Question

What is the equation of the line that is normal to `f(x)= ln(sin^2x)` at `x=(4pi)/3`?

Answer 1

`y(x) =ln(3/4)-sqrt(3)/2(x-(4pi)/3)`

Explanation:

The equation of the line normal to the curve `y=f(x)` at `x=barx`, provided that `f(x)` is differentiable is:

`y(x) = f(barx)-1/(f'(barx))(x-barx)`

In our case:

`f(x) = ln(sin^2x)`
`f'(x) = (2sinxcosx)/(sin^2x) = 2cotx`

and

`f((4pi)/3) = ln(sin^2((4pi)/3) )=ln((sqrt(3)/2)^2) =ln(3/4)`

`f'((4pi)/3) = 2 cot ((4pi)/3) = 2/sqrt(3)`

So the normal line is:

`y(x) =ln(3/4)-sqrt(3)/2(x-(4pi)/3)`