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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{If x}=\sqrt{\text{a}^{\sin^{-1}}\text{t}},\text{y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}},\text{ Show that}\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
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Answer

The given equations are $\text{x}=\sqrt{\text{a}^{\sin{-1}}\text{t}}\text{ and y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}}$
$$$\text{x}=\sqrt{\text{a}^{\sin^{-1}}\text{t}}\text{ and y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}}$
$\Rightarrow\ \text{x}=\Big({\text{a}^{\sin^{-1}}\text{t}\Big)}^{\frac{1}{2}}\text{ and y}=\Big({\text{a}^{\cos^{-1}}\text{t}\Big)^{\frac{1}{2}}}$
$\Rightarrow\ \text{x}=\text{a}^{\frac{1}{2}\sin^{-1}\text{t}}\text{ and y}=\text{a}^{\frac{1}{2}\cos^{-1}\text{t}}$
Consider $\text{x}=\text{a}^{\frac{1}{2}\sin^{-1}\text{t}}$
Taking logarithm on both the sides, we obtain
$\log\text{x}=\frac{1}{2}\sin^{-1}\text{t}\log\text{a}$
$\therefore \frac{1}{\text{x}}.\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\log\text{a}.\frac{\text{d}}{\text{dt}}(\sin^{-1}\text{t})$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{x}}{2}\log\text{a}.\frac{1}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{x}\log\text{a}}{2\sqrt{1-\text{t}^2}}$
Then, consider $\text{a}^{\frac{1}{2}\cos^{-1}\text{t}}$
Taking logarithm on both the sides, we obtain
$\log\text{y}=\frac{1}{2}\cos^{-1}\text{t}\log\text{a}$
$\therefore \frac{1}{\text{y}}.\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\log\text{a}.\frac{\text{d}}{\text{dt}}(\cos^{-1}\text{t})$
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{-\text{y}\log\text{a}}{2}.\Big(\frac{1}{\sqrt{1-\text{t}^2}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{-\text{y}\log\text{a}}{2\sqrt{1-\text{t}^2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dx}}\Big)}=\frac{\Big(\frac{-\text{y}\log\text{a}}{2\sqrt{1-\text{t}^2}}\Big)}{\Big(\frac{\text{x}\log\text{a}}{2\sqrt{1-\text{t}^2}}\Big)}=-\frac{\text{y}}{\text{x}}.$
Hence, proved.

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