# Math Help - rate of pressure increase

1. ## rate of pressure increase

Okay so this one I thought was going to be more straight forward because I was given an equation that relates that which dependent upon time t.

when a certain gas expands or contracts adiabatically, it obeys the law $PV^1.4 = K$ where P is pressure, V is volume and K is a constant.
At a certain instant the pressure is 40 N/cm^2, the volume is 32cm^3, and the volume is increasing at a rate of 5 cm^3 per second. At what rate is the pressure changing at this instant.

So I thought the given equation was the one I would use. I solve it to find K

$40(32^1.4) = 5120$

then I differentiated the equation

$\frac{d}{dt} [PV^1.4=k] \Rightarrow V^1.4\frac{dP}{dt} + (1.4V^.4)P\frac{dV}{dt} = 0$

I can't figure out how to make exponents with decimal points work in latex.. the exponents are 1.4 and .4 after the differentiation.

anyway, Am I on the right track?

2. ## Re: rate of pressure increase

there appears to be something wrong in the expression. Pl check if it is PV^(1.4) = k

3. ## Re: rate of pressure increase

I don't understand what you mean.

Yes that is the right equation given by the book if that is what you mean.

4. ## Re: rate of pressure increase

I checked the answer and it comes out to $-8.75 g/cm^2/s$

5. ## Re: rate of pressure increase

okay I have solved this and I did it two separate ways

$\frac{d}{dt} [PV^(1.4)=K] \Rightarrow V^(1.4)\frac{dP}{dt} + 1.4V^(0.4)P\frac{dV}{dt} = \frac{dK}{dt}\Rightarrow\frac{dP}{dt} = \frac{-[1.4V^(0.4)(40)(5)}{32^(1.4)}$

$\Rightarrow -8.75$

$\frac{d}{dt} [P = \frac{K}{V^(1.4)} \Rightarrow \frac{dP}{dt}=K[-1.4V^(-2.4)\frac{dV}{dt}]$

$\Rightarrow\frac{dP}{dt} = 5120[-1.4(32^(-2.4)(5)]\Rightarrow -8.75$

Okay so for one I needed the value of K and the other I needed the derivative of K being zero.

I wouldn't have known the g/cm^2/s part though so hopefully I learn that later on when I am required to take some physics.

also, is there a way to use exponents that have decimal points in them in LaTeX? Because, what I had to do makes it look a little bit confusing.