Question

The radius of the surface of the water in the tank is increasing at a rate of 0.25m/min. at what rate is the area of the surface changing when the radius is 1.2m?

Answer 1

Answer:

1.88 m^2/min to the nearest hundredth.

Step-by-step explanation:

Area = π r^2

Using Calculus notation:

dA/dr = 2π r

dr/dt = 0.25  (given)

So the rate of change of area

dA/dt = dA/dr * dr/dt

So when r = 1.2

dA/dt = 2 π * 1.2 * 0.25

= 0.6π  m^2/min.

= 1.88 m^2/min to the nearest hundredth.