Activity 13.
A potato is placed in an oven whose temperature is 350 degrees Fahrenheit, and the potato’s temperature \(F\) (in degrees Fahrenheit) is recorded in the following table. Time \(t\) is measured in minutes.
1
Thanks to Nick Owad of Hood College for conducting experiments with actual potatoes in his oven in order to generate the data for this activity.
| \(t\) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) |
| \(F(t)\) | \(65\) | \(94.7\) | \(141.7\) | \(167.6\) | \(182.4\) | \(197.9\) | \(209.3\) |
(a)
Use a central difference to estimate the instantaneous rate of change of the potato’s temperature at \(t= 20\text{.}\) Include units on your answer.
Hint.
Think about quantities such as \(\frac{F(30)-F(20)}{30-20}\text{.}\)
Answer.
\begin{equation*}
F'(20) \approx \frac{F(30)-F(10)}{30-10} = \frac{167.6-94.7}{20} = 3.645
\end{equation*}
degrees per minute.
Solution.
Using the central difference, we find that
\begin{equation*}
F'(20) \approx \frac{F(30)-F(10)}{30-10} = \frac{167.6-94.7}{20} = 3.645
\end{equation*}
degrees per minute.
(b)
Use a central difference to estimate the instantaneous rate of change of the potato’s temperature at \(t= 40\text{.}\) Include units on your answer.
Hint.
See the note in (a).
Answer.
\begin{equation*}
F'(40) \approx \frac{F(50)-F(30)}{50-30} = \frac{197.9-167.6}{20} = 1.515
\end{equation*}
degrees per minute.
Solution.
Using the central difference, we find that
\begin{equation*}
F'(40) \approx \frac{F(50)-F(30)}{50-30} = \frac{197.9-167.6}{20} = 1.515
\end{equation*}
degrees per minute.
(c)
Without doing any calculation, which do you expect to be greater: \(F'(50)\) or \(F'(60)\text{?}\) Why?
Hint.
Is \(F\) changing faster at \(t = 50\) or at \(t = 60\text{?}\)
Answer.
We expect \(F'(50) \gt F'(60)\text{.}\)
Solution.
As time goes on, we see that it’s usually the case that the amount of increase in the potato’s temperature gets less and less, thus we expect the value of \(F'(t)\) to get smaller and smaller as time goes on. We therefore expect \(F'(50) \gt F'(60)\text{.}\)
(d)
Suppose we know that \(F(46) = 192.5\) and \(F'(46) = 1.39\text{.}\) What are the respective units on these two quantities? What do you expect the temperature of the potato to be when \(t = 47\text{?}\) when \(t = 48\text{?}\) Why?
Hint.
Remember that the units on \(F'\) will be “degrees Fahrenheit per minute.”
Answer.
Because at time \(t = 46\) the potato’s temperature is increasing at 1.3941 degrees per minute, we expect that at \(t = 47\text{,}\) the temperature will be about 1.3941 degrees greater than at \(t = 46\text{.}\)
Solution.
The value \(F(46) = 192.5\) is the temperature of the potato in degrees Fahrenheit at time 46, while \(F'(46) = 1.3941\) measures the instantaneous rate of change of the potato’s temperature with respect to time at the instant \(t = 46\text{,}\) and its units are degrees per minute. Because at time \(t = 46\) the potato’s temperature is increasing at 1.3941 degrees per minute, we expect that at \(t = 47\text{,}\) the temperature will be about 1.3941 degrees greater than at \(t = 46\text{,}\) or in other words \(F(47) \approx 192.5 + 1.3941 = 193.8941\text{.}\) Similarly, at \(t = 48\text{,}\) two minutes have elapsed from \(t = 48\text{,}\) so we expect an increase of \(2 \cdot 1.3941\) degrees: \(F(48) \approx 192.5 + 2 \cdot 1.3941 = 195.2882\text{.}\)
(e)
Write a couple of careful sentences that describe the behavior of the temperature of the potato on the time interval \([10,40]\text{,}\) as well as the behavior of the instantaneous rate of change of the potato’s temperature on the same time interval.
Hint.
Be careful to distinguish between the temperature, \(F\text{,}\) and the rate of change of temperature, \(F'\text{,}\) in your commentary.
Answer.
We might say that on the interval \([10,40]\)“the temperature of the potato is increasing, but at a decreasing rate.”
Solution.
Throughout the time interval \([10,40]\text{,}\) the temperature \(F\) of the potato is increasing. But as time goes on, the rate at which the temperature is rising appears to be decreasing. That is, while the values of \(F\) continue to get larger as time progresses, the values of \(F'\) seem to be getting smaller (while still remaining positive). We thus might say that on the interval \([10,40]\)“the temperature of the potato is increasing, but at a decreasing rate.”
