We have 2 equations to specify the location of the object and we desire the velocity. In order to get that, we simply need to calculate the first derivative of each location equation. So: X = 2 cos(t) X' = 2 (-sin(t)) X' = -2 sin(t) Y = sin(t) Y' = cos(t) So the velocity vector at time t is (-2sin(t), cos(t)). But you want the velocity. So using the Pythagorean theorem we can get that by calculating the square root of the sum of the squares. So: V = sqrt((-2sin(t))^2 + cos^2(t)) V = sqrt(4sin^2(t) + cos^2(t)) Speed at t = 1, is V = sqrt(4sin^2(1) + cos^2(1)) V = sqrt(2.832293673 + 0.291926582) V = sqrt(3.124220255) V = 1.767546394 And t=3: V = sqrt(4sin^2(3) + cos^2(3)) V = sqrt(0.079659427 + 0.980085143) V = sqrt(1.05974457) V = 1.029438959 Now asking for velocity as a function of P, we have a bit of a complication. As shown above, it's trivial to calculate velocity as a function of t. But if all you're given is the X and Y coordinates of the object, we have a bit more work to do. The below equations will be using the trigonometric identity of cos^2(a) + sin^2(a) = 1 for any angle a. X = 2 cos(t) X' = -2 sin(t) We want to get from X which is 2cos(t) to X'^2 which is 4sin^2(t). So: X/2; We now have cos(t) (X/2)^2: We now have cos^2(t) 1-(X/2)^2: We now have sin^2(t) 4(1-(X/2)^2): We now have 4sin^2(t) which is what we want. Time to simplify 4(1 - (X/2)^2) 4(1 - (X^2/4)) 4 - 4(X^2/4) 4 - X^2 Now we need to get from Y to Y'^2. Will do the same as for X to X'^2, but without all the comments. Y = sin(t) Y' = cos(t) Y'^2 = 1 - Y^2 So the equation for the velocity as a function of X,Y we get V = sqrt(4 - X^2 + 1 - Y^2) V = sqrt(5 - X^2 - Y^2) In summary: Position at time t = (2cos(t), sin(t)) Velocity vector at time t = (-2 sin(t), cos(t)) Velocity as function of t is: V = sqrt(4sin^2(t) + cos^2(t)) Velocity as function of P is: V = sqrt(5 - X^2 - Y^2) Is object traveling at constant speed? NO Velocity at t = 1 is: V = 1.767546394 Velocity at t = 2 is: V = 1.029438959