# The length L of a rectangle is decreasing at the rate of 2 cm/sec while the width W is increasing...

## Question:

The length L of a rectangle is decreasing at the rate of 2 cm/sec while the width W is increasing at the rate of 2 cm/sec. When L = 12 and W = 2, find the rate of change of:

a) the area

b) the perimeter

c) the lengths of the diagonals

## Rate of Change:

The rate of change of area can be found by differentiating the area with respect to time that is we will keep the length as it is and times it by the change of width.

## Answer and Explanation:

To find the rate of change of area we will proceed as

{eq}l=12\\ w=2\\ \frac{\mathrm{d} l}{\mathrm{d} t}=-2\\ \frac{\mathrm{d} w}{\mathrm{d} t}=2 {/eq}

Now writing the formula for the area

{eq}\frac{\mathrm{d} A}{\mathrm{d} t}=lw\\ =l\frac{\mathrm{d} w}{\mathrm{d} t}+w\frac{\mathrm{d} l}{\mathrm{d} t}\\ =12(2)+2(-2)\\ 20 {/eq}

b) Now let us find the rate of change of perimeter

{eq}P=2(l+w)\\ \frac{\mathrm{d} P}{\mathrm{d} t}=2(\frac{\mathrm{d} l}{\mathrm{d} t}+\frac{\mathrm{d} w}{\mathrm{d} t})\\ =2(-2+2)\\ =0 {/eq}

c ) Now let us find the rate of change of diagonals

{eq}D=2\sqrt{l^{2}+w^{2}}\\ \frac{\mathrm{d} D}{\mathrm{d} t}=\frac{1}{2\sqrt{l^{2}+w^{2}}}\left ( 2l\frac{\mathrm{d} l}{\mathrm{d} t}+2w\frac{\mathrm{d} w}{\mathrm{d} t} \right )\\ =\frac{12(-2)+2(2)}{\sqrt{144+4}}\\ =\frac{-20}{\sqrt{148}} {/eq}

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from Glencoe Pre-Algebra: Online Textbook Help

Chapter 8 / Lesson 7