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Implicit diff (2 var), error or what?

  1. Feb 28, 2005 #1
    i am working on a homework assignment. it's easy, or, so i think....

    [tex] 3x^2 - xy^3 + sin(x^3 - y) = 4 [/tex]

    Find [tex] \frac{dy}{dx} [/tex]

    not a problem. i ended up with

    [tex] \frac{dy}{dx} = \frac {6x - y^3 + 3x^2 cos(x^3 - y)}{3xy^2 + cos(x^3 - y)} [/tex]

    using implicit differention.

    now, "What is the slope of the curve defined by the equation at the point [tex] (0, \pi) [/tex]?"

    at first, i plugged it into the derivative, ended up with [tex] \pi^3 [/tex], but, upon inspection, i noticed that the point [tex] (0, \pi) [/tex] does not exist in the original equation. am I to find the slope of the curve of the derivative, or is this a typo, or am i looking at this problem incorrectly? thanks.
  2. jcsd
  3. Feb 28, 2005 #2
    wouldn't you just use point slope form: [itex] y-y_{1} = m(x-x_{1}) [/itex]? Just plug in the values into the derivative and work off from the above form?
  4. Feb 28, 2005 #3
    um, point slope form looks messy, but that's besides the point. perhaps this wasn't clear enough. the point [tex] (0, \pi) [/tex] doesn't exist in the original equation. not on the graph. [tex] sin(-\pi) [/tex] does not equal 4. that's the crux of my problem.

    IF the point was on the original graph, no problem, plug the values into the derivative and what I get is [tex] \pi^3 [/tex].
    Last edited: Feb 28, 2005
  5. Feb 28, 2005 #4
    Take the limit as x-> 0 and y -> Pi or look at the graph?
  6. Feb 28, 2005 #5
    I think it is a typo. There are in infinite many tangent lines to a point... so the point would have to be on the curve in order for the derivative you found to be relavent.
  7. Feb 28, 2005 #6
    looking at the graph is what caused me to realize the point isn't on there. it's not even close. now, as x approches 0 from the left, y approaches 2pi. the slope would also be increasing without bound. but that's for 2pi, which has no bearing on my problem.

    it's becoming obvious to me that this must be a typo, though, it's so obviously incorrect i just dont' know how it got by
  8. Feb 28, 2005 #7
    thank you, jameson
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