Chapter 5.6, Problem 43E

a.

To determine

To find: the value of $\frac{dc}{dt}$ , $\frac{dr}{dt}$ and $\frac{dp}{dt}$ .

Answer to Problem 43E

The value of $\frac{dc}{dt}$ is $0.3$ , $\frac{dr}{dt}$ is $0.9$ and $\frac{dp}{dt}$ is $0.6$ .

Explanation of Solution

Given information:

  $r\left(x\right)=9\left(x\right)$ .

  $c\left(x\right)=x^3-6x^2+15x$

  $\frac{dx}{dt}=0.1$

  $x=2$

All variables are differentiable functions of t .

Calculation :

We have to calculate the value of $\frac{dc}{dt}$ , $\frac{dr}{dt}$ and $\frac{dp}{dt}$ .

  $c\left(x\right)=x^3-6x^2+15x$

Differentiate with respect to t .

  $\begin{array}{l}\frac{dc}{dt}=3x^2\frac{dx}{dt}-6.2x\frac{dx}{dt}+15x\frac{dx}{dt}\\ \frac{dc}{dt}=\left(3x^2-12x+15\right)\frac{dx}{dt}\end{array}$

Putting the value of $x=2$ and $\frac{dx}{dt}=0.1$ in above equation,

  $\begin{array}{l}\frac{dc}{dt}=\left[3{\left(2\right)}^2-12\left(2\right)+15\right]0.1\\ \frac{dc}{dt}=\left[3\left(4\right)-24+15\right]0.1\\ \frac{dc}{dt}=\left[12-24+15\right]0.1\\ \frac{dc}{dt}=\left[27-24\right]0.1\\ \frac{dc}{dt}=3\times 0.1\\ \frac{dc}{dt}=0.3\end{array}$

Again,

  $r\left(x\right)=9x$

Differentiate with respect to t .

  $\frac{dr}{dt}=9\frac{dx}{dt}$

Putting the value of $\frac{dx}{dt}=0.1$ in above equation,

  $\begin{array}{l}\frac{dr}{dt}=9\left(0.1\right)\\ \frac{dr}{dt}=0.9\end{array}$

Therefore,

  $\begin{array}{l}\frac{dp}{dt}=\frac{dr}{dt}-\frac{dc}{dt}\\ \frac{dp}{dt}=0.9-0.3\\ \frac{dp}{dt}=0.6\end{array}$

Hence, the value of $\frac{dc}{dt}$ is $0.3$ , $\frac{dr}{dt}$ is $0.9$ and $\frac{dp}{dt}$ is $0.6$ .

b.

To determine

To find: the value of $\frac{dc}{dt}$ , $\frac{dr}{dt}$ and $\frac{dp}{dt}$ .

Answer to Problem 43E

The value of $\frac{dc}{dt}$ is $-1.5625$ , $\frac{dr}{dt}$ is $3.5$ and $\frac{dp}{dt}$ is $5.0625$ .

Explanation of Solution

Given information:

  $r\left(x\right)=70\left(x\right)$ .

  $c\left(x\right)=x^3-6x^2+45x$

  $\frac{dx}{dt}=0.5$

  $x=1.5$

All variables are differentiable functions of t .

Calculation :

We have to calculate the value of $\frac{dc}{dt}$ , $\frac{dr}{dt}$ and $\frac{dp}{dt}$ .

  $\begin{array}{l}c\left(x\right)=x^3-6x^2+\frac{45}{x}\\ c\left(x\right)=x^3-6x^2+{45}^{-1}\end{array}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{dc}{dt}=3x^2\frac{dx}{dt}-2.6x\frac{dx}{dt}+45\cdot \left(-1\right)x^{-1-1}\frac{dx}{dt}\\ \frac{dc}{dt}=\left[3x^2-12x-45x^2\right]\frac{dx}{dt}\\ \frac{dc}{dt}=\left[3x^2-12x-\frac{45}{x^2}\right]\frac{dx}{dt}\end{array}$

Putting the value of $x=1.5$ and $\frac{dx}{dt}=0.05$ in above equation,

  $\begin{array}{l}\frac{dc}{dt}=\left[3{\left(1.5\right)}^2-12\left(1.5\right)-\frac{45}{{\left(1.5\right)}^2}\right]0.05\\ \frac{dc}{dt}=\left[3{\left(2.25\right)}^2-18-\frac{45}{2.25}\right]0.05\\ \frac{dc}{dt}=\left[6.75-18-20\right]0.05\\ \frac{dc}{dt}=\left[6.75-38\right]0.05\\ \frac{dc}{dt}=\left(-31.25\right)0.05\\ \frac{dc}{dt}=-1.5625\end{array}$

Again,

  $r\left(x\right)=70x$

Differentiate with respect to t .

  $\frac{dr}{dt}=70\frac{dx}{dt}$

Putting the value of $\frac{dx}{dt}=0.05$ in above equation,

  $\begin{array}{l}\frac{dr}{dt}=70\left(0.05\right)\\ \frac{dr}{dt}=3.5\end{array}$

Therefore,

  $\begin{array}{l}\frac{dp}{dt}=\frac{dr}{dt}-\frac{dc}{dt}\\ \frac{dp}{dt}=3.5-\left(-1.5625\right)\\ \frac{dp}{dt}=3.5+1.5625\\ \frac{dp}{dt}=5.0625\end{array}$

Hence, the value of $\frac{dc}{dt}$ is $-1.5625$ , $\frac{dr}{dt}$ is $3.5$ and $\frac{dp}{dt}$ is $5.0625$ .