Derivative of ##\theta## w.r.t. ##a##: A Detailed Explanation

[a.mlw.walker]

Big derivative, just want to make sure I am doing correct thing here.

a is the only changing dimension, r and l are constants

$\theta=\arccos\left(\frac{r^{2}+\left(r+l-a\right)^{2}-l^{2}}{2r\left(r+l-a\right)}\right)$

I want to differentiate $\frac{d\theta}{da}$

So what I did was using

$\theta=\arccos a$

$cos\theta=a$ differentiate this$-sin\theta \frac{d\theta}{da}=1$therefore with trig identities and a rearrange:

$\frac{d\theta}{da}=\frac{-1}{\sqrt{1-a^{2}}}$

where $a = \left(\frac{r^{2}+\left(r+l-a\right)^{2}-l^{2}}{2r\left(r+l-a\right)}\right)$

So the derivative of theta with respect to a is

$\frac{d \theta}{da}=\frac{-a'} {\sqrt{1-a^2}}$

This involves the quotient rule, and I end up the expression below: I took out a factor of 4 top and bottom of the derivative of a, hence the 3/2 coefficient

$\frac{d\theta}{da}= \frac{ra^2 - (2r^2 - rl )a + (2r^{3}+3lr^{2}$
$+l^{2}r)}{ra^{3}-\frac{3}{2}r^{2}a^{2}+(2r^{3}+3lr^{2}+rl^{2})a-(r^{4}-2r^{3}l-r^{2}l^{2})}$
$\sqrt{1-(\frac{a^{2}-2a(r-l )+(2r^{2}+2rl)}{-2ra + 2r^{2} +2rl})$

Thanks

 

[Mark44]

In a nutshell, for the derivative you need to use the following rules, in this order:

1. Use the chain rule form of the derivative with respect to u of $\arccos u$, where u is the rational expression in the parentheses. $\frac d{dx}\left(\arccos(u)\right) = \frac {-1}{\sqrt{1 - u^2}}\frac {du}{dx}$
2. Use the quotient rule to take the deriviative with respect to a of the rational expression. This will be a little bit messy, but not too bad if you're careful.

In the previous post, the last LaTeX was broken. I attempted to fix it, but gave up.