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A rectangle is inscribed in a circle with a radius 4 inches. If the length of the rectangle is...

Question:

A rectangle is inscribed in a circle with a radius 4 inches. If the length of the rectangle is decreasing at the rate of 3 inches per second, how fast is the area changing at the instant when the length is 4 inches?

Area of Rectangle:


Area of rectangle is defined as the product of the length and width of rectangle.

{eq}A = x\times y\\ {/eq}

We will find the rate of change of area by differentiating the area(A) with respect to time(t).

Answer and Explanation:


To find the rate of change of area

Given

Radius (r) = 4 inch

The length of the rectangle is decreasing at the rate {eq}\frac{dx}{dt} = 3 \ inch/sec {/eq}

Length of rectangle (x) = 4 inch

Let x be the length and y be the width of rectangle.

Area of rectangle is

{eq}A = x\times y\\ {/eq}

Apply phythargeon theroem in {eq}\triangle ABC {/eq} using

{eq}x^2 + y^2 = (8)^2 \\ y ^2 = 64 - x^2 \\ y = \sqrt{64 - x^2} \\ {/eq}

Substituting value of y in area A

{eq}A = x\times \sqrt{64 - x^2} \\ \text{Differentiate with respect to t} \\ \frac{dA}{dt} = (\frac{x(-2x)}{2\sqrt{64-x^2}} + \sqrt{64 - x^2})\frac{dx}{dt} \\ \frac{dA}{dt} = (\frac{-x^2}{\sqrt{64-x^2}} + \sqrt{64 - x^2})\frac{dx}{dt} \\ \text{Substituting value of x and } \ \frac{dx}{dt} \\ \frac{dA}{dt} = (\frac{-(4)^2}{\sqrt{64-(4)^2}} + \sqrt{64 - (4)^2})3 \\ \frac{dA}{dt} = (\frac{-16}{\sqrt{48}} + \sqrt{48})3 \\ \frac{dA}{dt} = (\frac{32}{\sqrt{48}})3 \\ \frac{dA}{dt} = \frac{96}{\sqrt{48}} \\ \frac{dA}{dt} = \frac{24}{\sqrt{3}} \\ \frac{dA}{dt} = 8\sqrt{3} \ inch^2/sec \\ {/eq}


Learn more about this topic:

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Measuring the Area of a Rectangle: Formula & Examples

from Geometry: High School

Chapter 8 / Lesson 7
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