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posted by  basic integration on 11/3/2009 4:19:46 PM  |  status: Closed  |  Earned Karma: 25

Slope of a tangent line

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Calculus N/A N/A N/A 11/3/2009 at 10:00:00 PM
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can someone help me out with this?

find the slope of the tangent line to the curve -x2+2xy+2y3=2 at the point (0,1)

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posted by tweaker_ on 11/3/2009 4:49:29 PM  |  status: Live
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Start off with your given equation:

f(x) = -x2 + 2xy + 2y3 = 2

Then find the derivative of that:

f'(x) = -2x + 2y + 2xy' + 6y2y' = 0

Keep all of your terms containing y' on one side and move anything else to the other side:

2xy' + 6y2y' = 2x - 2y

Then factor out y' from the left side:

y' [2x + 6y2] = 2x - 2y

And then finally divide the right side by the terms inside the brackets:

y' = [2x - 2y] / [2x + 6y2]

From here you should be able to determine your slope, cause y' is the same thing as your slope (m). Then from there you can work out the equation of the tangent line. Hope this helped! :)
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posted by anonymousxyz on 11/3/2009 4:54:28 PM  |  status: Live
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\begin{align*} & {-x^2}+2xy+2y^3 \;\;=\;\; 2 &\Rightarrow\\\\& \frac{\text{d}}{\text{d}x}\left({-x^2}+2xy+2y^3\right) \;\;=\;\; \frac{\text{d}}{\text{d}x}\left(2\right) &\Rightarrow\\\\& {-2x} + 2y + 2xy' + 6y^2y' \;\;=\;\; 0&\Rightarrow\\\\& 2xy' + 6y^2y' \;\;=\;\; 2x - 2y &\Rightarrow\\\\& (2x + 6y^2)y' \;\;=\;\;2x - 2y &\Rightarrow\\\\& y' \;\;=\;\; \frac{2x - 2y}{2x + 6y^2} \;\;=\;\; \frac{x - y}{x + 3y^2} \end{align*}


\begin{align*} & \text{The slope of the line tangent to the curve at the point } (0, 1) \text{ is:} \\\\& m \;\;=\;\; y'( 0, 1 ) \;\;=\;\; \frac{0 - 1}{0 + 3} \;\;=\;\; -\frac{1}{3} \end{align*}
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