How does these two equations relate?

[Zerius]

Hi, I'm in first year engineering and I have trouble understanding where this formula comes from in determining the relationship between the coefficient of volume expansion and the coeff of linear expansion.

dV=(dV/dL)*dL = 3L^2*dL

now i know that they changed the dV in the brackets to dL^3, but then I don't understand how dL^3/dL became 3L^2? Is there a part of calculus I am missing? does it have to do with a rate of change divided by a rate of change?

Thanks!

 

[George Jones]

Can you express V as a function of L?

 

[Zerius]

Volume = L^3, for a cube? i think that's the relationship there

 

[George Jones]

Volume = L^3, for a cube? i think that's the relationship there

What is dV/dL?

 

[Zerius]

delta V / delta L times delta L = delta V the delta L's cancel. they do it on wikipedia too like all of a sudden http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion it becomes 3Lo^2 and I don't understand

 

[George Jones]

If y = x^3, what is dy/dx?

 

[Zerius]

then it is 3x^2, that's would be dy/dx. :S how about dX^3/dX? is it the same? OH or is it like dy/dx is the same as d/dX so doesn't matter what it is at the top?

 

[George Jones]

Right, dy/dx = 3x^2.

Now treat dy and dx as symbols small quantities that can be manipulated like other quantities.

Multiplying both sides of dy/dx = 3x^2 by dx gives?

 

[Zerius]

umm dy = (3x^2)dx?

 

[George Jones]

umm dy = (3x^2)dx?

Right. Now change y to V and x to L throughout.

Even though engineers and physicists routinely do these types of manipulations, they make pure mathematicians cringe.

 

[Zerius]

>.< ok that makes more sense now. I will try to hatch this in my brain. Thanks George Jones!