Function of distance problem

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daniel_i_l

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Why does:
(dV(x)/dx)(dx/dt) = d(V^2(x)/2)/dx ? (V(x) is speed as a function of distance?
I know that the derivative of V^2(x)/2 if (dV(x)/dx)V(x) but I don't think that V(x) equals (dx/dt), that equal V(t)?

Thanks in advance!
 

siddharth

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Are you sure you wrote the equation down correctly? It should be d(V^2(x)/2)/dt on the RHS shouldn't it?
 

HallsofIvy

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daniel_i_l said:
Why does:
(dV(x)/dx)(dx/dt) = d(V^2(x)/2)/dx ? (V(x) is speed as a function of distance?
I know that the derivative of V^2(x)/2 if (dV(x)/dx)V(x) but I don't think that V(x) equals (dx/dt), that equal V(t)?
Thanks in advance!
V is, by definition, dx/dt.

Yes, if you know V(x), and x as a function of t, you could write V as a function of t: V(t) but it would still be V.

And, as siddharth said, that should be (dV(x))/dx)(dx/dt)= d(V^2(x)/2)/dt
 

daniel_i_l

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Thanks.
I know that a(x) = F(x) and I have to prove that V^2(x)/2 is the antiderivative of F(x).
That would mean that I have to prove that:
d(V^2(x)/2)/dx = F(x) right?

So if I know that F(x) = dV(X)/dt = dV(X)/dx * dx/dt
How do I prove that d(V^2(x)/2)/dx = F(x)?
 

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