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Thread: Equation of tangent to a curve.

  1. #1
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    Equation of tangent to a curve.

    Hi, I have a question that asks: Write an equation of the tangent to each curve at the given point. $\displaystyle y = \sqrt{3x^3} $

    I've been trying to do the question, but I end up with decimals in the equation for the tangent. The answer is supposed to be $\displaystyle 9x -2y - 9 = 0 $. Thanks in advance.

    EDIT: Sorry about that, the given point is $\displaystyle P(3,9)$
    Last edited by darksoulzero; Jul 12th 2010 at 07:44 PM.
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  2. #2
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    Start by finding $\displaystyle \frac{dy}{dx}$ at $\displaystyle x=3$
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Start by finding $\displaystyle \frac{dy}{dx}$ at $\displaystyle x=3$

    Thanks, but I think that something's wrong with my math. here's what I got:

    $\displaystyle \frac{dy}{dx} = \frac{9}{2} \sqrt{x}.$

    $\displaystyle f'(3) = \frac{9}{2} \sqrt{3}$

    Now I sub. f'(3) into the following equation:

    $\displaystyle y - y_{1} = m(x - x_{1})$

    $\displaystyle y - 9 = \frac{9}{2} \sqrt{3}x -\frac{9}{2} \sqrt{3} (3) $

    $\displaystyle y - 9 - \frac{9}{2} \sqrt{3}x + \frac{9}{2} \sqrt{3}(3) = 0 $

    From here, I'm stumped as to how to get to$\displaystyle 9x -2y - 9 = 0$ .
    Last edited by darksoulzero; Jul 12th 2010 at 08:18 PM.
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  4. #4
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    Nevermind, I got the answer.
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