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Quick implicit diff. question

  1. May 23, 2004 #1
    find the turning points on the curve with equation

    y^3 + 3xy^2 - x^3 = 3



    I'm confused bout the 3xy^2 mostly, i kno that xy goes to (y+x dy/dx) but not sure what 2 do bout the power...
     
  2. jcsd
  3. May 23, 2004 #2

    AKG

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    Use chain and power rules.

    d(f(x)g(x))/dx
    = f(x)g'(x) + g(x)f'(x)

    Also, remember, dz/dx = (dz/dy)(dy/dx)

    This is all you need to know, you should be able to figure it out yourself.
     
  4. May 23, 2004 #3
    We have:

    [tex]y^3+3xy^2-x^3=3[/tex]

    To perform an implicit differentiation here, we will apply the operator d/dx to each term on both sides of the equation:

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

    The important thing to remember is that:

    [tex]y=y(x)[/tex]

    That's why this is called implicit differentiation, because it's implied that y is a function of x. If y were not a function of x, then it could be treated as a constant when differentiating.

    On to the first term. You must use the extended chain rule here. The rule is:

    [tex]\frac{d}{dx}(u^n)=nu^{n-1}\frac{du}{dx}[/tex]

    So you have:

    [tex]\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}[/tex]

    This is just like taking the derivative of something like (x^2 + 3)^2 . First you would do the power rule on the whole quantity (x^2 + 3), then you would multiply by the derivative of the "inside" to get 2(x^2 + 3)(2x) . Here (y) is the whole quantity, and the derivative of the "inside" is dy/dx .

    It's important to note that you leave the dy/dx part alone after you've differentiated this part. You can't reduce it any further, since you don't know the functional relationship between y and x. Just treat it as another variable to carry along. At the end of the problem, sometimes you will want to solve for dy/dx = f(x,y) .

    The second term is done like the above, in combination with the product rule. The power rule is:

    [tex]\frac{d}{dx}(fg) = f\frac{dg}{dx} + g\frac{df}{dx}[/tex]

    Here, let 3x be f and y^3 be g:

    [tex]\frac{d}{dx}[(3x)(y^2)]=
    3x\frac{d}{dx}(y^2) + y^2\frac{d}{dx}(3x)[/tex]

    With this you should have enough information to finish the derivative.
     
  5. May 23, 2004 #4
    Thanx a lot! Think I got it sorted now then, I'm off 2 try it out
     
  6. May 23, 2004 #5
    Hmm...k well...I tried usin ur advice n I got... dy/dx = (3x^2 - 3y^2) / (3y^2 + 6xy) for the whole thing.....is that right? cos if so I'm not 2 sure bout how to find exact turning points, i kno it has to equal 0 but if u multiply through by the denominator, ur left with 0 = 3x^2 - 3y^2 ...so then 3x^2 = 3y^2...which is a lotta points? ...*doesn't have a clue* lol
     
  7. May 23, 2004 #6

    AKG

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    Is it? Basically, x^2 = y^2. However, notice that the denominator of the derivative cannot be zero (an undefined derivative is not a turning point, correct?). 3y^2 + 6xy <> 0 must also be satisfied. By inspection, y cannot be zero. Now, we can see that y + 2x <> 0 (divided by 3 and by "y", which is okay because we know y can't be zero). y <> -2x. When would y ever be -2x? If y^2 = x^2, then y could never be -2x anyways (except if y = 0, which is already thrown out) so we simply need to find all pairs, (x,y) such that y is not zero, x^2 = y^2 and the original equation is satisfied. So, plug in y^2 = x^2 into the original equation and solve, and throw out any answer where y = 0.
     
  8. May 23, 2004 #7
    DavioSenBoo, I think you have the right answer. I also get:

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

    I also agree with AKG's reasoning about the number of points given by the zero derivative condition. Looking at the numerator by itself and ignoring the denominator for the moment, we have:

    [tex]
    \begin{align*}
    y^2&=x^2\\
    y&=\pm x\\
    \end{align*}
    [/tex]

    Let's substitute +x in for y in the original equation to see what points we come up with:

    [tex]
    \begin{align*}
    y^3+3xy^2-x^3&=3\\
    x^3+3x^3-x^3&=3\\
    x&=1
    \end{align*}
    [/tex]

    From the first zero slope condition result, y = x, we see that when x is 1, y is 1, so we have our first important point, (1,1) . Now I will substitue -x in for y:

    [tex]
    \begin{align*}
    y^3+3xy^2-x^3&=3\\
    -x^3+3x^3-x^3&=3\\
    x&=3^{1/3}\approx 1.44
    \end{align*}
    [/tex]

    Since we used y = -x here, we have a second point, (1.44,-1.44) . These are the critical points, and it seems neither of them is dangerous with respect to the denominator of the derivative expression.

    I have made an implicit contour plot of the original equation in MATLAB with these two points labeled. With it, you can see the behavior quite well:

    http://omega.uta.edu/~tal0701/pf1.gif
     
    Last edited: May 23, 2004
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