Activity 15.
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 Table 152. Time \(t\) is measured in minutes.
2
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.
| \(t\) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) |
| \(F(t)\) | \(65\) | \(94.7\) | \(141.7\) | \(167.6\) | \(182.4\) | \(197.9\) | \(209.3\) |
| \(t\) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) |
| \(F'(t)\) | NA | \(3.805\) | \(3.645\) | \(2.065\) | \(1.515\) | \(1.35\) | NA |
In Activity 13, we used this data to compute approximations to \(F'(20)\) and \(F'(40)\) using central differences. Those values are provided in Table 153, along with several others computed in the same way.
(a)
What are the units on \(F'(t)\text{?}\) What is the precise meaning of the value \(F'(20) = 3.645\text{?}\)
Hint.
Remember that the derivative’s units are “units of output per unit of input.”
Answer.
Degrees Fahrenheit per minute.
Solution.
\(F'(t)\) has units measured in degrees Fahrenheit per minute.
(b)
Use a central difference to estimate the value of \(F''(30)\text{.}\)
Hint.
To estimate \(g'(a)\text{,}\) we can use
\begin{equation*}
g'(a) \approx \frac{g(a+h)-g(a-h)}{2h}
\end{equation*}
for an appropriate choice of \(h\text{.}\)
Answer.
\(F''(30) \approx -0.0516\text{.}\)
Solution.
Using a central difference,
\begin{equation*}
F''(30) \approx \frac{F'(40)-F'(20)}{20} = \frac{1.515-3.645}{20} \approx -0.0516\text{.}
\end{equation*}
(c)
What is the meaning of the value of \(F''(30)\) that you have computed in (b) in terms of the potato’s temperature? Write several careful sentences that describe the overall behavior of the potato’s temperature at this point in time. In particular, you should cite the values of \(F(30)\text{,}\) \(F'(30)\text{,}\) and \(F''(30)\text{,}\) each with appropriate units. Be sure to explicitly discuss what you expect to happen in the minute that transpires from \(t = 30\) to \(t = 31\text{.}\)
Hint.
For each of the values \(F'(30)\) and \(F''(30)\text{,}\) think about what they tell you about expected upcoming behavior in \(F(t)\) and \(F'(t)\text{,}\) respectively.
Answer.
At the moment \(t = 30\text{,}\) the temperature of the potato is \(167.6\) degrees; its temperature is rising at an instaneous rate of \(2.605\) degrees Fahrenheit per minute; and the rate at which the temperature is rising is falling at a rate of \(0.0516\) degrees Fahrenheit per minute per minute. Over the minute from \(t = 30\) to \(t = 31\text{,}\) we expect the temperature of the potato to rise about \(2.605\) degrees Fahrenheit and for the rate at which its temperature is increasing to drop by about \(0.0516\) degrees Fahrenheit per minute. We expect that \(F(31) \approx 169.7\) degrees Fahrenheit, and \(F'(31) \approx 2.01343\) degrees Fahrenheit per minute.
Solution.
The value \(F''(30) \approx -0.0516\text{,}\) which is measured in degrees Fahrenheit per minute per minute. Along with the other data, this tells us that at the moment \(t = 30\text{,}\) the temperature of the potato is \(167.6\) degrees, that its temperature is rising at a rate of \(2.065\) degrees Fahrenheit per minute, and that the rate at which the temperature is rising is falling at a rate of \(0.0516\) degrees Fahrenheit per minute per minute. That is, while the temperature is rising, it is rising at a slower and slower rate. At \(t = 31\text{,}\) we expect that \(F(31) \approx 167.6 + 2.065 = 169.7\) degrees Fahrenheit, and \(F'(31) \approx 2.065 - 0.0516 = 2.01343\) degrees Fahrenheit per minute.
(d)
On the interval from \(t = 10\) to \(t = 40\text{,}\) is the potato’s temperature increasing at an increasing rate, increasing at a constant rate, or increasing at a decreasing rate? Why?
Hint.
Think concavity.
Answer.
Increasing at a decreasing rate.
Solution.
On the interval \([10,40]\text{,}\) the potato’s temperature increasing at a decreasing rate because the values of the first derivative of \(F\) are falling. Equivalently, this is because the value of \(F''(t)\) is negative throughout the given time interval.
