If x= a\sec ^{3}\theta \, and \, y= a\tan ^{3}\theta ,\: find\; \frac{d^{2}y}{dx^{2}}\cdot

 

 

 

 
 
 
 
 

Answers (1)

If x= a\sec ^{3}\theta \; \; y= a\tan ^{3}\theta
\frac{dy}{dx}= \left ( \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }} \right )      y= a\tan ^{3}\theta              \tan \theta = t
                             y= at^{3}                       \sec ^{2}\theta = \frac{dt}{d\theta }
                     \frac{dy}{d\theta }= a\frac{dt^{3}}{dt}\: \: \frac{dt}{d\theta }
           = a\, 3t^{2}\cdot \sec ^{2}\theta
           = 3a\tan ^{2}\theta \sec ^{2}\theta
x= a \sec ^{3}\theta          
                    \sec \theta = p,\frac{dp}{d\theta }= \sec \theta \tan \theta
x= ap^{3}
\frac{dx}{d\theta }= \frac{dx}{dp}\cdot \frac{dp}{d\theta }\Rightarrow 3ap^{2}\sec \theta \tan \theta
                          \Rightarrow 3\, a\sec ^{3}\theta \tan \theta
\frac{dy}{dx}= \frac{3a\tan ^{2}\theta \sec ^{2}\theta }{3a\sec ^{3}\theta \tan \theta }= \frac{\tan \theta }{\sec \theta }= \frac{\sin \theta }{\cos \theta }\cdot \frac{\cos \theta }{1}
                                              \frac{dy}{dx}=\sin \theta
\frac{dy}{dx}=\sin \theta
\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{d\theta }\left ( \sin \theta \right )}{\frac{dx}{d\theta }}= \frac{\cos \theta }{3a\sec ^{3}\theta \tan \theta }
       = \frac{\cos \theta \cdot \cos ^{3}\theta }{3a}\times \frac{\cos \theta }{\sin \theta }= \frac{\cos ^{5}\theta }{3a\sin \theta } 

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