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प्रश्न
Write the equation of the normal to the curve y = x + sin x cos x at \[x = \frac{\pi}{2}\] ?
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उत्तर
\[\text { Here,} \]
\[y = x + \sin x \cos x\]
\[\text { On differentiating both sides w.r.t.x, we get }\]
\[\frac{dy}{dx} = 1 + \cos^2 x - \sin^2 x\]
\[\text { Now,} \]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{2}} {= 1+cos}^2 \frac{\pi}{2} {-sin}^2 \frac{\pi}{2}= 1-1=0\]
\[\text { When }x=\frac{\pi}{2},y=\frac{\pi}{2}+sin\frac{\pi}{2}\cos\frac{\pi}{2}=\frac{\pi}{2}\]
\[ \therefore \left( x_1 , y_1 \right) = \left( \frac{\pi}{2}, \frac{\pi}{2} \right)\]
\[\text { Equation of the normal }\]
\[ = y - y_1 = \frac{- 1}{\text { Slope of the tangent }}\left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{\pi}{2} = \frac{- 1}{0}\left( x - \frac{\pi}{2} \right)\]
\[ \Rightarrow x = \frac{\pi}{2}\]
\[ \Rightarrow 2x = \pi\]
