Question

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

Answer 1

`y=-18/(6*sqrt(3)*pi+pi^2)*x+pi^2/(6*sqrt(3))+6/(6*sqrt(3)+pi)`

Explanation:

We have given

`f(x)=x^2sin(x)`
then we get by the product rule

`(uv)'=u'v+uv'`
`f'(x)=2xsin(x)+x^2cos(x)`

`f'(pi/3)=2*pi/3*sin(pi/3)+(pi/3)^2cos(pi/3)`
`f'(pi/3)=pi/sqrt(3)+pi^2/18`
Note that `sin(pi/3)=sqrt(3)/2,cos(pi/3)=1/2`

so the slope of the normal line is given by

`m=-18/(6*sqrt(3)*pi+pi^2)`
We have
`f(pi/3)=pi^2/(6*sqrt(3))`
the searched equation hase the form

`y=mx+n`
We know `m` and `n` is given by

`pi^2/(6*sqrt(3))+6/(6*sqrt(3)+pi)=n`