Calculus: Related rates, how do you find the rate at which the area of the circle is increasing?

Radius, r = 2 in.

dr/dt = 1 in./min.

Area = A

Find dA/dt:

A = π r²

Differentiating Implicitly Over Time,

dA/dt = 2π r (dr/dt)

dA/dt = 2π (2) (1)

dA/dt = 4π

dA/dt = 12.57

The area is increasing at a rate of 12.57 sq. in./min.

V= (4/3)pir^3. take by-product v'=4pir^2(dr/dt) V'=4pi(9)^2(-0.2) V'=-sixty 4.8pi cm/min^3. if the diameter is reducing at a definite cost meaning the radius is reducing at a definite cost. I used 9cm because of the fact the diameter is eighteen and used -0.2 because of the fact the diameter cost is reducing if the diameters cost replaced into increasing then the fee would be 0.2.

A = pi * r^2

dA/dt = 2 * pi * r * dr/dt

We know dr/dt = 1

We need dA/dt when r = 2

So plug it in.

dA/dt = 2 * pi * 2 * 1 = 4pi