Chapter 3.8, Problem 33E

Patients undergo dialysis treatment to remove urea from their blood when their kidneys are not functioning properly. Blood is diverted from the patient through a machine that filters out urea. Under certain conditions, the duration of dialysis required, given that the initial urea concentration is c > I, is given by the equation

$t=\ln \left(\frac{3c+\sqrt{9c^2-8c}}{2}\right)$

Calculate the derivative of t with respect to c and interpret it.

To determine

To find: The derivative of t with respect to c with the equation $t=\ln \left(\frac{3c+\sqrt{9c^2-8c}}{2}\right)$ and write an interpretation statement based on the result.

Answer to Problem 33E

The derivative of t with respect to c is given as $\left[\frac{3}{\left[3c+\sqrt{9c^2-8c}\right]}+\frac{\left(9c-4\right)}{\left[3c\sqrt{\left(9c^2-8c\right)}+9c^2-8c\right]}\right]$.

Explanation of Solution

Given:

The given equation is as below.

$t=\ln \left(\frac{3c+\sqrt{9c^2-8c}}{2}\right)$ (1)

Calculation:

Use chain rule to differentiate the equation (1).

Assume the variable u as $\left(\frac{3c+\sqrt{9c^2-8c}}{2}\right)$, and it implies that variable t as $\ln u$.

Use the expression below.

$\begin{array}{c}\frac{dt}{dc}=\frac{dt}{du}\times \frac{du}{dc}\\ \frac{dt}{dc}=\frac{1}{u}\times \frac{du}{dc}\end{array}$

Substitute $\left(\frac{3c+\sqrt{9c^2-8c}}{2}\right)$ for $u$.

$\frac{dt}{dc}=\frac{1}{\frac{3c+\sqrt{9c^2-8c}}{2}}\times \frac{d}{dc}\left(\frac{3c+\sqrt{9c^2-8c}}{2}\right)$

$\frac{dt}{dc}=\frac{2}{\left[3c+\sqrt{9c^2-8c}\right]}\times \left[\frac{3}{2}+\frac{1}{2}\frac{d}{dc}\left(\sqrt{9c^2-8c}\right)\right]$ (1)

Refer equation (1), separate the term $\sqrt{9c^2-8c}$.

Indicate it as $f\left(c\right)=\sqrt{9c^2-8c}$.

Differentiate $f\left(c\right)=\sqrt{9c^2-8c}$ using the chain rule.

Assume the variable l as $9c^2-8c$, and it implies that variable $f\left(c\right)$ as $\sqrt{l}$.

Use the expression below.

$\begin{array}{c}\frac{df\left(c\right)}{dc}=\frac{df\left(c\right)}{dl}\times \frac{dl}{dc}\\ \frac{df\left(c\right)}{dc}=\frac{1}{2\sqrt{l}}\times \frac{dl}{dc}\end{array}$

Substitute $9c^2-8c$ for $l$.

$\begin{array}{c}\frac{df\left(c\right)}{dc}=\frac{1}{2\sqrt{\left(9c^2-8c\right)}}\times \frac{d}{dc}\left(9c^2-8c\right)\\ =\frac{1}{2\sqrt{\left(9c^2-8c\right)}}\times \left(18c-8\right)\\ =\frac{\left(18c-8\right)}{2\sqrt{\left(9c^2-8c\right)}}\\ =\frac{\left(9c-4\right)}{\sqrt{\left(9c^2-8c\right)}}\end{array}$

Substitute $\frac{\left(9c-4\right)}{\sqrt{\left(9c^2-8c\right)}}$ for $\frac{d}{dc}\left(\sqrt{9c^2-8c}\right)$ in equation (1).

$\begin{array}{c}\frac{dt}{dc}=\frac{2}{\left[3c+\sqrt{9c^2-8c}\right]}\times \left[\frac{3}{2}+\frac{1}{2}\frac{\left(9c-4\right)}{\sqrt{\left(9c^2-8c\right)}}\right]\\ =\left[\frac{3}{\left[3c+\sqrt{9c^2-8c}\right]}+\frac{1}{\left[3c+\sqrt{9c^2-8c}\right]}\frac{\left(9c-4\right)}{\sqrt{\left(9c^2-8c\right)}}\right]\\ =\left[\frac{3}{\left[3c+\sqrt{9c^2-8c}\right]}+\frac{\left(9c-4\right)}{\left[3c\sqrt{\left(9c^2-8c\right)}+9c^2-8c\right]}\right]\end{array}$

Write the interpretation based on the result as below.

The rate of reaction will be fast with higher initial concentration value. The time taken for removing the unit concentration of the material will be lesser as the initial concentration increases.

The derivative of t with respect to c is given as $\left[\frac{3}{\left[3c+\sqrt{9c^2-8c}\right]}+\frac{\left(9c-4\right)}{\left[3c\sqrt{\left(9c^2-8c\right)}+9c^2-8c\right]}\right]$.