Question

What is the equation of the line that is normal to `f(x)=sinx-e^xcotx` at `x=pi/3`?

Answer 1

Equation of normal at `x=pi/3` is `(y-sqrt3/2+e^(pi/3)/sqrt3)(1/2+(4-sqrt3)/3e^(pi/3))=pi/3-x`

Explanation:

Normal is perpendicular to tangent at the pint on the curve.

The slope of tangent to curve `y=f(x)` at `x=x+` is given by `f'(x_0)` and hence slope of normal is `-1/(f'(x_0))`. Further itpasses through point `(x_0,f(x_0))` on the curve. This gives us point slope form of normal to the curve as `y-f(x_0)=-1/(f'(x_0))(x-x_0)`.

Here `y=f(x)=sinx-e^xcotx` and `x_0=pi/3`

Hence `f(pi/3)=sin(pi/3)-e^(pi/3)cot(pi/3)=sqrt3/2-e^(pi/3)/sqrt3`

Further `f'(x)=cosx+e^xcsc^2x-e^xcotx`

and `f'(pi/3)=cos(pi/3)+e^(pi/3)csc^2(pi/3)-e^(pi/3)cot(pi/3)`

= `1/2+e^(pi/3)[4/3-1/sqrt3]=1/2+(4-sqrt3)/3e^(pi/3)`

and hence equation of normal is

`y-sqrt3/2+e^(pi/3)/sqrt3=-1/(1/2+(4-sqrt3)/3e^(pi/3))(x-pi/3)`

or `(y-sqrt3/2+e^(pi/3)/sqrt3)(1/2+(4-sqrt3)/3e^(pi/3))=pi/3-x`