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SK asked in Science & MathematicsMathematics · 1 decade ago

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

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  • jimbot
    Lv 6
    1 decade ago
    Favorite 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

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