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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{3\text{at}}{1+\text{t}^2}\text{ and y}=\frac{3\text{at}^2}{1+\text{t}^2}$
✓

Answer

Here, $\text{x}=\frac{3\text{at}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotiont rule,
$\frac{\text{dx}}{\text{dt}}= \bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(3\text{at})-3\text{at}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(3\text{a})-3\text{at}(2\text{t})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(3\text{a})-3\text{at}(2\text{t})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\Big[\frac{3\text{a}-3\text{at}^{2}}{(1-\text{t}^{2})^{2}}\Big]$ And,
$\frac{\text{dx}}{\text{dt}}=\frac{3\text{a}(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}.....(\text{i})$
$\text{y}=\frac{3\text{at}^{2}}{(1+\text{t}^{2})}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(3\text{at}^{2})-3\text{at}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$\frac{\text{dy}}{\text{dt}}=\Big[\frac{(1+\text{t}^{2})(6\text{at})-(3\text{at}^{2})(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dt}}=\Big[\frac{6\text{at}+6\text{at}^{3}-6\text{at}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dt}}=\frac{6\text{at}}{(1+\text{t}^{2})^{2}}....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{6\text{at}}{(1+\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{3\text{a}(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{t}}{1-\text{t}^{2}}$

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