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Implicit Differentiation Problem

  1. Feb 23, 2006 #1
    Consider the curve given by X^2+4y^2=7+3xy
    a) show that dy/dx=3y-2x/8y-3x
    b) show that there is a point P with x-cooridnate 3 at which the line tangent to the curve at P is horizontal. Find the y-cooridnate of P.
    c)find the value of d^2y/dx^2 at the point P found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer.

    (a) is easy. All you do is find the derivative.

    For (b), I got the point (3,2) by plugging 3 into the original equation and got 2.

    For (c), the value I got was -2/7 by finding the second derivative and plugging (3,2) for the x's and y's. I need to know if this is right and if there are any max's or min's at this point. Thanks.
  2. jcsd
  3. Feb 23, 2006 #2


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    -2/7 is correct. You know that dy/dx is 0 at the point, so you know it has to be either a maximum, a minimum, or a saddle point. Intuitively what do you think it should be, given that d^2y/dx^2 is less than 0? If it is less than 0 it means that the slope is decreasing--the slope is changing to become more negative. What would that mean in terms of maximum or minimum?
  4. Feb 23, 2006 #3


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    Slope of dy/dx is decreasing and is 0 at x= 3. That is, dy/dx is positive for x< 3 and negative for x> 3. y itself is increasing for x< 3, and decreasing for x>3. What does that tell you?
  5. Feb 23, 2006 #4
    Since y is increasing when x<3 and decreasing when x>3, does that mean there is a local maximum?
  6. Feb 24, 2006 #5


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    Yes. Draw a picture of that situation.
  7. Feb 24, 2006 #6
    Alright, thanks a lot for all of your help. I believe this graph looks like a slanted ellipse.
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