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

Answer 1

Elijah Abram

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)#

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)#

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Answer 2

Isaiah Allen

The equation of the line normal to f(x)=x^2cot^2x at x=pi/3 is y = -3√3(x - π/3) + 3.

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Answer from HIX Tutor

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