Then
KH2 = OH2 − OK2,
(10)
OK2 sin2 θ = CC′2 = G2 − 2GG′ cos θ + G′2,
(11)
A2 sin2 θ (dθ/dt)2 = 2A2n2 (F − cos θ) sin2 θ − G2 + 2GG′ cos θ − G′2;
(12)
and putting cos θ = z,
| ( | dz | ) = 2n2 (F − z) (1 − z2) − (G2 − 2GG′z + G′2) /A2 |
| dt |
(13)
Then
KH2 = OH2 − OK2,
(10)
OK2 sin2 θ = CC′2 = G2 − 2GG′ cos θ + G′2,
(11)
A2 sin2 θ (dθ/dt)2 = 2A2n2 (F − cos θ) sin2 θ − G2 + 2GG′ cos θ − G′2;
(12)
and putting cos θ = z,
| ( | dz | ) = 2n2 (F − z) (1 − z2) − (G2 − 2GG′z + G′2) /A2 |
| dt |
(13)