Question

If \[x = \cos t \text{ and y }  = \sin t,\] prove that  \[\frac{dy}{dx} = \frac{1}{\sqrt{3}} \text { at } t = \frac{2 \pi}{3}\] ?

 

Answer 1

\[\text{ We have, x } = \cos t \text{ and y } = \sin t\]

\[\Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left( \cos t \right) \text { and } \frac{dy}{dt} = \frac{d}{dt}\left( \sin t \right)\]
\[ \Rightarrow \frac{dx}{dt} = - \sin t \text{ and } \frac{dy}{dt} = \cos t \]
\[ \therefore \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{- \sin t} = - \cot t \]
\[\text{ Now,} \left( \frac{dy}{dx} \right)_{t = \frac{2\pi}{3}} = - \cot \left( \frac{2\pi}{3} \right) = \frac{1}{\sqrt{3}} \]