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.)

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 $\boxed{{\displaystyle 383.2}}$ cubic centimeters per week.