Related Rates Question Concerning Gay Lussac's Law Gay Lussac's Law for enclosed gases states that if the volume is kept constant, the pressure $P$ and temperature $T$ are related by the equation $\dfrac{P}{T}=k$, where $k$ is a constant. Suppose that the rate of change of the temperature is $3$ degrees per hour. What is the rate of change of pressure when $T$ is $250$ and $P$ is $500$?
I know this is a related rates question involving differentiation, and that $\dfrac{dT}{dt}=3$ but I just can't figure out how to tie the values together! Please help, thanks!

I thought Boyle's law involved pressure and volume (not temperature). But anyway, yes, $\dfrac{dT}{dt}=3$. So start from the given 
$$ \dfrac{P}{T}=k$$
So both $P$ and $T$ are functions of time, $t$. $k$ is just a constant. So, differentiating, we need to use the quotient rule: 
$$\dfrac{ T \dfrac{dP}{dt} - P \dfrac{dT}{dt}}{T^2} = 0$$. 
Along the way you need to understand, why is this equation equated to 0? 
You know that $\dfrac{dT}{dt} = 3$, so the rest must be easy. 
