Related Rates with Melting Snowball Homework Help

Question

In my Calculus 1 class we are currently studying implicit differentiation and related rates. I have one last homework problem that I simply cannot get right. Our math homework is done on an online program which uses jsMath v3.4e, so I'm wondering if I had the correct answer at one point and it is not accepting my decimals to 8-10 decimal places. Either way, I have 3 tries left on this problem and I can't waste more tries.

Here is the question: "The sun is shining and a spherical snowball of volume 180 cubic feet is melting at a rate of 16 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 2 hours?"

My solution:

So from the question, we are given two things:

1. $$\frac {dV}{dt} = -16 \frac{ft^3}{hr}$$
2. $$V = 180 {ft^3}$$ the volume before it starts melting

So what I preceded to do was find the radius of this snowball.

1. $$V = \frac{4}{3}\pi r^3$$
2. $$180 = \frac{4}{3}\pi r^3$$
3. $$ r = \sqrt[3]{\frac {15}{\pi}} $$

Then I differentiated the Volume of a Sphere equation

$$\frac {dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$

I realized that after 2 hours, the $$\frac {dV}{dt} = -32\frac{ft^3}{hr} $$ because $$-16\frac{ft^3}{hr} * 2 hours = -32\frac{ft^3}{hr} $$

so then I plugged in everything into the differentiated formula

1. $$\frac {dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
2. $$-32 = 4\pi (\sqrt[3]{\frac {15}{\pi}})^2 $$
3. and I got $$\frac {dr}{dt} = -0.8980748271 \frac{ft}{hr}$$

I have tried looking this problem up several times and trying it different ways but with no avail. Any help would be greatly appreciated! Thank you!

Answer 1

$V(t)=180-16t=\frac{4\pi r(t)^3}{3}$

$r(t)=\left(\frac{3(180-16t)}{4\pi}\right)^\frac{1}{3}$

$\left.\frac{\operatorname{d}r(t)}{\operatorname{d}t}\right|_{t=2}=-\frac{4}{\sqrt[3]{111^2\pi}}\approx-0.11824824793$

Answer 2

One mistake I noticed is that you tried to find what the radius equals at time $t=0$, but you want to know what $r$ is when $t=2$. The volume at that time is $180-2(16) = 148$. Hence $r = \sqrt[3]{3(148)/(4\pi)} = \sqrt[3]{111/\pi}$.