00:01
All right, we have r of t equaling a times the cosine of omega t plus b sine omega t.
00:12
The derivative of r of t is negative omega sine omega t a plus omega cosine omega t b, which equals omega times negative sine omega t a plus cosine omega t b.
00:37
Start by taking the derivative of r of t.
00:40
The vectors a and b are constants, so they just come along for the ride.
00:45
The derivative of cosine omega t is negative omega -sign omega -t, and the derivative of sine omega -t is omega -t, so the derivative of r -t is another combination of a and b with those new coefficients and an overall factor of omega.
01:05
The cross product of r of c and the derivative of r of t would be a, cosine omega t plus b, sine omega t, by omega times negative sine omega t a plus cosine omega t b.
01:31
Pull out omega.
01:53
Let c be equal to cosine omega t and s be equal to sine and omega -t.
01:58
So the cross -project of r and the derivative of r would be omega times ac plus b -s and negative s -a plus c -b.
02:12
Now we need the cross -product and first i factor out the omega because it's just a scalar...