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Homework Help: Calculus - Derivative Help?

  1. Nov 27, 2007 #1
    [SOLVED] Calculus - Derivative Help??

    1. The problem statement, all variables and given/known data
    Consider the curve given by the equation y^3 + 3x^(2)y + 13 = 0. Find the minimum y - coordinate of any point on the curve. Justify your answer.

    2. Relevant equations

    I solved parts a&b:
    dy/dx = -6xy / (3y^2 + 3x^2)
    Equation for line tangent to curve at (2,-1): y + 1 = 4/5 (x-2)

    3. The attempt at a solution
    This is the other problem I cannot get. I tried to make up my own minimum to find a relative minimum, which my instructor said we could find, but I still couldn't get critical points. I also tried using the second derivative ... but that got too messy and I didn't know what to do.

    Any help would be fantastic! Thank you for your time and explanations!!
  2. jcsd
  3. Nov 27, 2007 #2


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    at the minimum point...the gradient of the tangent at that point is zero...i.e. gradient function=0 => dy/dx=0
  4. Nov 27, 2007 #3
    sorry, we haven't learned gradient functions yet. can you explain this concept to me? or is there another way to do it?
  5. Nov 27, 2007 #4


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    Well the gradient at any point (x,y) on a curve is given by dy/dx...this is called the gradient function. So to find the gradient at any point you just sub the x value and y value(if there is y in your expression fot dy/dx) and you'll get the gradient of the tangent at that point.

    At a min point, the tangent drawn to that point is a horizontal line i.e. the line has a gradient at zero. So to find this point you equate dy/dx to zero and solve...

    dy/dx = -6xy / (3y^2 + 3x^2)=0

    for this to be equal to zero, the numerator must be zero...so solve for x and y in here and you will get a stationary point...find the 2nd derivative and sub the min point you get and check the sign, If [itex]\frac{d^2y}{d^2x}[/itex] is positive,then it is a minimum point, if it is negative, it is a maximum point.
    Is this clear? If not I shall try to explain better
  6. Nov 27, 2007 #5
    ok, one more question. so i got (0,0) from the first deriv, now i plug that in to the second to find if it is a max or min? i believe it must be a max anyway since it is higher than the point (2,-1) which the problem gives.
  7. Nov 28, 2007 #6


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    Neither (0,0) nor (2, -1) is even on the curve y^3 + 3x^(2)y + 13 = 0!

    In order that the derivative be 0, either x= 0 or y= 0. If x= 0, what is y? If y= 0, what is x?
  8. Nov 28, 2007 #7
    ahhhh, i understand now. i was being dense! but the sad thing is i already handed my problem in. oh, well, at least i will get it right next time!

    thank you all for the wonderful attention and help!
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