Question

Write the equation of the normal to the curve y = x + sin x cos x at \[x = \frac{\pi}{2}\] ?

Answer 1

\[\text { Here,} \]

\[y = x + \sin x \cos x\]

\[\text { On differentiating both sides w.r.t.x, we get }\]

\[\frac{dy}{dx} = 1 + \cos^2 x - \sin^2 x\]

\[\text { Now,} \]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{2}} {= 1+cos}^2 \frac{\pi}{2} {-sin}^2 \frac{\pi}{2}= 1-1=0\]

\[\text { When }x=\frac{\pi}{2},y=\frac{\pi}{2}+sin\frac{\pi}{2}\cos\frac{\pi}{2}=\frac{\pi}{2}\]

\[ \therefore \left( x_1 , y_1 \right) = \left( \frac{\pi}{2}, \frac{\pi}{2} \right)\]

\[\text { Equation of the normal }\]

\[ = y - y_1 = \frac{- 1}{\text { Slope of the tangent }}\left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{\pi}{2} = \frac{- 1}{0}\left( x - \frac{\pi}{2} \right)\]

\[ \Rightarrow x = \frac{\pi}{2}\]

\[ \Rightarrow 2x = \pi\]