To find the rate of increase of the surface area of the sphere when its radius is 2 cm, we can use related rates in calculus.
First, let’s recall some formulas:
- The volume of a sphere, V, is given by the formula: V = (4/3)πr³, where r is the radius.
- The surface area of a sphere, S, is given by the formula: S = 4πr².
We know that the volume is increasing at a rate of dV/dt = 3 cm³/s. We want to find dS/dt when r = 2 cm.
Using the formula for volume, we find the derivative with respect to time:
dV/dt = 4πr²(dr/dt)
We can rearrange this to find dr/dt:
dr/dt = (dV/dt) / (4πr²)
Substituting the known values:
dr/dt = 3 / (4π(2)²) = 3 / (16π)
Now we can move on to the rate of change of the surface area:
Taking the derivative of the surface area with respect to time gives us:
dS/dt = 8πr(dr/dt)
Substituting for dr/dt and r:
dS/dt = 8π(2)(3 / (16π))
This simplifies to:
dS/dt = 8 * 2 * 3 / 16 = 3
So, the rate of increase of the surface area when the radius is 2 cm is 3 cm²/s.