Answered: "What is the equation of the line normal..."

Trent Fuller

Answered question

2022-04-10

What is the equation of the line normal to

f (x) = \sec 4 x - \cot 2 x

x = \frac{π}{3}

ncruuk7ikt

Beginner2022-04-11Added 12 answers

Explanation:
The normal is the perpendicular line to the tangent.

f (x) = \sec (4 x) - \cot (2 x)

f^{'} (x) = 4 \sec (4 x) \tan (3 x) + 2 \csc^{2} (2 x)

f^{'} (\frac{π}{3}) = 4 \sec (\frac{4 π}{3}) \tan (\frac{4 π}{3}) + 2 \csc^{2} (\frac{2 π}{3}) = \frac{8 - 24 \sqrt{3}}{3}

For normal,

m = - \frac{1}{f^{'} (\frac{π}{3})} = - \frac{3}{8 - 24 \sqrt{3}}

f (\frac{π}{3}) = \sec (\frac{4 π}{3}) - \cot (\frac{2 π}{3}) = \frac{\sqrt{3} - 6}{3}

\frac{\sqrt{3} - 6}{3} = - \frac{3}{8 - 24 \sqrt{3}} (\frac{π}{3}) + c

c = \frac{\sqrt{3} - 6}{3} + \frac{π}{8 - 24 \sqrt{3}} = \frac{152 \sqrt{3} - 120 + 3 π}{24 - 72 \sqrt{2}}

Normal:

y = - \frac{3 x}{8 - 24 \sqrt{3}} + \frac{152 \sqrt{3} - 120 + 3 π}{24 - 72 \sqrt{2}}; y = 0.089 x - 1.52

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