If velocity components are x'(t) = sqrt(t) and y'(t) = t, the speed is sqrt(t + t^2).

If velocity components are x'(t) = sqrt(t) and y'(t) = t, the speed is sqrt(t + t^2).

• A. sqrt(t + t^2)
• B. sqrt(t) + t
• C. t + sqrt(t^2)
• D. sqrt(t^2 + t)

Answer: (A)

The speed is the magnitude of the velocity vector, so v(t) = sqrt[(dx/dt)^2 + (dy/dt)^2]. With dx/dt = sqrt(t) and dy/dt = t, you square each component: (dx/dt)^2 = t and (dy/dt)^2 = t^2. Adding gives t + t^2, and taking the square root yields sqrt(t + t^2). Since a square root is real only for t ≥ 0 when these components are real, the speed is defined for t ≥ 0. This matches the expression obtained.

The speed is the magnitude of the velocity vector, so v(t) = sqrt[(dx/dt)^2 + (dy/dt)^2]. With dx/dt = sqrt(t) and dy/dt = t, you square each component: (dx/dt)^2 = t and (dy/dt)^2 = t^2. Adding gives t + t^2, and taking the square root yields sqrt(t + t^2). Since a square root is real only for t ≥ 0 when these components are real, the speed is defined for t ≥ 0. This matches the expression obtained.