How is the line integral ∫_C F · dr computed parametrically?
A∫_a^b |F(r(t))| dt
B∫_a^b F(r(t)) · r'(t) dt
C∫_a^b F(r(t)) × r'(t) dt
D∫_a^b F(r(t)) · r(t) dt
The line integral uses the dot product of the field F (evaluated along the path) with the tangent vector r'(t). The dot product extracts the component of F in the direction of travel. Option A ignores direction entirely (just integrates magnitude). Option C uses a cross product, which gives a vector not a scalar. Option D dots with the position vector instead of the tangent, which has no physical meaning here.
Question 2 True / False
Reversing the orientation of path C in ∫_C F · dr changes the sign of the result.
TTrue
FFalse
Answer: True
When the path is reversed, the parametrization runs the opposite direction, so the tangent vector r'(t) points the other way. This flips the sign of F · r'(t) at every point, making the entire integral negative. Physically: if F is a force and you walk the path backward, the work done is the negative of the original work.
Question 3 Short Answer
What physical quantity does the line integral ∫_C F · dr compute when F is a force field?
Think about your answer, then reveal below.
Model answer: The work done by the force field F on an object moving along the path C.
Work is the integral of force in the direction of displacement. At each infinitesimal step along the path, the displacement vector is dr, and the force is F. The work contribution is F · dr (the component of force along the motion direction times the step size). Integrating over the whole path gives total work. This is the central physical motivation for line integrals of vector fields.