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Boyle's law states that when a sample of gas is compressed at a constant temperature P and volume V are related by the equation $PV^{1.4}=C$, where C is a constant. Suppose that at a certain instant the volume is 400 $cm^3$ and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume decreasing at this instant?

I have no clue where to even start on this is so greatly appreciated :]

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Take derivatives. $P(1.4)V^{.4}V' + P'V^{1.4} = 0$. We know at the certain instant (call it $T$), the conditions above are satisfied. Plug it in. We get $$(1.4)(80)(400)^{.4}V' + (-10)(400)^{1.4} = 0$$ Solve for $V'$

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By the way, if the pressure is decreasing, the volume must increase, so your "At what rate is the volume decreasing at this instant" should read increasing. – Euler....IS_ALIVE Mar 20 '13 at 5:09

We have $PV^{1/4}=C$. Note that in this problem $P$ and $V$ are functions of time, and $C$ is a constant.

Differentiate with respect to $t$, immediately.

Note that $PV^{1/4}$ is a product of two functions, so we will use the Product Rule. And to differentiate $V^{1/4}$, we will use the Chain rule. So we get $$(P)\left(\frac{1}{4}\right)V^{-3/4}V' +V^{1/4}P'=0.$$

Now consider what happens at the instant when $V=400$, and $P=80$, and $P'=-10$. The above equation lets us find $V'$ at that instant.

Remark: There is almost a "recipe" for solving related rates problems. Usually we are told how fast some quantity is changing, and want to know how fast some other quantity is changing. To do this, usually we have to find an algebraic relationship between the two quantities, and then we differentiate. In this problem, the algebraic relationship was supplied to us: it is $PV^{1/4}=C$.

If I wanted to solve the problem, I would probably use a slightly different method. From $PV^{1/4}=C$, we get, using logarithms, that $\log P+\frac{1}{4}\log V=\log C$. Now differentiate. We get $$\frac{P'}{P}+\frac{1}{4}\frac{V'}{V}=0.$$

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