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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    If \[x=\sin t\] and \[y=\sin pt\], then the value of  \[\left( 1-{{x}^{2}} \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}+{{p}^{2}}y\]is equal to [Pb. CET 2002]

    A) 0

    B) 1

    C) -1

    D) \[\sqrt{2}\]

    Correct Answer: A

    Solution :

    • Given that \[x=\sin t\], \[y=\sin pt\]           
    • \[\frac{dx}{dt}=\cos t\], \[\frac{dy}{dt}=p\cos pt\]           
    • \ \[\frac{dy}{dx}=\frac{p\cos pt}{\cos t}=\frac{p\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\]                   
    • Again differentiate w.r.t. x,           
    • \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{p\sqrt{1-{{x}^{2}}}.\frac{1}{2\sqrt{1-{{y}^{2}}}}.(-2y)\frac{dy}{dx}-p\sqrt{1-{{y}^{2}}}.\frac{1}{2\sqrt{1-{{x}^{2}}}}(-2x)}{(1-{{x}^{2}})}\]           
    • \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=-py\frac{\sqrt{1-{{x}^{2}}}}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}+px\frac{\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\]            \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=-{{p}^{2}}y+x\frac{dy}{dx}\]                   
    • \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}+{{p}^{2}}y=0\]-


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