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Chapter 4: Problem 12

Temperature Change It took 20 sec for the temperature to rise from \(0^{\circ} \mathrm{F}\) to \(212^{\circ} \mathrm{F}\) when a thermometer was taken from a freezer and placed in boiling water. Explain why at some mo- ment in that interval the mercury was rising at exactly \(10.6^{\circ} \mathrm{F} / \mathrm{sec} .\)

Short Answer

Expert verified
The mercury was rising at exactly \(10.6^{\circ} F/sec\) at some point during the 20-second interval because of the Mean Value Theorem, which states that for any function that is continuous and differentiable over an interval, there is at least one point in the interval at which the derivative is equal to the average rate of change over that interval. In this case, the average rate of change is \(10.6^{\circ} F/sec\), so there must be at least one point where the temperature was rising at this rate.

Step by step solution

01

Understand the Mean Value Theorem

The Mean Value Theorem states that if a function is continuous over a closed interval and differentiable over the open interval, there exists at least one point in the open interval where the derivative (rate of change) of the function is equal to the average rate of change over the closed interval.
02

Apply the Mean Value Theorem to the problem

The problem tells us that the temperature changed from \(0^{\circ}F\) to \(212^{\circ}F\) over 20 seconds. So, the average rate of change of the temperature over this interval is \((212 - 0) / 20 = 10.6^{\circ}F/sec\). The temperature function, which represents the temperature as a function of time, should be continuous and differentiable over the interval of 20 seconds, since the temperature changes smoothly and does not jump or have sudden changes.
03

Conclude why the thermometer was rising exactly at \(10.6^{\circ} F/sec\)

According to the Mean Value Theorem, there must be at least one point in the 20-second interval at which the rate of change of the temperature (the speed at which the temperature is rising) is equal to the average rate of change of \(10.6^{\circ} F/sec\). That's why at some moment in that interval, the mercury was rising at exactly \(10.6^{\circ} F/sec\).

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