Question

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

Answer 1

Equation of the line normal to `f(x)` at `x=pi/3` will be
`(y-pi^2/27)=(27sqrt3)/(8pi^2-6sqrt3pi)(x-pi/3)`

Explanation:

As `f(x)=x^2cot^2x`, `f(pi/3)=(pi/3)^2cot^2(pi/3)` or

`f(pi/3)=(pi)^2/9xx(1/sqrt3)^2=pi^2/27`

Now `(df)/(dx)=2xcot^2x+x^2xx2cotx xx(-csc^2x)`

= `2xcot^2x-2x^2cotxcsc^2x`

At `x=pi/3`, `(df)/(dx)=2pi/3cot^2(pi/3)-2(pi/3)^2cot(pi/3)csc^2(pi/3)`

= `2pi/3(1/3)-2(pi/3)^2(1/sqrt3)(2/sqrt3)^2`

= `2pi/9-2pi^2/9xx(4/(3sqrt3))=2pi/9-(8pi^2)/(27sqrt3)`

= `(6sqrt3pi-8pi^2)/(27sqrt3)`

This will be the slope of tangent and hence slope of normal would be `(-27sqrt3)/(6sqrt3pi-8pi^2)` or `(27sqrt3)/(8pi^2-6sqrt3pi)`

Hence equation of normal would be

`(y-pi^2/27)=(27sqrt3)/(8pi^2-6sqrt3pi)(x-pi/3)`