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Math Help - Calculating partial derivatives by implicit differentiation of the Law of Cosines

  1. #1
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    Calculating partial derivatives by implicit differentiation of the Law of Cosines

    If a,b,c are sides of a triangle and A,B,C are the opposite angles, calculate бA/бa, бA/бb, бA/бc by implicit differentiation of the Law of Cosines.

    Any help would be appreciated!
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  2. #2
    Member Greengoblin's Avatar
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    the cosine rule is a^2=b^2+c^2-2bc\cos{A}

    1.) \frac{\partial A}{\partial a}

    \frac{\partial}{\partial a}a^2=\frac{\partial}{\partial a}(b^2+c^2-2bc\cos{A})

    <br />
2a=\frac{\partial}{\partial a} (-2bc\cos{A})

    2a=A'2bc\sin{A}

    A'=\frac{2a}{2bc\sin{A}} = \frac{a}{bc\sin{A}}



    2.) \frac{\partial A}{\partial b}

    0=\frac{\partial}{\partial b}(b^2+c^2-2bc\cos{A})

    First the tricky bit:

    let u=2bc \text{ and } v=-\cos{A}

    then \frac{\partial}{\partial b}(-2bc\cos{A})=v\frac{\partial u}{\partial b}+u\frac{\partial v}{\partial b} = -2c\cos{A}+A'2bc\sin{A}

    0=2b-2c\cos{A}+A'2bc\sin{A}

    -A'2bc\sin{A}=2b-2c\cos{A}

    A'=\frac{-2b+2c\cos{A}}{2bc\sin{A}}=\frac{-b+c\cos{A}}{bc\sin{A}}


    3.) \frac{\partial A}{\partial c}

    0=\frac{\partial}{\partial c}(b^2+c^2-2bc\cos{A})

    similar to number 2, we use the product rule again on -2bc\cos{A} but this time with respect to c...this time yielding:

    0=2c-2b\cos{A}+A'2bc\sin{A}

    -A'2bc\sin{A}=2c-2b\cos{A}

    A'=\frac{-2c+2b\cos{A}}{2bc\sin{A}}=\frac{-c+b\cos{A}}{bc\sin{A}}

    P.S. Someone more experienced might want to check this for you. I think it's ok though.
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