Chapter 5.6, Problem 9E

a.

To determine

To find: the rates of change of the area.

Answer to Problem 9E

The value of $\frac{d\text{A}}{dt}$ is ${\text{14 cm}}^2\text{/sec}$ .

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

  $\frac{dl}{dt}=-2\text{ cm/sec}$ , $\frac{dw}{dt}=2\text{ cm/sec}$

l = 12 cm, w = 5 cm.

Calculation :

To calculate the rate of change of the area,

Therefore,

Area of a rectangle dimension

  $\text{A}=lw$

l = length, w = width

Differentiate with respect to t .

  $\frac{d\text{A}}{dt}=\text{L}\frac{dw}{dt}+\text{W}\frac{dl}{dt}\left[\text{Using product rule}\text{.}\right]$

Putting the value of $\frac{dl}{dt}=-2$ , $\frac{dw}{dt}=2$ , l = 12 and w = 5 in above equation,

  $\begin{array}{l}\frac{d\text{A}}{dt}=12\left(2\right)+5\left(-2\right)\\ \frac{d\text{A}}{dt}=24-10\\ \frac{d\text{A}}{dt}=14{\text{cm}}^2\text{/sec}\end{array}$

Hence, the value of $\frac{d\text{A}}{dt}$ is ${\text{14 cm}}^2\text{/sec}$ .

b.

To determine

To find: the rates of change of the perimeter.

Answer to Problem 9E

The value of $\frac{d\text{P}}{dt}$ is $\text{0 cm/sec}$ .

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

  $\frac{dl}{dt}=-2\text{ cm/sec}$ , $\frac{dw}{dt}=2\text{ cm/sec}$

l = 12 cm, w = 5 cm.

Calculation :

To calculate the rate of change of the perimeter,

Therefore,

Perimeter of a rectangle dimension

  $\text{P}=2\left(l+w\right)$

l = length, w = width

Differentiate with respect to t .

  $\frac{d\text{P}}{dt}=\text{2}\left(\frac{dl}{dt}+\frac{dw}{dt}\right)$

Putting the value of $\frac{dl}{dt}=-2$ and $\frac{dw}{dt}=2$ in above equation,

  $\begin{array}{l}\frac{d\text{P}}{dt}=2\left(-2+2\right)\\ \frac{d\text{P}}{dt}=2\left(0\right)=0\text{cm}\text{./sec}\end{array}$

Hence, the value of $\frac{d\text{P}}{dt}$ is $\text{0 cm/sec}$ .

c.

To determine

To find: the rates of change of the length of a diagonal of a rectangle.

Answer to Problem 9E

The value of $\frac{ds}{dt}$ is $\frac{-14}{13}\text{ cm/sec}$ .

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

  $\frac{dl}{dt}=-2\text{ cm/sec}$ , $\frac{dw}{dt}=2\text{ cm/sec}$

l = 12 cm, w = 5 cm.

Calculation :

To calculate the rates of change of the length of a diagonal of a rectangle,

Therefore,

Length of a diagonal of the rectangle dimension

  $\text{S}=\sqrt{{\text{L}}^2{\text{+W}}^2}={\left({\text{L}}^2{\text{+W}}^2\right)}^{\frac{1}{2}}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{ds}{dt}=\frac{1}{2}{\left[{\text{L}}^2+{\text{W}}^2\right]}^{\frac{1}{2}-1}\cdot \left(2\text{L}\frac{dl}{dt}+2\text{W}\frac{dw}{dt}\right)\\ \frac{ds}{dt}=\frac{1}{2}{\left[{\text{L}}^2+{\text{W}}^2\right]}^{-\frac{1}{2}}\cdot \left(2\text{L}\frac{dl}{dt}+2\text{W}\frac{dw}{dt}\right)\\ \frac{ds}{dt}=\frac{1}{2\sqrt{{\text{L}}^2+{\text{W}}^2}}\cdot 2\left(\text{L}\frac{dl}{dt}+\text{W}\frac{dw}{dt}\right)\\ \frac{ds}{dt}=\frac{\text{L}\frac{dl}{dt}+\text{W}\frac{dw}{dt}}{\sqrt{{\text{L}}^2+{\text{W}}^2}}\end{array}$

Putting the value of $\frac{dl}{dt}=-2$ , $\frac{dw}{dt}=2$ , l = 12 and w = 5 in above equation,

  $\begin{array}{l}\frac{ds}{dt}=\frac{12\left(-2\right)+5\left(2\right)}{\sqrt{{\left(12\right)}^2+{\left(5\right)}^2}}\\ \frac{ds}{dt}=\frac{-24+10}{\sqrt{144+25}}\\ \frac{ds}{dt}=\frac{-14}{\sqrt{169}}\\ \frac{ds}{dt}=\frac{-14}{13}\text{cm}\text{./sec}\end{array}$

Hence, the value of $\frac{ds}{dt}$ is $\frac{-14}{13}\text{ cm/sec}$ .

d.

To determine

To find: the quantities that are increasing and that are decreasing.

Answer to Problem 9E

The area is increasing because $\frac{d\text{A}}{dt}\ge 0$ .

The perimeter is not changing because $\frac{d\text{P}}{dt}=0$ .

The length of the diagonal is decreasing because $\frac{ds}{dt}\le 0$ .

Explanation of Solution

Given information:

  $\frac{dl}{dt}=-2\text{ cm/sec}$ , $\frac{dw}{dt}=2\text{ cm/sec}$

l = 12 cm, w = 5 cm.

All variables are differentiable function of $t$ .

Calculation :

To explain which of these quantities are increasing and which are decreasing.

Therefore,

We have calculated,

  $\frac{d\text{A}}{dt}$ is ${\text{14 cm}}^2\text{/sec}$ in the above question.

Since $\frac{d\text{A}}{dt}\ge 0$ , so it is increasing.

Again, we have calculated

  $\frac{d\text{P}}{dt}$ is $\text{0 cm/sec}$ in the above question.

Since $\frac{d\text{P}}{dt}=0$ , so it is not changing.

Again, we have calculated

  $\frac{ds}{dt}$ is $\frac{-14}{13}\text{ cm/sec}$ in the above question.

Since $\frac{ds}{dt}\le 0$ , so it is decreasing.