Find the equation of a normal line to the curve y=3√x^2-1 at the point where x=3

Question

zepdrix · Answer

$\large\rm y=3\sqrt{x^2-1}$

Hey Jan,
remember how to find slope of a `tangent line`?
Finding the slope of a `normal line` is a similar process.

zepdrix · Answer

We'll find the slope of the line tangent to the curve,
and then the `negative reciprocal` of that value gives us the slope of our normal line.

zepdrix · Answer

So umm, for tangent slope, we need a derivative.
Any trouble there?

anonymous · Answer

So (x^2-1)^1/3

anonymous · Answer

Then 1/3(x^2-1)^-2/3*(2x)

zepdrix · Answer

Ok great, let's clean it up a little bit.$$\large\rm y'=\frac{2x}{3(x^2-1)^{2/3}}$$Next, we want to evaluate this derivative at x=3,
$\large\rm y'(3)=?$

anonymous · Answer

So 1

zepdrix · Answer

Sorry I ran off :(
The site is acting up today.. I can't stay connected for some reason.

$\large\rm y'(3)=\dfrac{2(3)}{3(3^2-1)^{2/3}}$
Hmm I think it works out to 1/2 right?