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Use of derivatives to find coordinates

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the coordinates of the point(s) on the following curves where the second derivative is as stated.


    2. Relevant equations
    [itex]y= \frac{x^3}{12} and \frac{d^{2}y}{dx^{2}} = 1.5[/itex]


    3. The attempt at a solution
    I'm used to working with the first derivative. Would I need to use integration to convert the function to the first derivative then treat it like a usual gradient - derivative = gradient? If I do need to continue to work with the second derivative, could I receive some hints on how?
     
  2. jcsd
  3. Feb 27, 2013 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    Well you just need to calculate y'' explicitly and then find x such that y''=1.5.
     
  4. Feb 27, 2013 #3
    Yeah it does seem to be the case. Though I'm getting really large numbers. I'm using the quotient rule twice to differentiate. Is that correct?
     
  5. Feb 27, 2013 #4

    Mark44

    Staff: Mentor

    You're making this problem much more difficult than it actually is. For your function you should NOT use the quotient rule. It's not wrong to do so, but it's a more complicated method that is more likely to lead to errors.

    You should never use the quotient rule to differentiate a quotient with a constant in the denominator. Instead write the function as (1/k) * f(x) and use the constant multiple rule.

    d/dx(1/k * f(x)) = 1/k * f'(x)

    The same thinking holds for functions of the form y = k * f(x). Although this is a product, the natural tendency would be to use the product rule. The easier rule to use would be the constant multiple rule here as well.
     
  6. Feb 28, 2013 #5
    Yeah that's true. I'm known for unintentionally doing so.

    From what I'm gathering I write it like this: [itex]\frac{x^{3}}{12} [/itex] to [itex] x^{3}12^{-1}[/itex] and then go from their by normally differentiating twice over to get the second derivative - consequently finding x for the coordinate to substitute into the original to get the y for the final coordinates?
     
  7. Feb 28, 2013 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Why in the world write (1/12) as [itex]12^{-1}[/itex]? Surely you know that (Cf(x))'= Cf'(x) for any constant C? Just differentiate [itex]x^3[/itex] twice and divide that by 12.
     
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