Question

the volume of a cantaloupe is approximated by ( v=\frac{4}{3}pi r^{3} ). the radius is growing at the rate of ( 0.5 mathrm{~cm}/mathrm{week} ), at a time when the radius is ( 6.3 mathrm{~cm} ). how fast is the volume changing at that moment?
the volume is changing at a rate of about
(round to one decimal place as needed.)

Explanation:

Step1: Differentiate the volume formula

The volume formula of a sphere is \(V=\frac{4}{3}\pi r^{3}\). Differentiate both sides with respect to time \(t\) using the chain - rule.
By the power rule \(\frac{dV}{dt}=\frac{4}{3}\pi\times3r^{2}\frac{dr}{dt}\), which simplifies to \(\frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}\).

Step2: Substitute the given values

We are given that \(\frac{dr}{dt}=0.5\space cm/week\) and \(r = 6.3\space cm\).
Substitute these values into the formula \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
So \(\frac{dV}{dt}=4\pi\times(6.3)^{2}\times0.5\).
First, calculate \((6.3)^{2}=39.69\).
Then \(4\pi\times39.69\times0.5=2\pi\times39.69\).
\(2\pi\times39.69 = 79.38\pi\approx79.38\times3.14 = 249.2532\approx249.3\) (rounded to one decimal place).

Answer:

\(249.3\)