Chapter 4.2, Problem 12E

To determine

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

Answer to Problem 12E

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

Explanation of Solution

Given information:

  $20\sec$ for the temperature to rise from $0^{\circ }\text{F}$ to ${212}^{\circ }\text{F }$ .

A thermometer taken from a freezer and placed in boiling water.

Explain:

According to the Mean Value Theorem, there must be at least one $c$ in $(a,b)$ such that

  $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

Given a function $f(x)$ that is continuous on $(a,b)$ and differentiable on $(a,b)$ .

Define $f(t)$ as the temperature in degrees Fahrenheit at time $t$ seconds. Assume that $f(t)$ is continuous on $(0,20)$ and differentiable on $(0,20)$ .

Hence,

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

By the MVT there must be at least one $c$ such that $f^{\prime }(c)={10.6}^{\circ }\text{F}/\text{sec}$

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