A weather balloon rises through the air at a rate of $500 \ \text{m}/\text{min}$. Every $1000 \ \text{m}$, the decrease of air pressure outside the balloon causes its radius to increase by $8 \ \text{cm}$. How rapidly is the volume increasing at the instant its radius is $90 \ \text{cm}$?
I'm not sure how the height of the balloon has anything to do with the calculation. So far I have:
$$ V = \frac{4}{3} \pi r^3 $$
$$ V'(t) = 4 \pi r^2 r'(t) = 4 \pi (90)^2 (8) = 259200 \pi \ \frac{\text{cm}^3}{\text{min}} $$
The answer is supposed to be: $0.1296 \ \text{m}^3/\text{min}$ (of course I can change my answer to $\text{m}^3$ if needed but it still doesn't match).
How do I use the height to answer the question?
Best Answer
The problem arises because you've assumed $r'(t) = 8$ when in fact it does not. What you have is not $r'$ as a function of time, but of height. Read it closely:
(Emphasis mine). In essence, you know $r'(h)=8/1000$ centimeters per meter. To get this as a function of $t$ instead, you do know that $h'(t)=500$ meters per minute, so if we have $r(h(t))$, then $r'(t)=r'(h)h'(t)$ by the chain rule, giving $r'(t)=\frac{8}{1000}\cdot 500=\frac{4000}{1000}=4$ in units $\frac{cm}{m}\cdot\frac{m}{s}=\frac{cm}{s}$. Using this correction in your calculation yields the desired result. (Assuming the answer should be $0.1296\pi$ not just $0.1296$ as you've written?)