The volume of a cantaloupe is approximated by $V=\frac{4}{3} \pi r^{3}$. The radius is growing at the rate of $0.7 \mathrm{~cm} /$ week, at a time when the radius is $6.6 \mathrm{~cm}$. 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.)
Answer
Rounding to one decimal place, we find that the volume of the cantaloupe is changing at a rate of approximately \(\boxed{383.2}\) cubic centimeters per week.
Step 1 ：We are given a problem of related rates in calculus. The rate of change of the radius, \(\frac{dr}{dt}\), is given and we are asked to find the rate of change of the volume, \(\frac{dV}{dt}\), when the radius is 6.6 cm.
Step 2 ：We know that the volume of a sphere is given by the formula \(V=\frac{4}{3} \pi r^{3}\).
Step 3 ：We can differentiate both sides of this equation with respect to time \(t\) to get \(\frac{dV}{dt} = 4\pi r^{2} \frac{dr}{dt}\).
Step 4 ：Substituting the given values into this equation, we have \(r = 6.6\) and \(\frac{dr}{dt} = 0.7\).
Step 5 ：Calculating \(\frac{dV}{dt}\) gives us approximately 383.1737727730398.
Step 6 ：Rounding to one decimal place, we find that the volume of the cantaloupe is changing at a rate of approximately \(\boxed{383.2}\) cubic centimeters per week.
