Question

The area spherical balloon of radius 6 cm increases at the rate of 2 then find the rate of increase in the volume.

Answer 1

We can use the formulas for the surface area and volume of a sphere to solve this problem.

The surface area of a sphere with radius r is given by: A = 4πr^2 And the volume of a sphere with radius r is given by: V = (4/3)πr^3

We are given that the surface area is increasing at the rate of 2 cm^2/sec.

That is, dA/dt = 2 cm^2/sec.

We want to find the rate of change of the volume when the radius is 6 cm. Using the formulas for A and V, we can find the relationship between the rate of change of surface area and the rate of change of volume: dA/dt = 8πr(dr/dt) dV/dt = 4πr^2(dr/dt)

Here, dr/dt is the rate of change of the radius, which we don't know. However, we know that the radius is constant with respect to time, so dr/dt = 0. Therefore, we have: dV/dt = 4πr^2(dr/dt) = 4πr^2(0) = 0

This means that the volume is not changing with respect to time when the radius is constant. However, we are given that the radius is increasing at the rate of 2 cm/sec. That is, dr/dt = 2 cm/sec. So, the rate of change of the radius is positive.

Therefore, the volume is increasing, but the rate of increase is zero when the radius is constant. In summary, when the radius of the spherical balloon is 6 cm and is increasing at the rate of 2 cm/sec, the rate of increase in the volume is 0 cubic cm/sec.