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Math Help - Slope of the tangent

  1. #1
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    Slope of the tangent

    The slope of the tangent to the curve y^3x + y^2x^2 = 6 at (2,1) is:

    -3/2, -1, -5/14, -3/14, 0

    What am I suppose to do? Implicit differentiation? I tried that and I ended up getting 3y' + 8y' = 0

    What do I do??
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  2. #2
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    Quote Originally Posted by Reefer View Post
    The slope of the tangent to the curve y^3x + y^2x^2 = 6 at (2,1) is:

    -3/2, -1, -5/14, -3/14, 0

    What am I suppose to do? Implicit differentiation? I tried that and I ended up getting 3y' + 8y' = 0

    What do I do??
    y^3x+y^2x^2=6

    Note in the following explanation \frac{dy}{dx}=y'. I prefer using \frac{dy}{dx} instead of y'. Old habits die hard...

    Implicit differentiation yields:

    3y^2\frac{dy}{dx}(x)+y^3(1)+2y\frac{dy}{dx}(x^2)+y  ^2(2x)=0

    (remember you have to use product rule...x and y are both variables...so when differentiating the product we have to use product rule)

    We want all the terms with \frac{dy}{dx} on one side so we can factor out that \frac{dy}{dx} and solve for it. So the previous equation turns into:

    3y^2(x)\frac{dy}{dx}+2y(x^2)\frac{dy}{dx}=-y^3-2xy^2

    Factoring out \frac{dy}{dx} on the left hand side yields:

    \frac{dy}{dx}(3y^2x+2yx^2)=-y^3-2xy^2

    Now, to get \frac{dy}{dx} by itself, we must divide through by 3y^2x+2yx^2 which results in:

    \frac{dy}{dx}=y'=\frac{-y^3-2xy^2}{3y^2x+2yx^2}

    You want to find the slope of the tangent line at (x,y)=(2,1). We have solved for \frac{dy}{dx} which gives the slope of the tangent line at any point on the original function. So now, you just need to substitute x=2 and y=1 into the equation for \frac{dy}{dx}.

    Does that make sense?
    Last edited by elizsimca; November 20th 2008 at 07:33 PM. Reason: clarification
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  3. #3
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    Oh I get it.. Didn't know we had to use the product rule.
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  4. #4
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    Well, it's good that you ran across that in your homework as opposed to an in-class exam! I figured that was the issue..I differentiated it disregarding the x and came up with the same thing you did.

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