Chapter 4.4, Problem 26E

To determine

To Graph:The several members of the family.

Answer to Problem 26E

  

Explanation of Solution

Given: The family of functions $f(t)=C(e^{-at}-e^{-bt})$ where $a,b,c$ are the positive numbers $b>a$ has been used to model the concentration of a drug injected into the blood-stream at a time $t=0$ .Graph the several members of the family. What do they have in common? For fixed values of $C,a$ discover graphically what happens as $b$ increases. Then use calculus to prove what you have discovered.

Given: $f(t)=C(e^{-at}-e^{-bt})$

The graph for the function as $b$ increases is

  

Differentiating

  $\frac{d}{dt}(C(e^{-at}-e^{-bt})=C(b{e}^{-bt}-a{e}^{-at})$

Thus

  $a{e}^{-a{t}_0}=b{e}^{-b{t}_0}$

Taking log both sides

  $\begin{array}{l}\ln a{e}^{-a{t}_0}=\ln b{e}^{-b{t}_0}\\ \ln \frac{b}{a}=(b-a)t_0\\ t_0=\frac{\ln \frac{b}{a}}{b-a}\end{array}$

Differentiating with respect to $b$

  $\begin{array}{l}\frac{d{t}_0}{db}=\frac{\frac{b-a}{b}-\ln \frac{b}{a}}{{(b-a)}^2}\\ -\frac{1}{ab}<0\\ \frac{d{t}_0}{db}<0\end{array}$

Here the function reaches at the maximum point.

Now,

  $\begin{array}{l}f(t_0)=C\left({\left(\frac{b}{a}\right)}^{-\frac{a}{b-a}}-{\left(\frac{b}{a}\right)}^{-\frac{b}{b-a}}\right)\\ then\\ \frac{df(t_0)}{db}=\frac{{\left(\frac{b}{a}\right)}^{-\frac{b}{b-a}}\ln \left(\frac{b}{a}\right)}{b-a}\end{array}$

That means it is always greater than zero as $b>a$ so the value of the function as its maximum increases as $b$ increases.