Which of the following best explains why the vector r′(a) is tangent to the curve traced by the differentiable vector-valued function r(t)=⟨x(t),y(t)⟩ at the point corresponding to t=a?
a) r′(a) is always perpendicular to the position vector r(a), so it must lie along the normal to the radius and therefore along the curve's tangent.
b) Integrating r′(t) from 0 to a produces the total displacement vector; since that displacement lies along the path, the integrand r′(a) must also point tangent.
c) It gives the instantaneous rate of change of position, so its direction coincides with the path at that instant.
d) r′(a) measures how quickly the particle's speed is changing at t=a; changes in speed occur only along the direction of motion, so the vector must be tangent.