What is the equation of the line normal to # f(x)=xtanx# at # x=pi/3#?

1 Answer
Jul 28, 2016

#x+(sqrt3+(4pi)/9)y-(4pi)/3(1+pi/(3sqrt3))=0#

Explanation:

Let us find the slope of the tangent, which should give us slope of normal (as it is perpendicular to tangent and hence product of their slopes would be #-1#).

Slope of tangent is given by first derivative and this is

#f'(x)=1xxtanx+x xxsec^2x# and as we need slope of tangent at #x=pi/3#,

#f'([i/3)=tan(pi/3)+pi/3xxsec^2(pi/3)=sqrt3+pi/3xx(2/sqrt3)^2#

= #sqrt3+(4pi)/9#

and slope of normal is #-1/(sqrt3+(4pi)/9)#

As normal is needed at #(x,y)# with #x=pi/3#, #(x,y)# is #(pi/3,pi/3xxtan(pi/3))# or #(pi/3,pi/sqrt3)#

Hence, using point slope form of equation #(y-y_1)=m(x-x_1)#

Equation of normal is #(y-pi/sqrt3)=-1/(sqrt3+(4pi)/9)(x-pi/3)# or

#(sqrt3+(4pi)/9)(y-pi/sqrt3)=pi/3-x#

#sqrt3y-pi+(4piy)/9-(4pi^2)/(9sqrt3)-pi/3+x=0# or

#x+(sqrt3+(4pi)/9)y-(4pi)/3(1+pi/(3sqrt3))=0#