Question

Find $\frac{dy}{dx}$, when

$x=\frac{2t}{1+t^2}\mathrm{and}y=\frac{1-t^2}{1+t^2}$

Answer 1

$\mathrm{We}\mathrm{have},x=\frac{2t}{1+t^2}$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=\left[\frac{\left(1+t^2\right){\displaystyle \frac{d}{dt}}\left(2t\right)-2t{\displaystyle \frac{d}{dt}}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\right]\left[\mathrm{using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{\left(1+t^2\right)\left(2\right)-2t\left(2t\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{2+2t^2-4t^2}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{2-2t^2}{{\left(1+t^2\right)}^2}\right]...\left(\mathrm{i}\right) \\ \mathrm{and}, \\ y=\frac{1-t^2}{1+t^2} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dt}=\left[\frac{\left(1+t^2\right){\displaystyle \frac{d}{dt}}\left(1-t^2\right)-\left(1-t^2\right){\displaystyle \frac{d}{dt}}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dt}=\left[\frac{\left(1+t^2\right)\left(-2t\right)-\left(1-t^2\right)\left(2t\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dt}=\left[\frac{-4t}{{\left(1+t^2\right)}^2}\right]...\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}, \\ \frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{-4t}{{\left(1+t^2\right)}^2}\times \frac{{\left(1+t^2\right)}^2}{2\left(1-t^2\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{-2t}{1-t^2} \\ \Rightarrow \frac{dy}{dx}=-\frac{x}{y}\left[\because \frac{x}{y}=\frac{2t}{1+t^2}\times \frac{1+t^2}{1-t^2}=\frac{2t}{1-t^2}\right] \end{gathered}$$