Given a sphere and its dV/dt, and the pressure on the sphere along with the dP/dt, how could I find the volume of the sphere?

Question

I'm given a sphere (of gas) which is being squished and told its volume is decreasing at a rate 10 cubic inches per second, which I know means dV/dt = -10. I'm also told that the pressure on the sphere is currently 8 lbs./sq. in., but it's increasing at a rate of 2 lbs./sq. in. per second which I understand means dP/dt = 2.

What I don't understand is how I could ever relate the pressure on the sphere to its volume. I could easily solve the problem if I knew the relation, but I just can't see how these two measurements are at all related. I've been messing around with it strenuously for a while now, I tried to relate the pressure to the surface area and I think that's on the right track, but I've had no luck because I always get stuck since I don't know the radius or how the radius changes with time. At the end of the problem, it also mentions that the product of a gas's volume and pressure is a constant when the temperature is constant (which it is in this scenario), which only serves to confuse me further because I don't see where I would multiply the volume and pressure in the problem. I've also thought about using an integral, but discarded the idea because I don't know the upper limit of the radius.

I'd prefer a hint to set me on the right track because I already know the answer as supplied by the textbook. What the textbook doesn't explain is how you actually get the answer, which is frustrating, so I'd appreciate a nudge in the right direction as I'm probably just overlooking something extremely simple. The textbook is a calculus textbook, and I believe the subject of the problem is "more complicated" related rates.

Answer 1

PV is a constant.
$\frac{d}{dt} PV=0$
$P\frac{dV}{dt}+V\frac{dP}{dt}=0$

Answer 2

First use $$ V = \frac{4}{3} \pi r^3 \rightarrow dV = 4 \pi r^2 dr \rightarrow V_t = \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$ Note that $V_t$ does not depend on time according to your problem.

From $P=F/A$, it follows $$ dP = \frac{-F}{A^2} dA = \frac{-2F}{A^2} \cdot 4\pi r dr = \frac{-2FV_t}{rA^2} dt \rightarrow P_t = \frac{dP}{dt} = \frac{-2FV_t}{rA^2} $$ Note that $V_t$ does not depend on time according to your problem.

So eliminating $F$ yields $$ P_0 = \frac{F}{A_0} =\frac{-P_t r_0 A_0}{2 V_t} $$ from which you can deduce $r_0$. Please let me know what you find and if it matches the result of your textbook.