Understanding the Differentiation of the Cos Law Equation

  • Thread starter Thread starter mholland
  • Start date Start date
  • Tags Tags
    Cos Law
mholland
Messages
5
Reaction score
0
I'm struggling here as I've not done diferentiation in a few years.
The cos law states:
c=(a2+b2-2abcosθ)1/2

I'm trying to figure out how to differentiate this, so if c were a length, what the velocity with which c grows as θ increases (ie c dot)

Any pointers would be great!

If its easier, a and b are fixed lengths, so the equation simplifies to
c=(a-bcosθ)1/2
 
Last edited:
Physics news on Phys.org
is it just
c dot=0.5(a-bcosθ)-1/2.(bsinθ) ?
 
Yes, it's just an application of the basic differentiation laws, particularly the chain rule.

The deriviative of u^{1/2}, with respect to u, is (1/2)u^{-1/2}. The derivative of 1- v, with respect to v is -1, and, finally, the derivative of bcos(\theta), with respect to \theta, is -bsin(\theta).

Putting those together, using the chain rule, the derivative of c is
(1/2)(a- bcos(\theta)^{1/2}(-1)(-bsin(\theta)]<br /> which gives what you say.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top