Thread: equation of normal line

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?

Originally Posted by Mr_Green

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?

The tangent line has a slope equal to the derivative of the function at that point. So find the slope. Then you can get the slope of the line perpendicular to this. Now you have the slope of your normal line and you know a point that is passes through.

-Dan

am i finding the derivative of 3rdroot(x^2-1) ??

Originally Posted by Mr_Green

am i finding the derivative of 3rdroot(x^2-1) ??

yes, that's the function you were given correct? it's the same procedure in finding the tangent line, except you use the negative reciprocal of the slope of the tangent line

so for the derivative i got:
3(x^2 - 1) * 2x =

6x(x^2 - 1)

is this correct?

if so, now what do i do?

Originally Posted by Mr_Green

so for the derivative i got:
3(x^2 - 1) * 2x =

6x(x^2 - 1)

is this correct?

if so, now what do i do?

You have


The derivative of this is


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 for x = 3.

-Dan

im getting y +12x=38, which is also one of my multiple choice responces, so I hope its right.

Originally Posted by Mr_Green

im getting y +12x=38, which is also one of my multiple choice responces, so I hope its right.

There are some serious problems here.

The slope of the tangent line at x = 3 is 1/2. So the slope of the normal line is -2. You know this line passes through the point (3, 2). I get that the normal line is


-Dan