Question

Find the angle of intersection between the curves $y^2=\frac{2x}{\pi }$ and $y=\sin x$ at $x=\frac{\pi }{2}$.

• A. ${\cos }^{-1}\pi$
• B. ${\cos }^{-1}\frac{\pi }{2}$
• C. ${\cos }^{-1}2\pi$
• D. ${\cos }^{-1}\frac{\pi }{3}$

Answer 1

The correct option is A ${\cos }^{-1}\pi$

$y^2=\frac{2x}{\pi }$ and $y=\sin x$
$\therefore {\sin }^{2}x=\frac{2x}{\pi }\Rightarrow x=\frac{\pi }{2}{\sin }^{2}x,y=\sin \left(\frac{\pi }{2}{\sin }^{2}x\right)$
Now $y^2=\frac{2x}{\pi }\Rightarrow 2y\frac{dy}{dx}=\frac{2}{\pi }\Rightarrow \frac{dy}{dx}=\frac{1}{\pi y}$
And $y=\sin x\Rightarrow \frac{dy}{dx}=\cos x$
$\tan \theta =\left|\frac{\frac{1}{\pi y}-\cos x}{1-\frac{1}{\pi y}\cos x}\right|=\left|\frac{\frac{1}{\pi \left(\frac{\pi }{2}{\sin }^{2}x\right)}-\cos \left(\frac{\pi }{2}{\sin }^{2}x\right)}{1+\frac{1}{\pi \sin \left(\frac{\pi }{2}{\sin }^{1}x\right)}\cos \left(\frac{\pi }{2}{\sin }^{2}x\right)}\right|$
$=\left|\frac{1-\pi \cos \left(\frac{\pi }{2}{\sin }^{2}x\right)\sin \left(\frac{\pi }{2}{\sin }^{2}x\right)}{1+\pi \cos \left(\frac{\pi }{2}{\sin }^{2}x\right)\sin \left(\frac{\pi }{2}{\cos }^{2}x\right)}\right|$
$\Rightarrow \theta ={\cos }^{-1}\pi$