Arc Length and Speed of Parametric Curves
Compute arc length L = ∫√((dx/dt)² + (dy/dt)²) dt and speed |v(t)| = √((x'(t))² + (y'(t))²) for parametric curves. Interpret speed physically in the context of particle motion problems.
A lesson of AP Calculus BC, in Parametric Equations, Polar Coordinates, and Vectors, about 10 minutes.
Parametric Arc Length and Speed
By the end of this lesson, you will be able to calculate the arc length of a parametric curve and determine a particle's speed and distance traveled.
Arc Length
For a parametric curve $x = f(t)$, $y = g(t)$ from $t = a$ to $t = b$:
$$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}, dt$$
Derivation: the arc length of a small piece from $t$ to $t+\Delta t$ is approximately $\sqrt{(\Delta x)^2+(\Delta y)^2} = \sqrt{(x'(t))^2+(y'(t))^2},\Delta t$.
Speed
For a particle with position $(x(t), y(t))$, the speed is:
$$\text{speed} = \sqrt{(x'(t))^2+(y'(t))^2}$$
This is the magnitude of the velocity vector $\langle x'(t), y'(t)\rangle$.
Total distance traveled from $t = a$ to $t = b$ is the arc length:
$$d = \int_a^b \sqrt{(x'(t))^2+(y'(t))^2}, dt$$
Note: this differs from net displacement $\sqrt{(x(b)-x(a))^2+(y(b)-y(a))^2}$.
Example
A particle traces a curve with $x = 3\cos t$, $y = 3\sin t$ on $[0, 2\pi]$.
$x'(t) = -3\sin t$, $y'(t) = 3\cos t$
Speed $= \sqrt{9\sin^2 t + 9\cos^2 t} = \sqrt{9} = 3$ (constant speed)
$L = \int_0^{2\pi} 3, dt = 6\pi$ (circumference of a circle with radius 3 ✓)
Summary
Arc length: $L = \int_a^b\sqrt{(dx/dt)^2+(dy/dt)^2}, dt$
Speed $= \sqrt{(x'(t))^2+(y'(t))^2}$ (magnitude of velocity vector)
Total distance = arc length = $\int_a^b \text{speed}, dt$