# Equation of tangent to a curve.

• Jul 12th 2010, 07:34 PM
darksoulzero
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 $\displaystyle P(3,9)$
• Jul 12th 2010, 07:47 PM
pickslides
Start by finding $\displaystyle \frac{dy}{dx}$ at $\displaystyle x=3$
• Jul 12th 2010, 08:04 PM
darksoulzero
Quote:

Originally Posted by pickslides
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$\displaystyle 9x -2y - 9 = 0$ .
• Jul 12th 2010, 08:23 PM
darksoulzero