- #1
ryank614
- 8
- 0
It is a simple math question, but I am stuck.
The law of cosines is
R^2=h^2 + r^2 - 2 h r cos (theta). Theta is of course the angle facing R.
To differentiate, I first set
r^2 - 2hr cos\Theta+ (h^2-R^2) = 0
2 r \dot{}r - 2 h cos\Theta \dot{}r + (h^2-R^2) = 0.
Is there any way to put h^2-R^2 in terms of \dot{}\theta or d\theta/dt
The law of cosines is
R^2=h^2 + r^2 - 2 h r cos (theta). Theta is of course the angle facing R.
To differentiate, I first set
r^2 - 2hr cos\Theta+ (h^2-R^2) = 0
2 r \dot{}r - 2 h cos\Theta \dot{}r + (h^2-R^2) = 0.
Is there any way to put h^2-R^2 in terms of \dot{}\theta or d\theta/dt
