Chapter 4.3, Problem 57E

To determine

To described:

The level of medication in the bloodstream after a drug is administered.

Answer to Problem 57E

Inflection occurs at $t\approx 28.57,85.71$ .

Explanation of Solution

Given:

The function

  $S\left(t\right)=A{t}^p{e}^{-kt}$

Concept used:

Using the quadratic formula

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $S\left(t\right)=A{t}^p{e}^{-kt}\mathrm{.......................}\left(1\right)$

Differentiating equation (1) with respect to $t$

  $\begin{array}{l}{S}^{\prime }\left(t\right)=Ap{t}^{p-1}{e}^{-kt}-kA{t}^p{e}^{-kt}\\ S^{\prime }\left(t\right)=A{t}^p{e}^{-kt}\left(\frac{p}{t}-k\right)\\ S^{\prime }\left(t\right)=S\left(t\right)\left(\frac{p}{t}-k\right)\mathrm{.................}\left(2\right)\end{array}$

Again Differentiating equation (2) with respect to $t$

  $S^{″}\left(t\right)=S^{\prime }\left(t\right)\left(\frac{p}{t}-k\right)+S\left(t\right)\left(-\frac{p}{t^2}-0\right)$

Replace the expression for $S^{\prime }\left(t\right)$

  $\begin{array}{l}{S}^{″}\left(t\right)=S\left(t\right)\left(\frac{p}{t}-k\right)\left(\frac{p}{t}-k\right)+S\left(t\right)\left(-\frac{p}{t^2}\right)\\ S^{″}\left(t\right)=S\left(t\right){\left(\frac{p}{t}-k\right)}^2+S\left(t\right)\left(-\frac{p}{t^2}\right)\\ S^{″}\left(t\right)=A{t}^p{e}^{-kt}\left[{\left(\frac{p}{t}-k\right)}^2-\frac{p}{t^2}\right]\\ S^{″}\left(t\right)=A{t}^p{e}^{-kt}\left[\left(\frac{p^2-p}{t^2}-\frac{2pk}{t}+k^2\right)\right]\end{array}$

  $\begin{array}{l}{S}^{″}\left(t\right)=0\\ A{t}^p{e}^{-kt}\left[\left(\frac{p^2-p}{t^2}-\frac{2pk}{t}+k^2\right)\right]=0\end{array}$

Using the quadratic formula

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \frac{1}{t}=\frac{-2pk\pm \sqrt{{\left(-2pk\right)}^2-4k^2\left(p^2-p\right)}}{2\left(p^2-p\right)}\\ \frac{1}{t}=\frac{2pk\pm 2k\sqrt{{\left(p\right)}^2-\left(p^2-p\right)}}{2\left(p^2-p\right)}\\ \frac{1}{t}=\frac{pk\pm k\sqrt{p}}{\left(p^2-p\right)}\end{array}$

Take reciprocal of both sides

  $\begin{array}{l}t=\frac{\left(p^2-p\right)}{pk\pm k\sqrt{p}}\\ t=\frac{\left(p^2-p\right)}{pk-k\sqrt{p}}\end{array}$

Substituting $k=0.07,p=4$

  $\begin{array}{l}t=\frac{\left(p^2-p\right)}{pk-k\sqrt{p}}\\ t=\frac{\left(4^2-4\right)}{4\left(0.07\right)+\left(0.07\right)\sqrt{4}},t=\frac{\left(4^2-4\right)}{4\left(0.07\right)-\left(0.07\right)\sqrt{4}}\\ t=28.57,85.71\end{array}$

Inflection occurs at $t\approx 28.57,85.71$