Chapter 1.4, Problem 37E

To determine

To Find: and interpret the sensitivity of the patient’s temperature to the dosage with $D=2mg$ .

Answer to Problem 37E

The sensitivity of the patient’s temperature to the dosage when $D=2mg$ is $-\frac{4}{9}$ degrees per mg.

Since the change is negative, the dosage $\Delta D$ will drop temperature by approximately $\frac{4}{9}\Delta D$ degree.

Explanation of Solution

Given information:

The function T as a function of the dosage D of a medicine is given by $T\left(D\right)=99+\frac{4}{\left(1+D\right)}$ .

Concept used:

Instantaneous rate of change of any function f at any point a is given by: $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

For sensitivity of the patient’s temperature to the dosage with $D=2mg$ , find instantaneous rate of change of $T\left(D\right)=99+\frac{4}{\left(1+D\right)}$ at $D=2mg$ .

Let $a=2$ .

  $\begin{array}{c}{T}^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{T\left(a+h\right)-T\left(a\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{T\left(2+h\right)-T\left(2\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{99+\frac{4}{\left(1+2+h\right)}-\left(99+\frac{4}{\left(1+2\right)}\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{99+\frac{4}{3+h}-99-\frac{4}{3}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{12-4\left(3+h\right)}{\left(3+h\right)3}}{h}\\ =\underset{h\to 0}{\lim }\frac{12-12-4h}{3h\left(3+h\right)}\\ =\underset{h\to 0}{\lim }\frac{-4h}{3h\left(3+h\right)}\\ =-4\underset{h\to 0}{\lim }\frac{1}{3\left(3+h\right)}\\ =-\frac{4}{9}\end{array}$

Hence the sensitivity of the patient’s temperature to the dosage when $D=2mg$ is $-\frac{4}{9}$ degrees per mg.

Since the change is negative, the dosage $\Delta D$ will drop temperature by approximately $\frac{4}{9}\Delta D$ degree.