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  • If x equals a raised to sin inverse t , y equals a raised to cos inverse t , show that dy by dx = minu y upon x

Q11  If x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}, show that dy/dx = - y /x

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Given equations are
x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}
 

xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\ sin^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c

differentiating with respect to x

x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x} 

 

Posted by

Gautam harsolia

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