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I am solving my first differential equation word problem and I first would like to understand it.

The problem goes like this:

Imagine a square column of air, $1\times1$ inch, $x$ feet tall, standing on the earth's surface.

  • $P_0$ = weight of the column, which is the atmospheric pressure at sea level

  • atmospheric pressure at sea level is $14.5$ lbs per sq inch

  • $\text{weight}(x)$ = weight of air first few $x$ feet

  • formula given is $P(x) = P_0 - \text{weight}(x)$

So to help me understand...I will create a function $P(x)$, that will tell me the atmospheric pressure of something that is $x$ feet tall.

In doing so, knowing that $\text{weight}(x)$ provides the weight for the first few $x$ feet will help develop this function?

Does my picture capture the word problem?


atmospheric pressure

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up vote 2 down vote accepted

If you think of a thin layer of air $dx$ high, the weight of that layer is $A\frac P{P_0} g\rho dx$, where $A$ is the cross sectional area (1 in^2 in your example, but it will divide out. I like to keep it in so the units are correct-it helps catch errors) and $\rho$ is the density of air at sea level and a pressure $P_0$. If the air is the same temperature, the density will be proportional to the pressure. To keep from sinking, the pressure on the bottom has to be just enough higher than the pressure on the top to balance gravity. So $AP(x)-AP(x+dx)=A\frac P{P_0}g\rho dx$, which leads to $\frac {dP}{dx}=-g\rho\frac P{P_0}$

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The key point here being that $\rho(x) = \frac{P(x)}{P(0)} \rho(0)$. – copper.hat Feb 21 '13 at 16:26

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