Final answer:
The velocity vector v(t) for the particle's elliptical motion r(t) = cos(t)i + sin(t)j + 0k is given by -sin(t)i + cos(t)j + 0k with a constant speed of 1 unit per second.
Step-by-step explanation:
To find the velocity, v(t), of the particle traveling along the path of an ellipse, we differentiate the position vector r(t) with respect to time, t. The position vector given is r(t) = cos(t)i + sin(t)j + 0k. The velocity vector is found by calculating the derivative dr/dt. The derivative of cos(t) with respect to t is -sin(t), and the derivative of sin(t) with respect to t is cos(t). Therefore, the velocity vector is:
v(t) = -sin(t)i + cos(t)j + 0k
The magnitude of the velocity vector (or speed) can be found using the Pythagorean theorem as the square root of the sum of the squares of the components:
|v(t)| = sqrt((-sin(t))^2 + cos(t)^2) = 1, since sin(t)^2 + cos(t)^2 = 1.
Thus, the particle travels at a constant speed of 1 unit per second in a direction tangent to the ellipse at any point in time.