Question

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

Answer 1

$$\text{ Here,}$$

$$y=x+\sin x\cos x$$

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

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

$$\text{ Now,}$$

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

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

$$\therefore (x_1,y_1)=(\frac{\pi }{2},\frac{\pi }{2})$$

$$\text{ Equation of the normal }$$

$$=y-y_1=\frac{-1}{\text{ Slope of the tangent }}(x-x_1)$$

$$\Rightarrow y-\frac{\pi }{2}=\frac{-1}{0}(x-\frac{\pi }{2})$$

$$\Rightarrow x=\frac{\pi }{2}$$

$$\Rightarrow 2x=\pi$$