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Mathematics Question on Continuity and differentiability

For a>0,t∈(0,π2),a > 0 , t \in\left(0, \frac{\pi}{2}\right) , let x=asin⁡−1tx = \sqrt{a^{\sin^{-1}t}} and y=acos⁡−1t,y = \sqrt{a^{\cos^{-1}t}}, Then 1+(dydx)21+ \left(\frac{dy}{dx}\right)^{2} equals :

A

x2y2\frac{x^2}{y^2}

B

y2x2\frac{y^2}{x^2}

C

x2+y2y2\frac{x^2 + y^2 }{y^2}

D

x2+y2x2\frac{x^2 + y^2 }{x^2}

Answer

x2+y2x2\frac{x^2 + y^2 }{x^2}

Explanation

Solution

Let x=asin⁡−1tx = \sqrt{a^{\sin^{-1}t }} ⇒x2=asin⁡−1t⇒2log⁡x=sin⁡−1t.log⁡a\Rightarrow x^{2} = a^{\sin^{-1}t} \Rightarrow 2 \log x = \sin^{-1} t . \log a ⇒2x=log⁡a1−t2.dtdx\Rightarrow \frac{2}{x} = \frac{\log a}{\sqrt{1-t^{2}}} . \frac{dt}{dx} ⇒21−t2xlog⁡a=dtdx\Rightarrow \frac{2\sqrt{1-t^{2}}}{x \log a} = \frac{dt}{dx} ....(1) Now, let y=acos⁡−1ty = \sqrt{a^{\cos^{-1}t}} ⇒2log⁡y=cos⁡−1t.log⁡a\Rightarrow 2 \log y = \cos^{-1} t . \log a ⇒2y.dydx=−log⁡a1−t2.dtdx\Rightarrow \frac{2}{y} . \frac{dy}{dx} = \frac{-\log a}{\sqrt{1-t^{2}}} . \frac{dt}{dx} ⇒2y.dydx=−log⁡a1−t2×21−t2xlog⁡a\Rightarrow \frac{2}{y} . \frac{dy}{dx} = \frac{-\log a}{\sqrt{1-t^{2}} } \times \frac{2\sqrt{1-t^{2}}}{x \log a} (from (1) ⇒dydx=−yx\Rightarrow \frac{dy}{dx} = - \frac{y}{x} Hence , 1+(dydx)2=1+(−yx)2=x2+y2x21+ \left(\frac{dy}{dx}\right)^{2} = 1 + \left(\frac{-y}{x}\right)^{2} = \frac{x^{2} +y^{2}}{x^{2}}