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Math Help - Finding derivative of an xy function containing points

  1. #1
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    Finding derivative of an xy function containing points

    This question shouldn't be entirely that hard but for some reason when I check with my peers, all of us have completely varying or different answers...


    Let f(x) be the continuous function that satisfies the equation x^4 - 5x^2y^2 + 4y^2 = 0 and whose graph contains the points (2,1) and (-2,-2). Let [tex]\ell[m/ath] be the line tangent to the graph of f at x=2.

    A) find the expression for y '
    B) write an equation for line \ell
    C) give the coordinates of a point that is on the graph of f but is not on line \ell
    D) give coordinates of a point that is on line \ell but is not on the graph of f


    So for PART A, I did this.

    y' = 4x^3 - 10x^2y\frac{dy}{dx} - 10xy^2 + 16y^3\frac{dy}{dx} = 0

     -10x^2y\frac{dy}{dx} +16y^3\frac{dy}{dx} = -4^3 + 10xy^2

    \frac{dy}{dx}(16y^3 - 10x^2y) = 10xy^2 - 4x^3

    \frac{dy}{dx} = \frac{10xy^2 - 4x^3}{16y^3 - 10x^2y} = \frac{5xy^2 - 2x^3}{8y^3 - 5x^2y}


    Then for PART B, I plugged in (2,1) into my dy/dx and I got the slope to be 1/2 and thus my tangent line equation for \ell was y = \frac{1}{2}x + 0


    Then for PART C, I set the equation to y(2) = 0 for x=2 and I got that y = 1, -1, 2, -2. But I'm not sure what to do with this...


    So this is the work I did so far and I'm not sure on how to figure out part C or D of the question... If anyone could correct or check my initial work to see that it's correct or help me with Part C and D, then I would be really grateful! Because with some peers I was checking my work with, they got different derivatives or for part B got slopes like -3/14 or -13/14.
    Last edited by choi_siwon; April 13th 2009 at 01:26 AM.
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by choi_siwon View Post
    This question shouldn't be entirely that hard but for some reason when I check with my peers, all of us have completely varying or different answers... Here is the question. (Scanned the question from my worksheet)



    So for PART A, I did this.

    y' = 4x^3 - 10x^2y\frac{dy}{dx} - 10xy^2 + 16y^3\frac{dy}{dx} = 0

     -10x^2y\frac{dy}{dx} +16y^3\frac{dy}{dx} = -4^3 + 10xy^2

    \frac{dy}{dx}(16y^3 - 10x^2y) = 10xy^2 - 4x^3

    \frac{dy}{dx} = \frac{10xy^2 - 4x^3}{16y^3 - 10x^2y} = \frac{5xy^2 - 2x^3}{8y^3 - 5x^2y}


    Then for PART B, I plugged in (2,1) into my dy/dx and I got the slope to be 1/2 and thus my tangent line equation for \ell was y = \frac{1}{2}x + 0


    Then for PART C, I set the equation to y(2) = 0 for x=2 and I got that y = 1, -1, 2, -2. But I'm not sure what to do with this...


    So this is the work I did so far and I'm not sure on how to figure out part C or D of the question... If anyone could correct or check my initial work to see that it's correct or help me with Part C and D, then I would be really grateful! Because with some peers I was checking my work with, they got different derivatives or for part B got slopes like -3/14 or -13/14.
    Where is the original problem?
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  3. #3
    MHF Contributor matheagle's Avatar
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    You should spend some time trying to reduce this before you do any calculus.

     x^4-5x^2y^2+4y^4=(x^2-4y^2)(x^2-y^2)=(x-2y)(x+2y)(x-y)(x+y)
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  4. #4
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    Original problem is:

    Let f(x) be the continuous function that satisfies the equation x^4 - 5x^2y^2 + 4y^2 = 0 and whose graph contains the points (2,1) and (-2,-2). Let [tex]\ell[m/ath] be the line tangent to the graph of f at x=2.

    A) find the expression for y '
    B) write an equation for line \ell
    C) give the coordinates of a point that is on the graph of f but is not on line \ell
    D) give coordinates of a point that is on line \ell but is not on the graph of f

    Sorry if the image didn't show.

    ------

    matheagle: I thought it would be okay just find the derivative using the power rule, while using product rule uv'+vu' for the - 5x^2y^2. Since I separated it as

    - 5x^2y^2 = (-5x^2)(y^2)
     uv' + vu' = (-5x^2)(2y\frac{dy}{dx}) + (y^2)(-10x) = -10x^2y\frac{dy}{dx} - 10xy^2 <-- thus how I got middle part

    I'm not sure how to utilize the reduced form of the equation to help me on this problem, could you please explain it a little to me? (Sorry, I'm not so good at calculus . )
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