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Math Help - Implicit Differentiaton

  1. #1
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    Implicit Differentiaton

    Hi I had the following question:

    Find the coordinates of the stationary points of the curve with equation x^3 + y^3 - 3xy = 48 and determine their nature.

    So the first derivative gives y=x^2 so the turning points are at x=2 and x= cube root of -6

    Now why is that some of the points that the derivative gives don't lie on the original equation, and I have to substitute y=x^2 into the original equation?

    For instance, when i want to find the nature of the turning points, I tried looking at the sign of the first deravative (rather than the second) around the stationary point, but that gives dy/dx = 0 for -2, which isnt a solution?

    Thanks
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  2. #2
    Super Member Matt Westwood's Avatar
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    I think you may be having trouble differentiating the equation in the first place.

    You differentiate it term by term, using the usual rules:

    \frac d {dx} x^3 = 3 x^2 which is straightforward.

    Then \frac d {dx} y^3 needs two things, the power rule (as above) and also the chain rule: \frac d {dx} f(y) = \frac d {dy} f(y) \times \frac {dy} {dx}, which gives 3 y^2 \frac {dy} {dx}. In this context f(y) = y^3. (Differentiating with respect to y is the same as w.r.t x except the name of the variable is different.)

    Then you have the product rule and the chain rule to use:

    \frac d {dx} xy = x \frac d {dx} y + y \frac d {dx} x

    which works out as:

    \frac d {dx} xy = x \frac d {dy} y \frac {dy} {dx}+ y

    = x \frac {dy} {dx} + y

    Now put it all together, gather the \frac {dy}{dx} onto one side and solve for it equalling zero.
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  3. #3
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    Hi, thanks I understand that. Differentiating gives:

    At the stationary points and we are left with
    To find which of these points lie on the curve, we substitute into the equation for the curve. This gives:
    So either
    or,


    This was my first question, why is it necessary to input what I get from differentiating back into the original equation? Shouldn't differentiating the original equation provide with solutions of the stationary points straight away?



    So the stationary points are at (2, 4) and . Now we determine the nature of the stationary points. Taking a second derivative of our equation gives:
    At the stationary points , which gives:
    Now we just check the sign of the second derivative at each stationary point.
    At (2, 4), maximum point;
    At
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    And the second question is, why is it necessary to take the 2nd derivative, and the other method of comparing the gradient (first derivative) on either side of the stationary point now work?

    Like, if I try to check the value of the gradient at say, x = -2, that gives 0 since that is once again a solution to y=x^2 , what we got from the first derivative.
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  5. #5
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    Quote Originally Posted by Aquafina View Post
    Hi I had the following question:

    Find the coordinates of the stationary points of the curve with equation x^3 + y^3 - 3xy = 48 and determine their nature.

    So the first derivative gives y=x^2 so the turning points are at x=2 and x= cube root of -6

    Now why is that some of the points that the derivative gives don't lie on the original equation, and I have to substitute y=x^2 into the original equation?

    For instance, when i want to find the nature of the turning points, I tried looking at the sign of the first deravative (rather than the second) around the stationary point, but that gives dy/dx = 0 for -2, which isnt a solution?

    Thanks
    I don't know why you have the "double" formulas in your posts but it makes them very hard to read!
    You are correct (I think!) that differentiating  x^3 + y^3 - 3xy = 48 gives 3x^2+ 3y^2 (dy/dx)- 3y- 3x(dy/dx)= 0 and at a stationary point dy/dx= 0 so that reduces to 3x^- 3y= 0 or y= x^2. So stationary points on the original curve must satisfy both  x^3 + y^3 - 3xy = 48 and y= x^2. Replacing y by x^2 in that gives x^3+ x^6- 3x^3= x^6- 2x^3= 48. Setting z= x^3, that becomes z^2- 2z- 48= (z- 8)(z+ 6)= 0. x^3= 8 givex x= 2 and x= -\sqrt[3]{6} just as you say.

    However, I don't understand your statement that dy/dx= 0 when x= -2 also. As you say, if dy/dx= 0 then y must be x^2. That means that when x= -2, y must be 4. but putting x= -2, y= 4 into the original equation,  x^3 + y^3 - 3xy = 48 gives -8+ 64+ 24= 80\ne 48.

    You don't have to use the second derivative test, but it is simpler to evaluate the second derivative at a single point than to determine the sign of the first derivative at every point in a set.
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  6. #6
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    Oh thank you Sorry about the "double" formulae, I copied the formulae from the mark scheme which made them come out like that on the post for some reason
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