Question

If $x=asin2t(1+cos2t)$ and $y=bcos2t(1-cos2t),$ then find the values of $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$ and $t=\frac{\pi }{3}.$

Answer 1

$x=a\sin 2t(1+\cos 2t)$

$\frac{dx}{dt}=a[\sin 2t\frac{d}{dt}(1+\cos 2t)+(1+\cos 2t)\frac{d}{dt}\left(\sin 2t\right)]$

$\frac{dx}{dt}=a[-2{\sin }^{2}2t+(1+\cos 2t)\left(2\cos 2t\right)]$

$\frac{dx}{dt}=a[-2{\sin }^{2}2t+2\cos 2t+2{\cos }^{2}2t]$

$\frac{dx}{dt}=a[2\cos 2t+2({\cos }^{2}2t-{\sin }^{2}2t)]$

$\frac{dx}{dt}=a[2\cos 2t+2\cos 4t]$

$\frac{dx}{dt}=2a(\cos 2t+\cos 4t)$

$y=b\cos 2t(1-\cos 2t)$

$\frac{dy}{dt}=b[\cos 2t\frac{d}{dt}(1-\cos 2t)+(1-\cos 2t)\frac{d}{dt}\left(\cos 2t\right)]$

$\frac{dy}{dt}=b[2\sin 2t\cos 2t-2\sin 2t(1-\cos 2t)]$

$\frac{dy}{dt}=b[2\sin 2t\cos 2t-2\sin 2t+2\sin 2t\cos 2t]$

$\frac{dy}{dt}=b[4\sin 2t\cos 2t-2\sin 2t]$

$\frac{dy}{dt}=2b[2\sin 2t\cos 2t-\sin 2t]$

$\frac{dy}{dt}=2b(\sin 4t-\sin 2t)$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2b(\sin 4t-\sin 2t)}{2a(\cos 2t+\cos 4t)}$

$\Rightarrow \frac{dy}{dx}=\frac{b(\sin 4t-\sin 2t)}{a(\cos 2t+\cos 4t)}$

$\Rightarrow {\frac{dy}{dx}}_{t=\frac{\pi }{4}}=\frac{b(\sin \frac{4\pi }{4}-\sin \frac{2\pi }{4})}{a(\cos \frac{2\pi }{4}+\cos \frac{4\pi }{4})}$

$=\frac{b(\sin \pi -\sin \frac{\pi }{2})}{a(\cos \frac{\pi }{2}+\cos \pi )}$

$=\frac{b(0-1)}{a(0-1)}=\frac{b}{a}$

$\Rightarrow {\frac{dy}{dx}}_{t=\frac{\pi }{3}}=\frac{b(\sin \frac{4\pi }{3}-\sin \frac{2\pi }{3})}{a(\cos \frac{2\pi }{3}+\cos \frac{4\pi }{3})}$

$=\frac{b(\sin (\pi +\frac{\pi }{3})-\sin (\pi -\frac{\pi }{3}))}{a(\cos (\pi -\frac{\pi }{3})+\cos (\pi +\frac{\pi }{3}))}$

$=\frac{b(-\sin \frac{\pi }{3}-\sin \frac{\pi }{3})}{a(-\cos \frac{\pi }{3}-\cos \frac{\pi }{3})}$

$=\frac{-2b\sin \frac{\pi }{3}}{-2a\cos \frac{\pi }{3}}$

$=\frac{b\tan \frac{\pi }{3}}{a}=\frac{\sqrt{3}b}{a}$