Arc Length

The arc length of a curve measures its actual length along the curve (not the straight-line distance between endpoints). Calculating arc length uses integration and is a key topic in AQA Further Mathematics.

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1 Arc Length in Cartesian Form

Derivation

Consider a small element of a curve y = f(x). Between x and x + delta x, the curve moves:

• Horizontally by delta x
• Vertically by delta y

The length of this small element is approximately:

delta s = sqrt((delta x)^2 + (delta y)^2) = sqrt(1 + (delta y / delta x)^2) * delta x

In the limit as delta x -> 0:

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

The total arc length from x = a to x = b is:

s = integral from a to b of sqrt(1 + (dy/dx)^2) dx

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2 Worked Example 1 — Parabola

Find the arc length of y = x^(3/2) from x = 0 to x = 4.

Solution:

dy/dx = (3/2) x^(1/2)

(dy/dx)^2 = (9/4) x

s = integral from 0 to 4 of sqrt(1 + 9x/4) dx

Let u = 1 + 9x/4, then du = 9/4 dx, so dx = 4/9 du.

When x = 0, u = 1. When x = 4, u = 10.

s = integral from 1 to 10 of sqrt(u) * (4/9) du

= (4/9) * [2u^(3/2)/3] from 1 to 10

= (8/27) * [10^(3/2) - 1]

= (8/27) * [10 sqrt(10) - 1]

= (8/27)(10 sqrt(10) - 1)

This is approximately 9.073.

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3 Worked Example 2 — Catenary

Find the arc length of y = cosh x from x = 0 to x = a.

Solution:

dy/dx = sinh x

(dy/dx)^2 = sinh^2 x

1 + sinh^2 x = cosh^2 x (fundamental hyperbolic identity)

s = integral from 0 to a of sqrt(cosh^2 x) dx = integral from 0 to a of cosh x dx

= [sinh x] from 0 to a

= sinh a

This elegant result is why the catenary (the shape of a hanging chain) plays such an important role in mathematics: its arc length has a particularly simple formula.

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4 Worked Example 3 — Straight Line Check

Verify the arc length formula for y = 3x + 2 from x = 0 to x = 4.

Solution:

dy/dx = 3, so (dy/dx)^2 = 9.

s = integral from 0 to 4 of sqrt(1 + 9) dx = integral from 0 to 4 of sqrt(10) dx = 4 sqrt(10)

The straight-line distance from (0, 2) to (4, 14) is sqrt(16 + 144) = sqrt(160) = 4 sqrt(10). Confirmed!

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5 Arc Length in Parametric Form

If the curve is given parametrically as x = x(t), y = y(t), then:

ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt

s = integral from t1 to t2 of sqrt((dx/dt)^2 + (dy/dt)^2) dt

Worked Example 4 — Circle

Verify the circumference of a circle of radius r using x = r cos t, y = r sin t for 0 <= t <= 2pi.

Solution:

dx/dt = -r sin t, dy/dt = r cos t

(dx/dt)^2 + (dy/dt)^2 = r^2 sin^2 t + r^2 cos^2 t = r^2

s = integral from 0 to 2pi of sqrt(r^2) dt = integral from 0 to 2pi of r dt = 2pi r

The circumference is 2pi r, as expected.

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Worked Example 5 — Cycloid

Find the arc length of one arch of the cycloid x = a(t - sin t), y = a(1 - cos t) for 0 <= t <= 2pi.

Solution:

dx/dt = a(1 - cos t), dy/dt = a sin t

(dx/dt)^2 + (dy/dt)^2 = a^2(1 - cos t)^2 + a^2 sin^2 t

= a^2(1 - 2 cos t + cos^2 t + sin^2 t)