All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Arc length

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

The length of a curve between two points is found by adding up tiny straight pieces, which gives an integral. F3 asks for arc lengths of curves given in cartesian or parametric form.

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Where the formula comes from

A tiny piece of curve is almost straight, with length δs where (δs)2 ≈ (δx)2 + (δy)2 (Pythagoras). Dividing by (δx)2 and letting δx tend to 0:

ds/dx = sqrt(1 + (dy/dx)2).

The formulae

Cartesian, from x = a to x = b: s = ∫ from a to b of sqrt(1 + (dy/dx)2) dx.

Parametric, from t = t1 to t = t2: s = ∫ from t1 to t2 of sqrt((dx/dt)2 + (dy/dt)2) dt.

(Polar arc length is not in this unit.)

Making the square root simplify

Exam curves are chosen so that 1 + (dy/dx)2 is a perfect square. Look for:

If it does not simplify, you may need a hyperbolic substitution.

Worked example

Find the length of y = cosh x from x = 0 to x = ln 2.

  1. dy/dx = sinh x, so 1 + (dy/dx)2 = cosh2 x
  2. s = ∫ from 0 to ln 2 of cosh x dx = [sinh x]
  3. sinh(ln 2) = (2 - 1/2)/2

Answer: 3/4

Key idea: s = ∫ sqrt(1 + (dy/dx)2) dx or s = ∫ sqrt((dx/dt)2 + (dy/dt)2) dt. Use identities to make the square root a perfect square.

Check yourself

  1. Find the length of y = cosh x from x = 0 to x = 1, to 3 decimal places.
    Show answer

    1.175. s = ∫ from 0 to 1 of cosh x dx = sinh 1 = 1.175.

  2. Find the exact length of y = (2/3)x3/2 from x = 0 to x = 3. Give your answer as a fraction in its simplest form.
    Show answer

    14/3. dy/dx = x1/2, so s = ∫ from 0 to 3 of sqrt(1 + x) dx = [(2/3)(1 + x)3/2] = (2/3)(8 - 1) = 14/3.

  3. Find the length of the curve x = cos3 t, y = sin3 t from t = 0 to t = π/2.
    Show answer

    1.5. dx/dt = -3cos2 t sin t and dy/dt = 3sin2 t cos t. The sum of squares is 9sin2 t cos2 t(cos2 t + sin2 t), so the integrand is 3 sin t cos t. s = [(3/2)sin2 t] from 0 to π/2 = 3/2.

  4. Find the length of one arch of the cycloid x = t - sin t, y = 1 - cos t, from t = 0 to t = 2π.
    Show answer

    8. (dx/dt)2 + (dy/dt)2 = (1 - cos t)2 + sin2 t = 2 - 2cos t = 4sin2(t/2). So s = ∫ from 0 to 2π of 2 sin(t/2) dt = [-4cos(t/2)] = 8.

  5. Which integral gives the arc length of a curve x = f(t), y = g(t)?
    • ∫ sqrt(1 + (dy/dt)2) dt
    • ∫ (dx/dt + dy/dt) dt
    • ∫ sqrt((dx/dt)2 + (dy/dt)2) dt
    • ∫ y (dx/dt) dt
    Show answer

    ∫ sqrt((dx/dt)2 + (dy/dt)2) dt. Pythagoras on the small changes δx and δy, divided by δt.

  6. Find the length of y = ln(cos x) from x = 0 to x = π/4, to 3 decimal places.
    Show answer

    0.881. dy/dx = -tan x, so the integrand is sec x. s = [ln(sec x + tan x)] from 0 to π/4 = ln(1 + sqrt(2)) = 0.881.

  7. Find the length of x = t2, y = (2/3)t3 from t = 0 to t = 1, to 3 decimal places.
    Show answer

    1.219. The integrand is sqrt(4t2 + 4t4) = 2t sqrt(1 + t2), giving [(2/3)(1 + t2)3/2] from 0 to 1 = (2/3)(2sqrt(2) - 1) = 1.219.

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Last updated 8 November 2026. Spotted a mistake? Tell us.