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Calculus: max and mins through derivatives...?
The potential energy, U, of a particle moving along the x-axis is give by:
U = b(a^2/x^2 - a/x)
where a and b are positive constants and x >0. What value of x minimizes the potential energy?
I took the derivative, but I don't think I did it right, and I'm not sure what to do after. Can someone show me step by step please? Thanks so much!
1 Answer
- jimbotLv 61 decade agoFavorite Answer
U = b(a^2/x^2 - a/x)
= ba²/x² - ba/x
U' = -2ba²/x³ + ba/x² = 0
ba/x² = 2ba²/x³
bax³ = 2ba²x²
bax = 2ba²
x = 2ba²/ba
= 2a