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].

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]

Because [latex]a=\frac{4.394}{\mathrm{ln}\left(9\right)}\approx 2[/latex], an appropriate model for the data is [latex]y=2\mathrm{ln}\left(x\right)[/latex]. To check the accuracy of the model, we graph the function together with the given points. We can conclude that the model is a good fit to the data. Compare the figure above to the graph of [latex]y=\mathrm{ln}\left({x}^{2}\right)[/latex] shown below.The graphs appear to be identical when 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.

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]