An easy question

[Desert Sleet]

your answer is correct (the ball does not change orientation), but I think that your explanation is incomplete. Drawing a mark on the north pole only demonstrates that the poles do not move, but it still leaves some doubt as to whether the moving ball might have rotated around the north-south axis relative to the fixed ball.

Just in case anyone's interested, here's an abbreviated solution, using quaternions. Quaternions are of the form a+bi+cj+dk where a,b,c,d are real; ij = k; jk = i; ki = j; ii = jj = kk = ijk = -1. The coolest thing about them is that rotations can be handled with them. A rotation of w radians around the unit vector (x,y,z) is expressed by q1 = cos(w/2) + sin(w/2)(xi+yj+zk). Multiple rotations are obtained by multiplying the quaternions together, in reverse order.

I will set my coordinates so that the first rotation is from the z axis towards the x axis. It's a rotation of pi radians, around the y axis. So q1 = cos(pi/2) + sin(pi/2)j = j.

The second rotation is from the x axis towards the y axis, around the z axis, some angle 2t. The quaternion q2 = cos(t) + sin(t)k.

The last one is a rotation of pi radians back to the top, around the vector (sin(t) -cos(t) 0). So q3 = cos(pi/2) + sin(pi/2)(sin(t)i-cos(t)j) = sin(t)i - cos(t)j.

Combining them
Q = q3q2q1
= (sin(t)i - cos(t)j) (cos(t) + sin(t)k) (j)
= (sin(t)i - cos(t)j) (cos(t)j - sin(t)i)
= sin(t)cos(t)k +sin(t)sin(t) + cos(t)cos(t) - cos(t)sin(t)k
= 1, which corresponds to a rotation of 0 radians.


---------------------------------------------
When The Going Gets Tough ... The Tough Gets Going ...