What is the equation of the line normal to f(x)=xsin⁡xtan⁡x at x=π3?

As f(x)=xsin⁡xtan⁡x, at x=π3,f(x)=π3×323=π6Hence,m we are seeking equation of normal at (π3,π6)As slope of tangent is given by f′(x)=cos⁡x−xsin⁡x and at x=π3,f′(π3)=cos⁡(π3)−π3×sin⁡(π3)=12−π3×32=12−π23=3−π23Hence slope of tangent is 3−π23and that of normal would be −13−π23=23π−3and equation of normal would be(y−π6)=23π−3×(x−π3)

"What is the equation of the line normal..." solved and explained

Quinn Moses

Answered question

2022-04-07

What is the equation of the line normal to

f (x) = \frac{x \sin x}{\tan x}

x = \frac{π}{3}

Answer & Explanation

firenzesunzc65

Beginner2022-04-08Added 16 answers

f (x) = \frac{x \sin x}{\tan x}

, at

x = \frac{π}{3}, f (x) = \frac{\frac{π}{3} \times \frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{π}{6}

Hence,m we are seeking equation of normal at

(\frac{π}{3}, \frac{π}{6})

As slope of tangent is given by

f^{'} (x) = \cos x - x \sin x

and at

x = \frac{π}{3}

f^{'} (\frac{π}{3}) = \cos (\frac{π}{3}) - \frac{π}{3} \times \sin (\frac{π}{3}) = \frac{1}{2} - \frac{π}{3} \times \frac{\sqrt{3}}{2}

= \frac{1}{2} - \frac{π}{2 \sqrt{3}} = \frac{\sqrt{3} - π}{2 \sqrt{3}}

Hence slope of tangent is

\frac{\sqrt{3} - π}{2 \sqrt{3}}

and that of normal would be

\frac{- 1}{\frac{\sqrt{3} - π}{2 \sqrt{3}}} = \frac{2 \sqrt{3}}{π - \sqrt{3}}

and equation of normal would be

(y - \frac{π}{6}) = \frac{2 \sqrt{3}}{π - \sqrt{3}} \times (x - \frac{π}{3})

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