Chapter 4.6, Problem 24E

(a)

To determine

To find: The value of $dy/dt$ at the point where $x=-3$ .

Answer to Problem 24E

The answer:

  $\frac{dy}{dt}=-46\text{cm}/\text{sec}$

Explanation of Solution

Given information:

  $\begin{array}{c}y=f(x)\\ =\frac{10}{1+x^2}\end{array}$

And,

  $dx/dt=3\text{cm}/\text{sec}$

Calculation:

  $\begin{array}{c}y=x^3-4x\\ \Rightarrow \frac{dy}{dt}=\left(3x^2-4\right)\frac{dx}{dt}\end{array}$

Put the given values

  $\begin{array}{c}x=-3,\\ \frac{dx}{dt}=-2\end{array}$

Then simplify:

  $\begin{array}{c}\frac{dy}{dt}=\left(3{(-3)}^2-4\right)(-2)\\ =-46\end{array}$

Therefore, the required value of $dy/dt$ at the point where $x=-3$ is $\frac{dy}{dt}=-46\text{cm}/\text{sec}$ .

(b)

To determine

To find: The value of $dy/dt$ at the point where $x=1$ .

Answer to Problem 24E

The answer:

  $2m/\sec$

Explanation of Solution

Given information:

  $\begin{array}{c}y=f(x)\\ =\frac{10}{1+x^2}\end{array}$

And,

  $dx/dt=3\text{cm}/\text{sec}$

Calculation:

  $\begin{array}{c}y=x^3-4x\\ \frac{dx}{dt}=-2\\ {\left.\frac{dy}{dt}\right|}_{x=1}\\ \frac{dy}{dt}=\left(3x^2-4\right)\frac{dx}{dt}\end{array}$

Put,

  $\begin{array}{c}x=1,\\ \frac{dx}{dt}=-2\end{array}$

Then simplify:

  $\begin{array}{c}\frac{dy}{dt}=(3\cdot 1-4)(-2)\\ =2\\ {\left.\frac{dy}{dt}\right|}_{x=1}=2\text{cm}/\text{sec}\end{array}$

Therefore, the required value of $dy/dt$ at the point where $x=1$ is $2m/\sec$ .

(c)

To determine

To find: The value of $dy/dt$ at the point where $x=4$ .

Answer to Problem 24E

The answer:

  $-88\text{cm}/\text{sec}$

Explanation of Solution

Given information:

  $\begin{array}{c}y=f(x)\\ =\frac{10}{1+x^2}\end{array}$

And,

  $dx/dt=3\text{cm}/\text{sec}$

Calculation:

  $\begin{array}{c}y=x^3-4x,\\ \frac{dx}{dt}=-2\\ {\left.\frac{dy}{dt}\right|}_{x=4}\\ \frac{dy}{dt}=\left(3x^2-4\right)\frac{dx}{dt}\end{array}$

Put, the given values,

  $\begin{array}{c}x=4,\\ \frac{dx}{dt}=-2\end{array}$

Then simplify:

  $\begin{array}{c}\frac{dy}{dt}=(3\cdot 16-4)(-2)\\ =-88\\ {\left.\frac{dy}{dt}\right|}_{x=4}=-88\text{cm}/\text{sec}\end{array}$

Therefore, the required value of $dy/dt$ at the point where $x=4$ is $-88\text{cm}/\text{sec}$ .