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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[u=-f(\theta )\sin \theta +f(\theta )\cos \theta \]and \[v=f(\theta )\cos \theta +f(\theta )\sin \theta \],then \[{{\int{\left[ {{\left( \frac{du}{d\theta } \right)}^{2}}+{{\left( \frac{dv}{d\theta } \right)}^{2}} \right]}}^{\frac{1}{2}}}d\theta \]is equal to

    A)  \[f(\theta )-f(\theta )+c\]          

    B)  \[f(\theta )+f(\theta )+c\]

    C)  \[f(\theta )+f(\theta )+c\]         

    D)  \[f(\theta )-f(\theta )+c\]

    E)  \[f(\theta )+f(\theta )+c\]         

    Correct Answer: B

    Solution :

    Given,                   \[u=-f(\theta )\sin \theta +f(\theta )\cos \theta \] and        \[v=f(\theta )\cos \theta +f(\theta )\sin \theta \] On differentiating w.r.t.\[\theta \]respectively, we get \[\frac{du}{d\theta }=-f\sin \theta -f(\theta )\sin \theta \]                                 \[-f(\theta )\sin \theta \] \[=-f(\theta )\sin \theta -f(\theta )\sin \theta \] and \[\frac{dv}{d\theta }=f(\theta )\cos \theta -f(\theta )\sin \theta \]                                 \[+f(\theta )\sin \theta +f(\theta )\cos \theta \]                 \[=f(\theta )\cos \theta +f(\theta )\cos \theta \] \[\therefore \] \[{{\left( \frac{du}{d\theta } \right)}^{2}}+{{\left( \frac{dv}{d\theta } \right)}^{2}}=[f{{(\theta )}^{2}}+{{[f(\theta )]}^{2}}\]       \[+2f(\theta )f(\theta )\]                                 \[=\int{[f(\theta )+f(\theta )]}d\theta \]                                 \[=f(\theta )+f(\theta )+c\]


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