A normal line to the graph of a function f at the point (x,f(x)) is defined to be the line perpendicular to the tangent line at that point. The equation of the normal line to the curve y = 3rdroot(x^2-1) at the point where x = 3 is?
A normal line to the graph of a function f at the point (x,f(x)) is defined to be the line perpendicular to the tangent line at that point. The equation of the normal line to the curve y = 3rdroot(x^2-1) at the point where x = 3 is?
You have
$\displaystyle \sqrt[3]{x^2 - 1} = (x^2 - 1)^{1/3}$
The derivative of this is
$\displaystyle \frac{d}{dx}(x^2 - 1)^{1/3} = \frac{1}{3}(x^2 - 1)^{-2/3} \cdot 2x$
Now do the rest I told you. Find the slope for x = 3, use this to find the slope of your normal line, then use that slope and the fact that the normal line passes through the point $\displaystyle (x, \sqrt[3]{x^2 - 1}$ for x = 3.
-Dan