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Homework Help: Implicit Differentiation

  1. Mar 5, 2010 #1
    The function of a line is y^2 + x^3 = 9. I calculated the slope of its tangent to be -3x^2/2y. The question asked us to find a point(s) so that the equation of its tangent line is y + 6x = 13. So it's slope must be -6 at that point.

    I got (2,1) as a point. Are there more than one, or is that the only one?
  2. jcsd
  3. Mar 5, 2010 #2


    User Avatar
    Homework Helper

    well you will have 3x2/2y =6 such that x2=4y

    So for each 'x' you pick, there will be a corresponding value of 'y' from that relation. Yes (2,1) is a point, so is (-2,1)
  4. Mar 5, 2010 #3
    But if the point is (-2,1), then the equation for the tangent line can't be y + 6x = 13, because (-2,1) isn't on the line y + 6x = 13.
  5. Mar 6, 2010 #4


    Staff: Mentor

    So here's what you know:
    For the curve y^2 + x^3 = 9 (it's not a line, as you stated), y' = -3x^2/(2y).
    The slope of the specified tangent line is -6.
    Equating those values gives x^2 = 4y ==> x = +/-2sqrt(y)

    Now substitute for x in the equation of the tangent line y = -6x + 13. This gives you two values of y, one of which is y = 1. The other is extraneous.

    Now substitute this y value in the equation of your curve y^2 + x^3 = 9, and you get only one x-value.

    (-2, 1) is not a point on your curve.
  6. Mar 6, 2010 #5
    So the only point is (2,1) as I stated in my first post, right?
  7. Mar 6, 2010 #6


    Staff: Mentor

    Right. (2, 1) and (2, -1) are both points on the graph of the curve, but only the first point has a tangent whose slope is -6.
  8. Mar 6, 2010 #7
    Thank you so much!
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