Chapter 4.6, Problem 9E

(a)

To determine

To find : The rates of changes of the area.

Answer to Problem 9E

The answer: $\frac{dA}{dt}=14{\text{cm}}^2/\text{sec}$

Explanation of Solution

Given information:

The length $l$ of rectangle is decreasing at the rate of $2cm/\sec$

The width $w$ of rectangle is increasing at the rate of $2cm/\sec$

  $\begin{array}{c}l=12cm\\ w=5cm\end{array}$

Calculation:

Indicate how a circle equation's area relates to $t$ implicitly. Then enter the value inputs.

  $\begin{array}{c}\frac{dL}{dt}=-2\\ \frac{dw}{dt}=2\\ L=12\\ w=5\end{array}$

So,

  $\begin{array}{c}A=Lw\\ \Rightarrow \frac{dA}{dt}=L\frac{dw}{dt}+w\frac{dL}{dt}\\ =12(2)+5(-2)\\ =14\end{array}$

Therefore the required rates of changes of the area is $\frac{dA}{dt}=14{\text{cm}}^2/\text{sec}$

(b)

To determine

To find: The rates of changes of the perimeter.

Answer to Problem 9E

The answer: $\frac{dP}{dt}=0\text{cm}/\text{s}$

Explanation of Solution

Given information:

The length $l$ of rectangle is decreasing at the rate of $2cm/\sec$

The width $w$ of rectangle is increasing at the rate of $2cm/\sec$

  $\begin{array}{c}l=12cm\\ w=5cm\end{array}$

Calculation:

Differentiate a perimeter $P$ equation's area with regard to $t$ implicitly. Then enter the value inputs.

  $\begin{array}{c}\frac{dL}{dt}=-2\\ \frac{dw}{dt}=2\\ L=12\\ w=5\end{array}$

So,

  $\begin{array}{c}P=2L+2w\\ =2(L+w)\\ \Rightarrow \frac{dP}{dt}=2\left(\frac{dL}{dt}+\frac{dw}{dt}\right)\\ =2(-2+2)\\ =0\text{cm}/\text{sec}\end{array}$

Therefore the required rates of changes of the perimeter is $\frac{dP}{dt}=0\text{cm}/\text{s}$

(c)

To determine

To find: The length of the diagonal of the rectangle.

Answer to Problem 9E

The answer: $\frac{ds}{dt}=-\frac{14}{13}\text{cm}/\text{sec}$

Explanation of Solution

Given information:

The length $l$ of rectangle is decreasing at the rate of $2cm/\sec$

The width $w$ of rectangle is increasing at the rate of $2cm/\sec$

  $\begin{array}{c}l=12cm\\ w=5cm\end{array}$

Calculation:

Implicitly distinguish between the diagonal $s$ equation's length and the time constant $t$ . Then enter the value inputs.

  $\begin{array}{c}\frac{dL}{dt}=-2\\ \frac{dw}{dt}=2\\ L=12\\ w=5\end{array}$

So,

  $\begin{array}{c}s=\sqrt{L^2+w^2}\\ \Rightarrow \frac{ds}{dt}=\frac{1}{2}{\left(L^2+w^2\right)}^{-1/2}\left(2L\frac{dL}{dt}+2w\frac{dw}{dt}\right)\\ ={\left(L^2+w^2\right)}^{-1/2}\left(L\frac{dL}{dt}+w\frac{dw}{dt}\right)\\ ={\left({(12)}^2+{(5)}^2\right)}^{-1/2}((12)(-2)+(5)(2))\\ =\frac{-14}{\sqrt{169}}\\ =\frac{-14}{13}\end{array}$

The required length of the diagonal of the rectangle is $\frac{ds}{dt}=-\frac{14}{13}\text{cm}/\text{sec}$ .

(d)

To determine

To find: The qualities that are decreasing and increasing.

Answer to Problem 9E

The answer:

The area is increasing at the moment because $\frac{dA}{dt}>0$

The length of the diagonal is decreasing at the moment because $\frac{ds}{dt}<0$

Explanation of Solution

Given information:

The length $l$ of rectangle is decreasing at the rate of $2cm/\sec$

The width $w$ of rectangle is increasing at the rate of $2cm/\sec$

  $\begin{array}{c}l=12cm\\ w=5cm\end{array}$

Calculation:

The area is increasing at the moment because $\frac{dA}{dt}>0$

Because the rate at which the width is growing is equal to the rate at which the length is shrinking, the perimeter is not changing.

The length of the diagonal is decreasing at the moment because $\frac{ds}{dt}<0$