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Thread: Slope of the tangent

  1. #1
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    Slope of the tangent

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

    $\displaystyle -3/2, -1, -5/14, -3/14, 0$

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

    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$\displaystyle y^3x + y^2x^2 = 6$ at (2,1) is:

    $\displaystyle -3/2, -1, -5/14, -3/14, 0$

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

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

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

    Implicit differentiation yields:

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

    (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 $\displaystyle \frac{dy}{dx}$ on one side so we can factor out that $\displaystyle \frac{dy}{dx}$ and solve for it. So the previous equation turns into:

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

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

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

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

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

    You want to find the slope of the tangent line at (x,y)=(2,1). We have solved for $\displaystyle \frac{dy}{dx}$ 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 $\displaystyle \frac{dy}{dx}$.

    Does that make sense?
    Last edited by elizsimca; Nov 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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