Chapter 5.6, Problem 24E

a.

To determine

To find: the value of $\frac{dy}{dt}$ at the point where $x=-3$ .

Answer to Problem 24E

The value of $\frac{dy}{dt}$ is $-46\text{ cm}\text{./sec}$ .

Explanation of Solution

Given information:

  $y=f\left(x\right)=x^3-4x$ .

  $\frac{dx}{dt}=-2\text{ cm}\text{./sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the value of $\frac{dy}{dt}$ at the point where $x=-3$ .

Therefore,

  $y=x^3-4x$ .

Differentiate with respect to t .

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

Putting the value of $x=-3$ and $\frac{dx}{dt}=-2$ ,

  $\begin{array}{l}\frac{dy}{dt}=\left[3{\left(-3\right)}^2-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left[3\left(9\right)-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left[27-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left(23\right)\left(-2\right)\\ \frac{dy}{dt}=-46\text{ cm}\text{./sec}\end{array}$

Hence, the value of $\frac{dy}{dt}$ is $-46\text{ cm}\text{./sec}$ .

b.

To determine

To find: the value of $\frac{dy}{dt}$ at the point where $x=1$ .

Answer to Problem 24E

The value of $\frac{dy}{dt}$ is $\text{2 cm}\text{./sec}$ .

Explanation of Solution

Given information:

  $y=f\left(x\right)=x^3-4x$ .

  $\frac{dx}{dt}=-2\text{ cm}\text{./sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the value of $\frac{dy}{dt}$ at the point where $x=-3$ .

Therefore,

  $y=x^3-4x$ .

Differentiate with respect to t .

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

Putting the value of $x=1$ and $\frac{dx}{dt}=-2$ ,

  $\begin{array}{l}\frac{dy}{dt}=\left[3{\left(1\right)}^2-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left[3\left(1\right)-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left(3-4\right)\left(-2\right)\\ \frac{dy}{dt}=\left(-1\right)\left(-2\right)\\ \frac{dy}{dt}=2\text{ cm}\text{./sec}\end{array}$

Hence, the value of $\frac{dy}{dt}$ is $\text{2 cm}\text{./sec}$ .

c.

To determine

To find: the value of $\frac{dy}{dt}$ at the point where $x=4$ .

Answer to Problem 24E

The value of $\frac{dy}{dt}$ is $-88\text{ cm}\text{./sec}$ .

Explanation of Solution

Given information:

  $y=f\left(x\right)=x^3-4x$ .

  $\frac{dx}{dt}=-2\text{ cm}\text{./sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the value of $\frac{dy}{dt}$ at the point where $x=4$ .

Therefore,

  $y=x^3-4x$ .

Differentiate with respect to t .

  $\frac{dy}{dt}=\left(3x^2-4\right)\frac{dx}{dt}$

Putting the value of $x=4$ and $\frac{dx}{dt}=-2$ ,

  $\begin{array}{l}\frac{dy}{dt}=\left[3{\left(4\right)}^2-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left[3\left(16\right)-4\right]\left(-2\right)\\ \frac{dy}{dt}=\left(48-4\right)\left(-2\right)\\ \frac{dy}{dt}=\left(44\right)\left(-2\right)\\ \frac{dy}{dt}=-88\text{ cm}\text{./sec}\end{array}$

Hence, the value of $\frac{dy}{dt}$ is $-88\text{ cm}\text{./sec}$ .