Question

A circle is inside a square. The radius of the circle is increasing at a rate of 4 meters per day and the sides of the square are increasing at a rate of 3 meters per day. When the radius is 3 meters, and the sides are 20 meters, then how fast is the AREA outside the circle but inside the square changing?

Answer 1

Answer:

The area is changing by the rate of 44.62 meters per sec.

Step-by-step explanation:

Let x be the side of the square and r be the radius of the circle,

Then, the area outside the circle but inside the square is,

V = Area of square - area of circle,

∵ Area of a square = side² and area of a circle = $\pi$ (radius)²,

Thus,

$V=x^2-\pi(r)^2$

Differentiating with respect to t ( time )

$\frac{dV}{dt}=2x\frac{dx}{dt} -2\pi r\frac{dr}{dt}$

We have,

x = 20 meters, r = 3 meters, $\frac{dx}{dt}=3\text{ m per sec}$ $\frac{dr}{dt}=4\text{ meters per sec}$

$\implies \frac{dV}{dt}=2(20)(3)-2\pi(3)(4)$

$=120-24\pi$

$=44.6017763138$

$\approx 44.62\text{ meter per sec}$