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How to differentiate this potential energy function?

  1. Aug 15, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    Given the potential energy function V(x,y)=V(ax-by) where a,b is an arbitrary constants differentiate with respect to x and y.

    2. Relevant equations
    Multivariavle differentiation

    3. The attempt at a solution
    The answer yields (d/dt)p1=-aV'(ax-by)
    (d/dt)p2=+bV'(ax-by). The right side is confusing. How did it get to that? Im not in my MultiCal class yet as im just in the middle of SingCal so i dont get it. Can you guide through the process?
     
  2. jcsd
  3. Aug 15, 2015 #2

    ShayanJ

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    Assuming ## z=ax-by ## and using chain rule, we have:
    ## \frac{dV}{dx}=\frac{dV}{dz}\frac{dz}{dx}=a\frac{dV}{dz}##. Now we change the name of ## \frac{dV}{dz} ## to ## V'##.
     
  4. Aug 15, 2015 #3

    diredragon

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    Ok, but why is it -aV'?
     
  5. Aug 15, 2015 #4

    ShayanJ

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    At first I didn't know what are p1 and p2 and what are those equations, but I thought you yourself know! Anyway, I think those are the x and y components of Newton's 2nd law and so they are components of ## \frac{d\vec p}{dt}=-[\frac{dV}{dx}\hat x+\frac{dV}{dy}\hat y]##. The extra minus sign is because of the minus sign in the formula itself.
     
  6. Aug 15, 2015 #5

    diredragon

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    Oh yes, i know what the equations are but have totally forgotten about the minus in front of the right side of lagrangian. Thanks
     
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