For the unit circle path r(t) = (cos t, sin t), the distance traveled from t=0 to t=π is represented by which integral?

For the unit circle path r(t) = (cos t, sin t), the distance traveled from t=0 to t=π is represented by which integral?

• A. ∫_0^π sqrt((dx/dt)^2 + (dy/dt)^2) dt
• B. ∫_0^π (dx/dt) dt
• C. ∫_0^π |dx/dt| dt
• D. ∫_0^π (dy/dt) dt

Answer: (A)

Distance traveled along a parametric curve is found by integrating the speed, which is the magnitude of the velocity vector. For a path x(t), y(t), the distance from t = a to t = b is ∫_a^b sqrt[(dx/dt)^2 + (dy/dt)^2] dt. Here x(t) = cos t and y(t) = sin t, so dx/dt = -sin t and dy/dt = cos t. The speed is sqrt(sin^2 t + cos^2 t) = 1, since sin^2 t + cos^2 t = 1. Therefore the distance traveled from t = 0 to t = π is ∫_0^π 1 dt = π, which is the arc length of a semicircle of unit radius. The other expressions don’t measure arc length: ∫_0^π (dx/dt) dt would give x(π) - x(0) = -2, the net horizontal change. ∫_0^π |dx/dt| dt gives the total horizontal distance moved, which would be 2, not the actual path length. ∫_0^π (dy/dt) dt gives y(π) - y(0) = 0, the net vertical change.

Distance traveled along a parametric curve is found by integrating the speed, which is the magnitude of the velocity vector. For a path x(t), y(t), the distance from t = a to t = b is ∫_a^b sqrt[(dx/dt)^2 + (dy/dt)^2] dt.

Here x(t) = cos t and y(t) = sin t, so dx/dt = -sin t and dy/dt = cos t. The speed is sqrt(sin^2 t + cos^2 t) = 1, since sin^2 t + cos^2 t = 1. Therefore the distance traveled from t = 0 to t = π is ∫_0^π 1 dt = π, which is the arc length of a semicircle of unit radius.

The other expressions don’t measure arc length: ∫_0^π (dx/dt) dt would give x(π) - x(0) = -2, the net horizontal change. ∫_0^π |dx/dt| dt gives the total horizontal distance moved, which would be 2, not the actual path length. ∫_0^π (dy/dt) dt gives y(π) - y(0) = 0, the net vertical change.