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Find the equation of the tangent and normal to the curves.$\;y=x-\sin x\cos x, $at$\;x=\large\frac{\pi}{2}$

Note: This is part 2nd of a 4 part question, split as 4 separate questions here.

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  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
$y=x -\sin x \cos x$ at $x=\large\frac{\pi}{2}$
Step 1:
When $x=\large\frac{\pi}{2}$$,y=\large\frac{\pi}{2}$$-\sin \large\frac{\pi}{2} $$\cos \large\frac{\pi}{2}=\frac{\pi}{2}$
We are required to find the equations of the tangent and normal at $\bigg(\large\frac{\pi}{2},\frac{\pi}{2}\bigg)$
Step 2:
Slope of the tangent at $\bigg(\large\frac{\pi}{2},\frac{\pi}{2}\bigg)$ is $\large\frac{dy}{dx} \bigg(\large\frac{\pi}{2},\frac{\pi}{2}\bigg)$
$y=x-\sin x \cos x$ or $y=x-\large\frac{1}{2} $$\sin 2x$
$\large\frac{dy}{dx}$$=1-\large\frac{1}{2} $$.2 \cos 2x=1-\cos 2x$
equation of the tangent $\bigg(\large\frac{\pi}{2},\frac{\pi}{2}\bigg)$
$\bigg(y-\large\frac{\pi}{2}\bigg)$$=2 (x-\large\frac{\pi}{2})$
$=> 2x -y-\large\frac{\pi}{2}$$=0$
Step 3:
Normal at $\bigg(\large\frac{\pi}{2},\frac{\pi}{2}\bigg)$ is $\perp$ to the tangent
Slope $=\large\frac{-1}{m}=\frac{-1}{2}$
Equation of the normal at $\bigg(\large\frac{\pi}{2},\frac{\pi}{2}\bigg)$
$\bigg(y-\large\frac{\pi}{2}\bigg)$$=\large\frac{-1}{2} $$\bigg(x-\large\frac{\pi}{2}\bigg)$
$2y-\pi=-x+\large\frac{\pi}{2}$ or
$x+2y-\large\frac{3 \pi}{2}$$=0$
answered Jul 23, 2013 by meena.p

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