Arc length of parametric curves

About this lesson

Calculus BC (AP-aligned) › Parametric equations, polar coordinates and vector-valued functions

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Arc length of parametric curves

Parametric equations, polar coordinates and vector-valued functions.

Warm-up

Three quick questions from earlier lessons.

Discover

In a tiny time dtdt, a point moves dxdx across and dydy up. How far does it travel?

  1. dx2+dy2\sqrt{dx^2 + dy^2}
  2. dx+dydx + dy

Key idea

Add up tiny hypotenuses.

Arc length of x(t)x(t), y(t)y(t) for a≤t≤ba \le t \le b: L=∫ab(dxdt)2+(dydt)2 dtL = \displaystyle\int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\,dt (if the curve is traced once).

Worked example

x=3tx = 3t, y=4ty = 4t, 0≤t≤20 \le t \le 2.

x′=3x' = 3, y′=4y' = 4
Rates.
L=∫025 dtL = \displaystyle\int_0^2 5\,dt
√(9 + 16) = 5.
=10= 10
Check: from (0, 0) to (6, 8), distance 10.

Your turn, guided

Your turn: the circle x=2cos⁡tx = 2\cos t, y=2sin⁡ty = 2\sin t, 0≤t≤2π0 \le t \le 2\pi.

L=4πL = 4\pi
√(4 sin² + 4 cos²) = 2; ∫₀^{2π} 2 dt.

Practice ★

  1. Find the length of the curve x=3cos⁡tx = 3\cos t, y=3sin⁡ty = 3\sin t, 0≤t≤2π0 \le t \le 2\pi.
  2. Find the length of the curve x=5cos⁡tx = 5\cos t, y=5sin⁡ty = 5\sin t, 0≤t≤2π0 \le t \le 2\pi.
  1. Find the length of the path x=3tx = 3t, y=6ty = 6t, 0≤t≤10 \le t \le 1. Give an exact value.
  2. Find the length of the path x=4tx = 4t, y=4ty = 4t, 0≤t≤20 \le t \le 2. Give an exact value.

Watch out!

Practice ★★

  1. Find the length of x=tx = t, y=23t3/2y = \tfrac{2}{3}t^{3/2} for 0≤t≤30 \le t \le 3.
  2. Find the length of x=tx = t, y=23t3/2y = \tfrac{2}{3}t^{3/2} for 0≤t≤150 \le t \le 15.

Exit ticket

Three questions to check you have got it.

  1. Find the length of the path x=3tx = 3t, y=6ty = 6t, 0≤t≤10 \le t \le 1. Give an exact value.
  2. Find the length of the path x=4tx = 4t, y=4ty = 4t, 0≤t≤20 \le t \le 2. Give an exact value.
  3. Find the length of the path x=2tx = 2t, y=5ty = 5t, 0≤t≤20 \le t \le 2. Give an exact value.

Summary card

Arc length

L=∫abx′(t)2+y′(t)2 dtL = \displaystyle\int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt.

Watch out: Pythagoras, not a sum.

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