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  • Try this! - Limit , continuity and differentiability - JEE Main-11

Let  x = (\theta + \cos\theta), y = (\theta - \sin\theta) then \frac{d^{2}y}{dx^{2}} equals ?

  • Option 1)

    \frac{\sin\theta + \cos\theta + 1}{(1-\sin\theta)^{3}}

  • Option 2)

    \frac{\sin\theta - \cos\theta + 1}{(1-\sin\theta)^{3}}

  • Option 3)

    \frac{\sin\theta - \cos\theta - 1}{(1-\sin\theta)^{3}}

  • Option 4)

    \frac{\sin\theta + \cos\theta - 1}{(1-\sin\theta)^{3}}

 

Answers (1)

best_answer

As we have learnt,

 

Second order derivative for parametric function -

When we find 

\frac{dy}{dx}=F(t)\:then\:\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\:F(t)}{\frac{dx}{dt}}

-

 

 

 

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{d}{d\theta}(\theta - \sin\theta)}{\frac{d}{d\theta}(\theta + \cos\theta)} = \frac{1-\cos\theta}{1 - \sin\theta}

\frac{d^2 y}{dx^2} = \frac{d}{dx}\left (\frac{1-\cos\theta}{1 -\sin\theta} \right ) =\frac{\frac{d}{d\theta}\left (\frac{1-\cos\theta}{1 -\sin\theta} \right )}{\frac{dx}{d\theta}}

\frac{d^2 y}{dx^2} = \frac{(1-\sin\theta)(\sin\theta)- (1-\cos\theta)(-\cos\theta)}{(1-\sin\theta)^3} = \frac{\sin\theta+\cos\theta -1 }{(1- \sin\theta)^3}


Option 1)

\frac{\sin\theta + \cos\theta + 1}{(1-\sin\theta)^{3}}

Option 2)

\frac{\sin\theta - \cos\theta + 1}{(1-\sin\theta)^{3}}

Option 3)

\frac{\sin\theta - \cos\theta - 1}{(1-\sin\theta)^{3}}

Option 4)

\frac{\sin\theta + \cos\theta - 1}{(1-\sin\theta)^{3}}

Posted by

Himanshu

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