What is the equation of the line normal to f(x)=sec4x-cot2xf(x)=sec4x−cot2x at x=pi/3x=π3?

1 Answer
1s2s2p
Apr 11, 2018

"Normal"=>y=-(3x)/(8-24sqrt3)+(152sqrt3-120+3pi)/(24-72sqrt2)=>y~~0.089x-1.52Normal⇒y=−3x8−24√3+152√3−120+3π24−72√2⇒y≈0.089x−1.52

Explanation:

The normal is the perpendicular line to the tangent.

f(x)=sec(4x)-cot(2x)f(x)=sec(4x)−cot(2x)
f'(x)=4sec(4x)tan(3x)+2csc^2(2x)

f'(pi/3)=4sec((4pi)/3)tan((4pi)/3)+2csc^2((2pi)/3)=(8-24sqrt3)/3

For normal, m=-1/(f'(pi/3))=-3/(8-24sqrt3)

f(pi/3)=sec((4pi)/3)-cot((2pi)/3)=(sqrt3-6)/3

(sqrt3-6)/3=-3/(8-24sqrt3)(pi/3)+c

c=(sqrt3-6)/3+pi/(8-24sqrt3)=(152sqrt3-120+3pi)/(24-72sqrt2)

"Normal":y=-(3x)/(8-24sqrt3)+(152sqrt3-120+3pi)/(24-72sqrt2);y=0.089x-1.52

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