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!
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!
the cosine rule is $\displaystyle a^2=b^2+c^2-2bc\cos{A}$
1.) $\displaystyle \frac{\partial A}{\partial a}$
$\displaystyle \frac{\partial}{\partial a}a^2=\frac{\partial}{\partial a}(b^2+c^2-2bc\cos{A})$
$\displaystyle
2a=\frac{\partial}{\partial a} (-2bc\cos{A})$
$\displaystyle 2a=A'2bc\sin{A}$
$\displaystyle A'=\frac{2a}{2bc\sin{A}} = \frac{a}{bc\sin{A}}$
2.) $\displaystyle \frac{\partial A}{\partial b}$
$\displaystyle 0=\frac{\partial}{\partial b}(b^2+c^2-2bc\cos{A})$
First the tricky bit:
let $\displaystyle u=2bc \text{ and } v=-\cos{A}$
then $\displaystyle \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} $
$\displaystyle 0=2b-2c\cos{A}+A'2bc\sin{A}$
$\displaystyle -A'2bc\sin{A}=2b-2c\cos{A}$
$\displaystyle A'=\frac{-2b+2c\cos{A}}{2bc\sin{A}}=\frac{-b+c\cos{A}}{bc\sin{A}}$
3.) $\displaystyle \frac{\partial A}{\partial c}$
$\displaystyle 0=\frac{\partial}{\partial c}(b^2+c^2-2bc\cos{A})$
similar to number 2, we use the product rule again on $\displaystyle -2bc\cos{A}$ but this time with respect to c...this time yielding:
$\displaystyle 0=2c-2b\cos{A}+A'2bc\sin{A}$
$\displaystyle -A'2bc\sin{A}=2c-2b\cos{A}$
$\displaystyle A'=\frac{-2c+2b\cos{A}}{2bc\sin{A}}=\frac{-c+b\cos{A}}{bc\sin{A}}$
P.S. Someone more experienced might want to check this for you. I think it's ok though.