Chapter 5.2, Problem 12E

To determine

To explain : Why at some moment in that interval the mercury was rising at exactly ${10.6}^0\text{F/sec}$ .

Answer to Problem 12E

The possible explanation is given below.

Explanation of Solution

Given information :

It took 20 sec for the temperature to rise from $0^0\text{F/sec}$ to ${212}^0\text{F/sec}$ when a thermometer was taken from a freezer and placed in boiling water.

Calculation :

The Mean Value Theorem states that if $f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there exists at least one $c$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .

Let $f\left(x\right)$ as the temperature in degree Fahrenheit at time $x$ seconds. Assume $f\left(x\right)$ as continuous on $\left[0,20\right]$ and differentiable on $\left(0,20\right)$ .

Then according to Mean Value Theorem there must be at least one $c$ such that

  $f\text{'}\left(c\right)=\frac{f\left(20\right)-f\left(0\right)}{20-0}$

Here $f\left(20\right)=212$ , therefore,

  $\begin{array}{l}f\text{'}\left(c\right)=\frac{f\left(20\right)-f\left(0\right)}{20-0}\\ f\text{'}\left(c\right)=\frac{212-f\left(0\right)}{20-0}\\ f\text{'}\left(c\right)=\frac{212}{20}\\ f\text{'}\left(c\right)={10.6}^0\text{F/sec}\end{array}$

Thus,

There was at least one moment when the temperature was rising exactly ${10.6}^0\text{F/sec}$ .