Question

Find $\frac{dy}{dx}$, when
i) $x=a{t}^2$ and $y=2at$
ii) $x=a(\theta +\sin \theta )$ and $y=a(1-\cos \theta )$.

Answer 1

i)

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{\frac{d}{dt}\left(2at\right)}{\frac{d}{dt}\left(a{t}^2\right)}=\frac{2a}{2at}=\frac{1}{t}$

ii)

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

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

$=\frac{a\sin \theta }{a(1+\cos \theta )}$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{1+2{\cos }^2\frac{\theta }{2}-1}$

$=\tan \frac{\theta }{2}$