Big differential

1. Oct 13, 2009

a.mlw.walker

Big differential, 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 $$\frac{\delta\Theta}{\delta\ a}$$

So what I did was using

$$\Theta=arccos a$$

$$cos\Theta=a$$ differentiate this

$$-sin\Theta \frac{\delta\Theta}{\delta\ a}=1$$

therefore with trig identities and a rearrange:

$$\frac{\delta\Theta}{\delta\ a}=\frac{-1}{\sqrt{1-a^{2}}}$$

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

So the differential of theta with respect to a is

$$\frac{\delta\Theta}{\delta\ a}=\frac{-a^{'}}{\sqrt{1-a^{2}}}$$

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

$$\frac{\delta\Theta}{\delta\ a}= \frac{ra^{2}-\left(2r^{2}-rl\right)a + \left(2r^{3}+3lr^{2}+l^{2}r\right)}{ra^{3}-\frac{3}{2}r^{2}a^{2}+\left(2r^{3}+3lr^{2}+rl^{2}\right)a-\left(r^{4}-2r^{3}l-r^{2}l^{2}\right)} / \sqrt{1-\left(\frac{a^{2}-2a\left(r-l\right)+\left(2r^{2}+2rl\right)}{-2ra + 2r^{2} +2rl}\right)$$

Thanks

Last edited: Oct 13, 2009