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#1




Finan 38.5
Gas molecules move about with varying velocity which has, according to the
Maxwell Boltzmann law, a probability density given by $ f(v) = cv^2e^{\beta v^2} $ The kinetic energy is given by $Y=E= \frac{1}{2}mv^2$ where m is the mass. What is the density function of Y ? I simplify the Y equation as $ \frac{2y}{m} = v^2 $ and then just sub it in with and multiple by the abs d/dx. But, I get a different solution. Is there something wrong with the method? In a pdf transformation, do I need to simplify $v^2$ to just $v$? Now that I think about I bet the square factor changes the derivation. Correct solution is: $\frac{c}{m}{(\frac{2y}{m})}^.5e^{\frac{2\beta y}{m}}$ Last edited by jchaney; 09162016 at 10:34 PM.. 
#2




It one to one so write v in terms of y and substitute. Then calculate dv/dy in terms on y and multiply.

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