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

The area outside the circle but inside the square is found by subtracting the area of the circle from the area of the square. The rate of change of this area can be calculated using the chain rule.

Step-by-step explanation:

The area outside the circle but inside the square is the difference between the area of the square and the area of the circle. The area of the square is found by multiplying the length of one side by itself, so when the sides are 20 meters, the area of the square is 20 * 20 = 400 square meters.

The area of the circle is found by multiplying the radius squared by π, so when the radius is 3 meters, the area of the circle is 3 * 3 * π = 9π square meters. However, we want to find the area outside the circle but inside the square, so we subtract the area of the circle from the area of the square: 400 - 9π square meters.

To find how the area is changing with respect to time, we need to use the chain rule. Let's call the area A and the time t. The rate of change of the area (∂A/∂t) is equal to the rate of change of the sides of the square (∂s/∂t) multiplied by the rate of change of the area (∂A/∂s).

Answer 2

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 )

`(dV)/(dt)=2x(dx)/(dt) -2\pi r(dr)/(dt)`

We have,

x = 20 meters, r = 3 meters,
`(dx)/(dt)=3\text{ m per sec}`
`(dr)/(dt)=4\text{ meters per sec}`

`\implies (dV)/(dt)=2(20)(3)-2\pi(3)(4)`

`=120-24\pi`

`=44.6017763138`

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