Find the equation of a line normal to the curve y=tan^-1(x/2) at x=3?

Question

You don't have to answer if you don't want to. A few good hints would be great!

Answer 1

hint: the normal line is perpendicular to the tangent line.
Find the slope of the tangent line first.
If m is the slope of the tangent line then the slope of the normal line is -1/m.

I hope that helps.

Answer 2

Find the slope at x=3 and then use the negative reciprocal to find the slope of the line perpendicular to that point (normal)

dy/dx = 1/ 2(x²/4 + 1)
so at x = 3 dy/dx = 2/13

thus the normal line should have slope -13/2
and it goes through the point (3, arctan(3/2))

so then use point slope to get the equation of the line
y - y1 = m (x-x1)

you can probably do it from where i left off

Answer 3

at x=3, y=tan^-1(3/2)

Slope of normal to the curve at point (x,y) is given by -dx/dy

dy/dx= 1/2 . 4/ (4+x^2) = 2/ (4+x^2)

slope of normal at the given point

=-(4+9)/2 = -13/2

Equation of normal

y-[tan^-1(3/2)] = -13/2 (x-3)