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.
Newton’s Law of Cooling
The temperature of an object, [latex]T[/latex], in surrounding air with temperature [latex]{T}_{s}[/latex] will behave according to the formula
[latex]T\left(t\right)=A{e}^{kt}+{T}_{s}[/latex]
where
[latex]t[/latex] is time
[latex]A[/latex] is the difference between the initial temperature of the object and the surroundings
[latex]k[/latex] is a constant, the continuous rate of cooling of the object
The 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{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}[/latex]
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 [latex]A[/latex] and [latex]k[/latex].
Substitute in the desired time to find the temperature or the desired temperature to find the time.
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 [latex]10[/latex] 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?
Because the surrounding air temperature in the refrigerator is [latex]35[/latex] degrees, the cheesecake’s temperature will decay exponentially toward [latex]35[/latex], following the equation
[latex]T\left(t\right)=A{e}^{kt}+35[/latex]
We know the initial temperature was [latex]165[/latex], so [latex]T\left(0\right)=165[/latex].
We were given another data point, [latex]T\left(10\right)=150[/latex], which we can use to solve for [latex]k[/latex].
[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 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}[/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 [latex]70[/latex] 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 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}[/latex]
It will take about [latex]107[/latex] minutes, or one hour and [latex]47[/latex] minutes, for the cheesecake to cool to [latex]70^\circ\text{F}[/latex].
Logistic Growth
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.
[latex]\dfrac{c}{1+a}[/latex] is the initial value
[latex]c[/latex] is the carrying capacity or limiting value
[latex]b[/latex] is a constant determined by the rate of growth.
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.
[latex]\\[/latex]
For example, at time [latex]t = 0[/latex] there is one person in a community of [latex]1,000[/latex] people who has the flu. So, in that community, at most [latex]1,000[/latex] people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is [latex]b = 0.6030[/latex].
[latex]\\[/latex]
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.
We substitute the given data into the logistic growth model
Because at most [latex]1,000[/latex] people, the entire population of the community, can get the flu, we know the limiting value is [latex]c = 1000[/latex]. To find [latex]a[/latex], we use the formula that the number of cases at time [latex]t = 0[/latex] is [latex]\frac{c}{1+a}=1[/latex], from which it follows that [latex]a = 999[/latex]. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}\approx 293.8[/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to [latex]294[/latex]. In the long term, the number of people who will contract the flu is the limiting value, [latex]c = 1000[/latex].
The graph of [latex]f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}[/latex].
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 logistic growth model is