Question

Write the equation of the normal to the curve y = x + sin x cos x at $x=\frac{\mathrm{\pi }}{2}$.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ y=x+\sin x\cos x \\ \mathrm{On}\mathrm{d}\text{ifferentiating both sides w.r.t.}x,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=1+{\cos }^{2}x-{\sin }^{2}x \\ \mathrm{Now}, \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{2}}{\text{= 1+cos}}^2\frac{\mathrm{\pi }}{2}{\text{-sin}}^2\frac{\mathrm{\pi }}{2}\text{= 1-1=0} \\ \text{When}\mathit{\text{x}}\text{=}\frac{\mathrm{\pi }}{2}\text{,}\mathit{\text{y}}\text{=}\frac{\mathrm{\pi }}{2}\text{+sin}\frac{\mathrm{\pi }}{2}\text{cos}\frac{\mathrm{\pi }}{2}\text{=}\frac{\mathrm{\pi }}{2} \\ \therefore \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \\ \text{Equation of the normal} \\ =y-y_1=\frac{-1}{\mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{tangent}}\left(x-x_1\right) \\ \Rightarrow y-\frac{\mathrm{\pi }}{2}=\frac{-1}{0}\left(x-\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow 2x=\mathrm{\pi } \end{gathered}$$