# Solve this

Question:

Find $\frac{d y}{d x}$, when

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

Solution:

We have, $x=\frac{3 a t}{1+t^{2}}$

Differentiating with respect to $t$,

$\frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}(3 a t)-3 a t \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$        [using quotient rule]

$\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right)(3 a)-3 a t(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\left[\frac{3 a+3 a t^{2}-6 a t^{2}}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\left[\frac{3 a-3 a t^{2}}{\left(1-t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$

and, $y=\frac{3 a t^{2}}{1+t^{2}}$                    $\ldots(i)$

Differentiating it with respect to t,

$\frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}\left(3 a t^{2}\right)-3 a t^{2} \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$           [using quotient rule]

$\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right)(6 a t)-3 a t^{2}(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\left[\frac{6 a t+6 a t^{3}-6 a t^{3}}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\frac{6 a t}{\left(1+t^{2}\right)^{2}}$          $\ldots(i i)$

Dividing equation (ii) by (i),

$\frac{\frac{d y}{d t}}{\frac{d x}{d x}}=\frac{6 a t}{\left(1+t^{2}\right)^{2}} \times \frac{\left(1+t^{2}\right)^{2}}{3 a\left(1-t^{2}\right)}=\frac{2 t}{1-t^{2}}$