Question

A spherical snowball is melting. its radius decreases at a constant rate of 3 cm per minute from an initial value of 100 cm. how fast is the volume decreasing 25 minutes later? give your answer in cubic centimeters/minute.

Answer 1

The function that would tell us how ball's radius is changing over time is:
$r(t)=100-3t$
We know that volume of a sphere is:
$V=\frac{4}{3}\pi r^3\\ V(t)=\frac{4}{3}\pi (100-3t)^3$
Now to find the rate of change we have to find the derivative with respect to time:
$\frac{dV(t)}{dt}=\frac{4}{3}\pi ((100-3t)^3)'\\ \frac{dV(t)}{dt}=\frac{4}{3}\pi\cdot 3(100-3t)^2\cdot(100-3t)'\\ \frac{dV(t)}{dt}=\frac{4}{3}\pi\cdot 3(100-3t)^2\cdot(-3)\\ \frac{dV(t)}{dt}=-12\pi\cdot (100-3t)^2$
Now we simply plug in t=25 to get our answer:
$\frac{dV(t)}{dt}=-12\pi\cdot (100-3\cdot 25)^2\\ \frac{dV(t)}{dt}=-23561.94\ \frac{cm^3}{min}$

Answer 2

The rate of decrement in volume after 25 minutes is $-23561.94 \;\rm cm^{3}/min$.

Given data:

The rate of decrement in radius is, $\dfrac{dr}{dt} = 3 \;\rm cm/min$.

The initial value of radius is, r = 100 cm.

The time interval is, t = 25 minutes.

The volume of snowball is,

$V = \dfrac{4}{3} \pi (100-3r)^{3}$

Differentiate the volume as,

$\dfrac{dV}{dt} = \dfrac{4}{3} \pi (3)(-3)(100-3t)^{2}\\\dfrac{dV}{dt} = \dfrac{4}{3} \pi (3)(-3)(100-3(25))^{2}\\\\\dfrac{dV}{dt} = -23561.94 \;\rm cm^{3}/min$

Thus, the rate of decrement in volume after 25 minutes is $-23561.94 \;\rm cm^{3}/min$.

Learn more about the volume of spherical entity here:

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