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Thread: derivative

  1. #1
    Member kjchauhan's Avatar
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    derivative

    Please help me to solve this
    If $\displaystyle sin^2 A+sin^2 B+sin^2 C=K$ then obtain $\displaystyle \frac{dB} {dC}$
    Thanks in advance
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  2. #2
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    Re: derivative

    let $\lambda = K-\sin^2(A)$

    assume $\dfrac{dA}{dC} = 0$

    $\sin^2(C) = \lambda - \sin^2(B)$

    $\dfrac{d}{dC} \sin^2(C) = \dfrac{d}{dC} (\lambda - \sin^2(B))$

    $2 \sin(C)\cos(C) = -2\sin(B)\cos(B)\dfrac{dB}{dC}$

    $\dfrac{dB}{dC} =- \dfrac{ \sin(C)\cos(C)}{\sin(B)\cos(B)}=-\dfrac{\sin(2C)}{\sin(2B)}$
    Thanks from topsquark and Vinod
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  3. #3
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    Re: derivative

    Use implicit differentiation. Differentiate both sides with respect to C.
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  4. #4
    Senior Member Vinod's Avatar
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    Re: derivative

    Quote Originally Posted by romsek View Post
    let $\lambda = K-\sin^2(A)$

    assume $\dfrac{dA}{dC} = 0$

    $\sin^2(C) = \lambda - \sin^2(B)$

    $\dfrac{d}{dC} \sin^2(C) = \dfrac{d}{dC} (\lambda - \sin^2(B))$

    $2 \sin(C)\cos(C) = -2\sin(B)\cos(B)\dfrac{dB}{dC}$

    $\dfrac{dB}{dC} =- \dfrac{ \sin(C)\cos(C)}{\sin(B)\cos(B)}=-\dfrac{\sin(2C)}{\sin(2B)}$
    Hello,
    Assumption of $\frac{dA}{dC}$ is necessary.Without that we cann't compute $\frac{dB}{dC}$
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  5. #5
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    Re: derivative

    Quote Originally Posted by Vinod View Post
    Hello,
    Assumption of $\frac{dA}{dC}$ is necessary.Without that we cann't compute $\frac{dB}{dC}$
    That is not true.

    $$\sin 2A\cdot \dfrac{dA}{dC}+\sin 2B\cdot \dfrac{dB}{dC}+\sin 2C=\dfrac{dK}{dC}$$

    $$\dfrac{dB}{dC} = \dfrac{\dfrac{dK}{dC}-\sin 2A\cdot \dfrac{dA}{dC}-\sin 2C}{\sin 2B}$$
    Thanks from topsquark and Vinod
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