Question

The temperature T, in degrees Fahrenheit, of a yam after sitting in a hot oven for t minutes is given by the function $T\left(t\right)=350-280e^{-0.2t}$

(a) What is the initial temperature of the yam, before it is put in the oven?

(b) Given that over time the temperature of the yam will approach the temperature inside the oven, use a limit to determine the temperature of the oven.

(c) How long will it take for the yam to be within 5 degrees Fahrenheit of the temperature of the oven?

(d) The first derivative of T(t) measures the rate of change of the temperature of the yam. The second derivative of T(t) measures the rate of change of the rate of change of the temperature of the yam. Use T'(t) and T''(t) to argue that the temperature of the yam increases at a decreasing rate. This statement is related to the odd saying “Cold water boils faster.” How?

Short answer

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)T\left(t\right)=70° \\ \mathrm{Part}\left(\mathrm{b}\right)T\left(t\right)=350° \\ \mathrm{Part}\left(\mathrm{c}\right)20\mathrm{minutes} \\ \mathrm{Part}\left(\mathrm{d}\right)\mathrm{the}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{yam}\mathrm{increases}\mathrm{at}\mathrm{a}\mathrm{decreasing}\mathrm{rate}. \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information

The given function is$T\left(t\right)=350-280e^{-0.2t}$

Part(a) Step 2. Calculation

Substitute t as 0 in the given function to find the initial temperature.

$$\begin{gathered} T\left(t\right)=350-280e^{-0.2t} \\ T\left(0\right)=350-280e^{-0.2\left(0\right)} \\ =350-280 \\ =70° \end{gathered}$$

Part(b) Step 1. Calculation

Substitute $t=\infty$in the given function.

$$\begin{gathered} T\left(t\right)=350-280e^{-0.2t} \\ T(\infty )=350-280e^{-0.2(\infty )} \\ =350-0 \\ =350° \end{gathered}$$

Part(c) Step 1. Calculation

Here, we have,

$$\begin{gathered} T\left(t\right)=350-280e^{-0.2t} \\ 350-5=350-280e^{-0.2t} \\ 345=350-280e^{-0.2t} \\ 280e^{-0.2t}=5 \\ {e}^{-0.2t}=\frac{5}{280} \\ -0.2t\ln \left(e\right)=\ln \left(\frac{5}{280}\right) \\ t=\frac{\ln \left(\frac{5}{280}\right)}{-0.2} \\ t=20minutes \end{gathered}$$

Part(d) Step 1. Calculation

The model for T'(t) is as follows,

$$\begin{gathered} T\text{'}\left(t\right)=\frac{d}{dt}\left(350-280e^{-0.2t}\right) \\ =-280(-0.2)e^{-0.2t} \\ =56e^{-0.2t} \end{gathered}$$

And the model for T''(t) is as follows,

$$\begin{gathered} T\text{'}\text{'}\left(t\right)=56\frac{d}{dt}{e}^{-0.2t} \\ =56(-0.2)e^{-0.2t} \\ =-11.2e^{-0.2t} \end{gathered}$$

As the coefficient is decreasing naturally, therefore we can conclude that the temperature of the Yam increases at a decreasing rate.