Question

find dy/dx if x = a (1-t^2/1+t^2), y= 2bt/1+t^2

(do it in {d/dt} form and send solution)​

Answer 1

Answer:

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Answer 2

Given,

$\[x = a\left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right)\]$ and $\[y = 2b\left( {\frac{t}{{1 + {t^2}}}} \right)\]$

Solution,

$\[y = 2b\left( {\frac{t}{{1 + {t^2}}}} \right)\]$

Differentiate with respect to t.

$\[ \Rightarrow \frac{{dy}}{{dt}} = 2b \times \frac{{\left( {1 + {t^2}} \right) \times 1 - t \times \left( {2t} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}\]$

$\[ \Rightarrow \frac{{dy}}{{dt}} = 2b \times \frac{{\left( {1 + {t^2}} \right) - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}\]$

$\[ \Rightarrow \frac{{dy}}{{dt}} = 2b \times \frac{{1 - {t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}\]$----------(1)

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

Differentiate with respect to t.

$\[ \Rightarrow \frac{{dx}}{{dt}} = a \times \frac{{\left( {1 + {t^2}} \right) \times \left( { - 2t} \right) - \left( {1 - {t^2}} \right) \times \left( {2t} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}\]$

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

$\[ \Rightarrow \frac{{dx}}{{dt}} = a \times \left( {\frac{{ - 4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right)\]$----------------(2)

Divide equation 1 by equation 2.

$\[\frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{2b \times \frac{{1 - {t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}}}{{a \times \left( {\frac{{ - 4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right)}}\]$

$\[\frac{{dy}}{{dx}} = \frac{{2b \times \left( {1 - {t^2}} \right)}}{{a \times \left( { - 4t} \right)}}\]$

$\[\frac{{dy}}{{dx}} = \frac{{b\left( {1 - {t^2}} \right)}}{{2at}}\]$

Hence the value is $\[\frac{{b\left( {1 - {t^2}} \right)}}{{2at}}\]$