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

• October 9th 2008, 09:18 PM
leungsta
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!
• October 10th 2008, 07:29 AM
Greengoblin
the cosine rule is $a^2=b^2+c^2-2bc\cos{A}$

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

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

$
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.