Related rates with snowball The sun is shining and a spherical snowball of volume 340 ft$^3$ is melting at a rate of $10$ cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after $2.5$ hours? 
And this answer is wrong? please help
$10\times 2.5 =25$
$340 - 25 = 315$
so
$r^3 = \dfrac{315\times 3 }{ 4\pi}$
$r= 4.2$ ft
 A: You need to start by relating $\frac{dV}{dt}$ to $\frac{dr}{dt}$. As you know, the equation for spherical volume is given by
$$V=\frac{4}{3}\pi r^3.$$
If we treat $V$ and $r$ as both being implicitly differentiable functions of $t$, then differentiating implicitly across $V$ gives,
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$$
Solving this for $\frac{dr}{dt}$ we have,
$$\frac{dr}{dt}=\frac{dV}{dt}\frac{1}{4\pi r^2}.$$
Now you were already given that $\frac{dV}{dt}=10$. All that is left is to find $r$ after $2.5$ hours. You have rightly concluded that in $2.5$ hours, $V=340-25=315$. To find $r$, try using your original formula for $V$ with this new volume for $V$. Solve that for $r$. Plug this new found value for $r$ into your equation for $\frac{dr}{dt}$, make sure the result makes sense, and have a nice day.
A: It looks like you have calculated the radius of a 315 ft^3 sphere correctly(though the precision is light).  You have not addressed how quickly it is decreasing, which is the question you have.  You need to take the derivative of the volume formula with respect to time to relate the decrease in volume to the decrease in radius
