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Finding Points on Curve where Derivative Doesn't Exist

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx on the curve

    1/3*y^3 + y*sin(x) = 1 - x^6

    At which points on the curve does this derivative does not exist? Find the slope of the line tangent to the curve at the point (0,3^(1/3)).

    2. Relevant equations

    dy/dx = - Fx/Fy

    3. The attempt at a solution

    I solved dy/dx implicitly using the above formula to get:

    dy/dx = - (y*cos(x) + 6x^5) / (y^2 + sin(x))

    To find the points where dy/dx doesn't exist, I let:

    y^2 + sin(x) = 0

    I've then tried rearranging the above expression in terms of y and x and substituing into the original curve equation, but can't complete the algebra by hand. Wolfram returned multiple answers depending how it's done. Any help?

    Thank you! :)

    EDIT: And for the last bit, I would just substitute those values for x and y into dy/dx which gives me something about -0.6 from memory?
     
  2. jcsd
  3. Apr 14, 2013 #2

    mfb

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    Staff: Mentor

    If sin(x)=-y^2,
    1/3*y^3 - y^3 = 1 - x^6
    -2/3 y^3 = 1-x^6
    y=...
    This can be put in y^2 + sin(x) = 0 again to get a (messy) function of x only.
    In the same way, you can solve for x and get a function of y only.
    I don't think you are supposed to give analytic solutions to those equations.
    The values where the derivative does not exist? That should not give a number for the derivative.
     
  4. Apr 14, 2013 #3
    So I can't get the values seen here:

    http://www.wolframalpha.com/input/?i=1/3*y^3+y*sin(x)=1-x^6,+y^2+sinx=0

    by hand? The algebra seems to reach a dead end every way I try it. How would I state my answer for the points of the curve that dy/dx is undefined then?

    And the values for the part of the question asking: "Find the slope of the line tangent to the curve at the point (0,3^(1/3))."

    I got-0.69, which looks right on the graph at that point?

    Thank you! :)

    EDIT: Also for showing the implicit derivative I should just be able to use dy/dx = -Fx/Fy, without showing how to arrive at that formula? Like I shouldn't have to do it with the chain rule?
     
    Last edited: Apr 14, 2013
  5. Apr 14, 2013 #4

    mfb

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    Staff: Mentor

    I would describe them as "the two solution of this set of equations".

    That is not a point, but if you mean x=0,3^(1/3), then it looks reasonable.

    Should be fine.
     
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