would force the argument to be negative), the function $y=\mathrm{ln}\left({x}^{2}\right)$ 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. $y=2{e}^{0.5x}$.

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 $10$ and $e$. In science and mathematics, the base e is often preferred. We can use properties of exponents and properties of logarithms to change any base to base e.

How To: Given a model with the form $y=a{b}^{x}$, change it to the form $y={A}_{0}{e}^{kx}$

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

Example: Changing to base $e$

Change the function $y=2.5{\left(3.1\right)}^{x}$ so that this same function is written in the form $y={A}_{0}{e}^{kx}$.

Answer: The formula is derived as follows

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

Try It

Change the function $y=3{\left(0.5\right)}^{x}$ to one having e as the base.

Answer: $y=3{e}^{\left(\mathrm{ln}0.5\right)x}$

Licenses & Attributions

CC licensed content, Shared previously

• Question ID 5801. Authored by: Lippman,David. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
• College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].