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• Question 10 ) Textbook Videos A circle is inside a square. The radius of the circle is decreasing at a rate of 4 meters per day and the sides of the square are increasing at a rate of 4 meters per day. When the radius is 2 meters, and the sides are 16 meters, then how fast is the AREA outside the circle but inside the square changing? The rate of change of the area enclosed between the circle and the square is square meters per day. Submit Question Jump to Answer MacBook Air

Answer 1

let r be the radius of the circle at any time t.

Let x be the side length of the square at any time t.

The area b/w square and circle is given by,

A= 2 - 4

diff wrt t,

$\frac{\mathrm{d} A}{\mathrm{d} t} = 2x\frac{\mathrm{d} x}{\mathrm{d} t}-2\pi r \frac{\mathrm{d} r}{\mathrm{d} t}$

plugin the values,

$\frac{\mathrm{d} A}{\mathrm{d} t} = 2(16)(4)-2\pi (2) (-4)$

$\frac{\mathrm{d} A}{\mathrm{d} t} = 128+8\pi$

I hope this answer helps,
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