Question

the volume of a sphere is increasing at a rate of 2cm^3/sec. Find the rate of change of its surface area when its volume is 256pi/3 cm^3.

Answer 1

The rate of change of the surface area of the sphere is 0.99 cm²/s.

Explanation:

The surface area (A) of a sphere is given by:

`A = 4\pi r^(2)`

If we derivate the above equation with respect to the time we have:

`(dA)/(dt) = 4\pi (2r)(dr)/(dt)`

`(dA)/(dt) = 8\pi r(dr)/(dt)` (1)

Where:

r: is the radius

We need to find
`(dr)/(dt)` and r.

From the volume we can find the radius:

`V = (4)/(3)\pi r^(3) = 256 (\pi)/(3) cm^(3)`

`r = \sqrt[3]{(3V)/(4\pi)} = \sqrt[3]{(3*256*(\pi)/(3))/(4\pi)} = 4 cm`

And by derivating the volume of the sphere with respect to the time we can calculate
`(dr)/(dt)`:

`(dV)/(dt) = (4)/(3)\pi(3r^(2))(dr)/(dt)`

`(dV)/(dt) = 4\pi r^(2)(dr)/(dt)`

`(dr)/(dt) = ((dV)/(dt))/(4\pi r^(2)) = (2)/(4\pi(4)^(2)) = 0.0099 cm/s`

Now, we can calculate the rate of change of the surface area (equation 1):

`(dA)/(dt) = 8\pi r(dr)/(dt) = 8\pi*4*0.0099 = 0.99 cm^(2)/s`

Therefore, the rate of change of the surface area of the sphere is 0.99 cm²/s.

I hope it helps you!