Find dy dx , when x= 2t 1+t^2 and y= 1-t^2 1+t^2 - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{2\text{t}}{1+\text{t}^2}\text{ and y}=\frac{1-\text{t}^2}{1+\text{t}^2}$

✓

Answer

Here, $\text{x}=\frac{2\text{t}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dx}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(2\text{t})-2\text{t}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(2)-2\text{t}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2+2\text{t}^{2}-4\text{t}^{2}}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2-2\text{t}^2}{(1+\text{t}^2)}\Big]$
$\frac{\text{dx}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}...(\text{i})$
And, $\text{y}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})-(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(-2\text{t})-(1-\text{t})^{2}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{-2\text{t}-2\text{t}^{3}-2\text{t}+2\text{t}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{-4\text{t}}{(1+\text{t}^{2})^{2}}\Big]...(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-4\text{t}}{(1-\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{2(1+\text{t}^{2})}$
$=\frac{-2\text{t}}{(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}} $
$\Big[\text{Since}\frac{\text{x}}{\text{y}}=\frac{2\text{t}}{1+\text{t}^{2}}\times\frac{1+\text{t}^{2}}{1-\text{t}^{2}}=\frac{2\text{t}}{1-\text{t}^{2}}\Big]$

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