Question

If $x=10\left(t-\sin t\right),y=12\left(1-\cos t\right),\mathrm{find}\frac{dy}{dx}.$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},x=10\left(t-\sin t\right)\mathrm{and}y=12\left(1-\cos t\right) \\ \Rightarrow \frac{dx}{dt}=\frac{d}{dt}\left[10\left(t-\sin t\right)\right]\mathrm{and}\frac{dy}{dt}=\frac{d}{dt}\left[12\left(1-\cos t\right)\right] \\ \Rightarrow \frac{dx}{dt}=10\frac{d}{dt}\left(t-\sin t\right)\mathrm{and}\frac{dy}{dt}=12\frac{d}{dt}\left(1-\cos t\right) \\ \Rightarrow \frac{dx}{dt}=10\left(1-\cos t\right)\mathrm{and}\frac{dy}{dt}=12\left[0-\left(-\sin t\right)\right]=12\sin t \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{12\sin t}{10\left(1-\cos t\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{12\times 2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}{10\times 2{\sin }^2{\displaystyle \frac{t}{2}}} \\ \Rightarrow \frac{dy}{dx}=\frac{6}{5}\cot \frac{t}{2} \end{gathered}$$