Question

The area of sphere increase at the rate of 0.5 CM square per second the rate at which its perimeter is increased and side of square is 10 cm

Answer 1

$\displaystyle \frac{\pi}{40}\text{ cm/s}$

Answer 2

1. Differentiate $A=\pi r^2$ to get $\frac{dr}{dt}=\frac{0.5}{2\pi r}$.

2. Differentiate $C=2\pi r$ to get $\frac{dC}{dt}=\frac{0.5}{r}$.

3. At $2\pi r=40$ (since square’s perimeter = 40), find $r=\frac{20}{\pi}$.

4. Substitute $r$ to get $\frac{dC}{dt}=\frac{0.5}{20/\pi}=\frac{\pi}{40}$ cm/s.