Question

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

$x=\frac{3at}{1+t^2},\mathrm{and}y=\frac{3a{t}^2}{1+t^2}$

Answer 1

$\mathrm{We}\mathrm{have},x=\frac{3at}{1+t^2}$
Differentiating with respect to t,
$$\begin{gathered} \frac{dx}{dt}=\left[\frac{\left(1+t^2\right){\displaystyle \frac{d}{dt}}\left(3at\right)-3at{\displaystyle \frac{d}{dt}}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\right]\left[\mathrm{using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{\left(1+t^2\right)\left(3a\right)-3at\left(2t\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{3a+3a{t}^2-6a{t}^2}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{3a-3a{t}^2}{{\left(1-t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\frac{3a\left(1-t^2\right)}{{\left(1+t^2\right)}^2}...\left(i\right) \\ \mathrm{and},y=\frac{3a{t}^2}{1+t^2} \end{gathered}$$
Differentiating it with respect to t,

$$\begin{gathered} \frac{dx}{dt}=\left[\frac{\left(1+t^2\right){\displaystyle \frac{d}{dt}}\left(3a{t}^2\right)-{\displaystyle 3a{t}^2\frac{d}{dt}}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\right]\left[\mathrm{using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{\left(1+t^2\right)\left(6at\right)-3a{t}^2\left(2t\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{6at+6a{t}^3-6a{t}^3}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\frac{6at}{{\left(1+t^2\right)}^2}...\left(ii\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{i}\right), \\ \frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{6at}{{\left(1+t^2\right)}^2}\times \frac{{\left(1+t^2\right)}^2}{3a\left(1-t^2\right)}=\frac{2t}{1-t^2} \end{gathered}$$