Question

A spherical snowball is melting in such a way that it maintains its shape. The snowball is decreasing in volume at a constant rate of 8 cubic centimeters per hour. At what rate, in centimeters per hour, is the radius of the snowball decreasing at the instant when the radius is 10 centimeters?

Answer 1

0.006366 cm/h

Explanation:

First we need to know that the volume of a sphere is calculated with the following formula:

V = (4/3)*pi*r^3

Where V is the volume and r is its radius.

Then, to find the rate of change, we need to derivate the equation with respect to the radius:

dV/dr = (4/3)*pi*(3r^2) = 4*pi*r^2

Then, we can write dV/dr as being (dV/dt)*(dt/dr), where dt is the change of time (variable t)

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

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

The rate of change of the volume is 8 cm3/h when the radius is 10, so using these values, we can find the rate of change of the radius (dr/dt):

8 = 4*pi*10^2*(dr/dt)

8 = 400*pi*(dr/dt)

dr/dt = 8/(400*pi)

dr/dt = 0.006366 cm/h

Answer 2

0.0628cm/hr

Explanation:

the snowball decreases in volume at a constant rate of 8 cubic cm per hour

Volume of a sphere = 4/3πr^3

Differentiate volume with respect to time(t)

dV/dt = 3(4/3πr^3)

= 4πr^2 . dr/dt

From the question, dV/dt = 8, r= 10

Substituting,

8 = 4π(10)^2 . dr/dt

8 = 400π . dr/dt

dr/dt = 8/400π

dr/dt =1/50π

dr/dt = 0.0628cm/hr