- #1
ryank614
- 10
- 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[tex]\Theta[/tex]+ (h^2-R^2) = 0
2 r [tex]\dot{}r[/tex] - 2 h cos[tex]\Theta[/tex] [tex]\dot{}r[/tex] + (h^2-R^2) = 0.
Is there any way to put h^2-R^2 in terms of [tex]\dot{}\theta[/tex] or d[tex]\theta[/tex]/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[tex]\Theta[/tex]+ (h^2-R^2) = 0
2 r [tex]\dot{}r[/tex] - 2 h cos[tex]\Theta[/tex] [tex]\dot{}r[/tex] + (h^2-R^2) = 0.
Is there any way to put h^2-R^2 in terms of [tex]\dot{}\theta[/tex] or d[tex]\theta[/tex]/dt