How would I do this problem I am not sure if I did it correctly.
The volume $V$ of a spherical balloon is increasing at a constant rate of 32 cubic feet per minute.
A. How fast is the radius r increasing when the radius is exactly 3 feet.
I know $\dfrac{dv}{dt}=32$ $,r=3$
the formula for volume is
$$v=\frac{4}{3}\pi r^3$$
$$32=\frac{4}{3}\pi3r^2$$
$$32=4\pi 9\frac{dr}{dt}$$
$$\frac{dr}{dt}=\frac{8\pi}{9}$$ I am not sure would this be correct?
My second question is
B.How fast is the surface area a increasing when the radius is 4 feet
I know
$$A=4\pi r^2$$
but then what would I do
$$\frac{dA}{dt}=4\pi2r \frac{dr}{dt}$$
but what would I plug in for $\dfrac{dr}{dt}$?
Best Answer
Your $\dfrac{dr}{dt}$ should have $\pi$ in the denominator: $$\frac{dr}{dt}=\frac{8}{9\pi}\tag{1}$$
Use $\dfrac{dr}{dt}$ from computing $dr/dt$ as you did for probem $1$ but using $r = 4$
$$32=\frac{4}{3}\pi3r^2 \implies 32 = \frac {3\cdot4}3 (4^2)\pi\frac{dr}{dt} \iff \frac{dr}{dt} = \frac {1\cdot 32}{64 \pi} = \frac 1{2\pi}$$
And simplify. Then substitute into the equation you found:
$$\frac{dA}{dt}=4\pi2r \frac{dr}{dt}$$