Question

A spherical snowball is melting in such a way that its radius is decreasing at rate of 0.1 cm/min. At what rate is the volume of the snowball decreasing when the radius is 14 cm

Answer 1

Answer:

$\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}$

Explanation:

Knowing that the volume of a sphere is V=(4/3)πr³ and $\frac{dr}{dt}=-0.1\frac{cm}{min}$

We must find $\frac{dV}{dt}=?$ when r=14cm

V=(4/3)πr³ ⇒

$\frac{dV}{dt}=\frac{4}{3}\pi3r^{2}\frac{dr}{dt}\\\frac{dV}{dt}=4\pi r^{2}(-0.1\frac{cm}{min})$

and r=14cm then

$\frac{dV}{dt}=4\pi(14cm)^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=4\pi196cm^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}$

Note: as you can see the relationship of change of r with respect to t is negative because it is a decrease, and also its volume ratio