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
Best Answer
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.