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Implicit/Explicit Differenciation, and Inflection point of a Graph

  1. Jan 25, 2007 #1
    I have two questions, one about differentiating, and the other is about finding the inflection point on a function. Any help would be greatly appreciated.

    Question 1
    1. The problem statement, all variables and given/known data

    If ay^3 = x^4, show that d^2y/dx^2 = 4x^2/9ay^2

    now, this question can be done by both implicit or explicit differentiation. I can leave the ay^3 = x^4 as it is and start simplifying right away, or I could isolate y by making the equation y = (x^4a^-1)^(1/3)

    2. Relevant equations
    knowledge of the rules of differentiating.

    3. The attempt at a solution

    I am going to list two different attempts, one implicit and one explicit.

    ay^3 = x^4
    a3y^2 dy/dx = 4x^3

    dy/dx = 4x^3/ a3y^2

    d^2y/dx^2 = (12x^2((a3y^2)-6ya(4x^3)(dy/dx)) / (3y^2a)^2
    d^2y/dx^2 = (36x^2y^2a - 24yax^3(dy/dx) ) / (3y^2a)^2
    *I then subbed the previous found value for dy/dx into dy/dx in this equation.
    d^2y/dx^2 = (36x^2y^2a - 96yax^6/3y^2a) / (3y^2a)^2
    d^2y/dx^2 = (36x^2y^2a - 32x^6/y) / 9y^4a^2

    I stopped here because I could find no way to make the a^2 at the bottom become just a, similarily the y^4 could not be reduced down to y^2 like in the question.


    y = (x^4a^-1)^(1/3)
    y' = (1/3)(x^4a^-1)^(-2/3)(4x^3a^-1)
    y' = [(x^4a^-1)^(-2/3)(4x^3a^-1)]/3
    y'' = [3(12x^2a^-1)^(-2/3)(x^4a^-1)^(-5/3)(4x^3a^-1)]/9
    y'' = [-2(48x^5(a^-2)(x^4a^-1)^(-5/3)]/9
    y'' = [-96xa^-1(x^4a^-1)(x^4a^-1)^(-5/3)]/9
    y'' = -96x / 9a [(x^4/a^1)^(1/3)]^2
    since (x^4/a^1)^(1/3) = y
    y'' = -96x / 9ay^2

    as you can see, I got a lot further by the final answer is still wrong... I have part of the answer but have a -96x instead of 4x...

    could someone please be kind enough to solve this question, or tell me where I went wrong... if it is possible please also provide a solution to the implicit way of solving this question.

    Question 2
    1. Find the point of inflection of y = x^4 - 4x^3 + 6x^2 + 12x

    at a glance, this question may seem very easy, but for some reason I can't get the right answer (-0.4 and 2.4)

    2. to find inflection point, you find the 2nd derivative of the function and solve for x/b]

    3. The attempt at a solution

    y = x^4 - 4x^3 + 6x^2 + 12x
    y' = 4x^3 - 12x^2 + 12x + 12
    y'' = 12x^2 - 24x + 12
    let y'' = 0
    0 = 12(x^2 - 2x + 1)
    0 = 12(x-1)(x-1)
    x = 1

    I put this function into a graphing calculator, it showed no apparent change in concavity at x = 1. I set the min y = -100 and max y = 100 while keeping min x = -20 and max x = 20 to get a better view. the graph seemed to be a parabola but like the answer suggested, the concavity is slightly different between -0.4 and 2.4.

    anyone wishing to see the graph can just use this one and paste in the function. remember the set the min y, max y, min x, max x values so that you can see the graph as described by me.

    Thanks for you time.
    Last edited: Jan 26, 2007
  2. jcsd
  3. Jan 26, 2007 #2
    Question 1:
    For your first attempts, you are stucked in the last step:

    [tex] \frac{d^2y}{dx^2} = \frac{(36x^2y^2a - 32x^6/y)}{ 9y^4a^2} [/tex]

    You could simplify it one step further to get

    [tex] \frac{d^2y}{dx^2} = \frac{4x^2}{ay^2} - \frac{32x^6}{9a^2y^5} [/tex]

    The first term on the RHS is very similar to what you wanna prove, but the second term mess up the whole thing, therefore you must find a way to get rid of the second term...
    Try to expand the [tex]x^6[/tex] into [tex] x^2x^4 [/tex] and substitude the [tex]x^4[/tex] in the second term by the original equation, [tex] x^4 = ay^3 [/tex]
    This will solve your problem....

    For the second attempt, try to do it in a simpler maner...
    [tex] y = (\frac{x^4}{a})^{1/3} = a^{-1/3}x^{4/3} [/tex]

    [tex] \frac{dy}{dx} = \frac{4}{3}a^{-1/3} x^{1/3} [/tex]

    [tex] \frac{d^2y}{dx^2} = \frac{4}{3} \frac{1}{3} a^{-1/3}x^{-2/3} = \frac{4}{9} a^{-1/3}\frac{x^2}{x^{8/3}} = \frac{4}{9} a^{-1/3}\frac{x^2}{a^{2/3}y^2} = \frac{4x^2}{9ay^2} [/tex]

    For question 2, Your solution is abosulotely correct, maybe the answer is wrong or you misunderstand the question..

    Good Luck
    Last edited: Jan 26, 2007
  4. Jan 26, 2007 #3
    thank you so much :D

    your second solution was so much easier than mine, but I'm curious of where I went wrong in my solution, did I make a arithmetic mistake? or does using the power rule in this equation eventually give me -96x at the top?
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