find the equation of normal to the curve: y=2 sin^2 3x at x= /6 - Maths - Application of Derivatives

Question

find the equation of normal to the curve: y=2 sin^2 3x at x=?/6

Himanshu Pandey · Answer

Given -> y = 2*sin23x
=> dy / dx = 4*sin3x*cos3x*3
= 6*sin6x [ 2sinxcosx = sin2x ]

[ dy / dx ] x = pi / 6 = 0

Equation of normal is
y - y0 = - dx / dy * ( x - x0 )

Now, x0 = pi / 6

=> y0 = 2*sin2[(pi)/2]
= 2
Hence, eq of normal is
y - 2 = - 1 / 0 * (x - pi/6 )
=> x = pi / 6

Hope this helps you,
Cheers!!!!