2.4 - Lengths of curves | Cambridge A-Level Further Mathematics

The distance along a curve is longer than the straight line between its endpoints. You will build its length from small straight segments, then calculate lengths from Cartesian, parametric and polar equations without counting a traced arc twice.

Cartesian arc length

How long is the curved path $y=x^2/2$ from $x=0$ to $x=1$ ? Its horizontal width is $1$ , but its length must also include the vertical movement.

A short chord has horizontal change $\Delta x$ , vertical change $\Delta y$ , and length $\sqrt{(\Delta x)^2+(\Delta y)^2}$ . As the chords become shorter, their slope approaches the derivative. Adding their lengths and taking a limit gives the arc-length integral.

[DIAGRAM: asset_name: A short chord builds the length element; asset_slug: cam9231-cartesian-length-element; caption: Pythagoras combines horizontal and vertical changes before the limiting integral; description: Exact graph of y equals x squared over two on zero to 1.4. Two points on the curve at x equals 0.6 and 1.1 are joined by a chord. A right triangle labels delta x, delta y and delta s. A separate panel gives the positive length element ds equals square root of one plus y prime squared times dx for increasing x.]
Pythagoras combines horizontal and vertical changes before the limiting integral

Cartesian arc length

For a continuously differentiable curve $y=f(x)$ , traversed once with $a\le x\le b$ ,

L=\int_a^b\sqrt{1+(f'(x))^2}\,\mathrm{d}x.

For $x=g(y)$ , use $L=\int_c^d\sqrt{1+(g'(y))^2}\,\mathrm{d}y$ , with $c<d$ . Split a piecewise smooth curve into suitable pieces.

Not in the formula booklet: learn it.

The positive square root matters: length cannot cancel when the gradient changes sign. These are lengths along the curve, rather than signed displacements or areas underneath it.

Worked example 1: a Cartesian parabola

Question. Find the exact length of $y=x^2/2$ from $x=0$ to $x=1$ .

Plan. Differentiate, then use $x=\sinh u$ to integrate the square root.

Since $y'=x$ , the required integral is $\int_0^1\sqrt{1+x^2}\,\mathrm{d}x$ . Under $x=\sinh u$ , $\mathrm{d}x=\cosh u\,\mathrm{d}u$ and $\sqrt{1+x^2}=\cosh u$ . Thus

\begin{aligned} L&=\int_0^{\sinh^{-1}1}\cosh^2u\,\mathrm{d}u \\ &=\frac12[u+\sinh u\cosh u]_0^{\sinh^{-1}1} \\ &=\frac12(\sinh^{-1}1+\sqrt2) \end{aligned}

Here $\cosh^2u=(1+\cosh2u)/2$ , and $\cosh(\sinh^{-1}1)=\sqrt2$ .

Answer. $L=\tfrac12(\sqrt2+\sinh^{-1}1)=\tfrac12(\sqrt2+\ln(1+\sqrt2))$ .

Check. The chord between $(0,0)$ and $(1,1/2)$ has length $\sqrt5/2=1.118\ldots$ . Moving horizontally then vertically has length $1.5$ . The result $1.14779\ldots$ lies between them, as a monotone curved path should.

Change only the upper endpoint. The antiderivative can now be reused, but both its algebraic and inverse-hyperbolic terms need the new input.

Worked example 2: a longer part of the same curve

Question. Find the exact length of $y=x^2/2$ from $x=0$ to $x=\sqrt3$ .

Plan. Use the same length integral and the antiderivative obtained by the substitution above.

\begin{aligned} L&=\int_0^{\sqrt3}\sqrt{1+x^2}\,\mathrm{d}x \\ &=\frac12[x\sqrt{1+x^2}+\sinh^{-1}x]_0^{\sqrt3} \\ &=\sqrt3+\frac12\sinh^{-1}\sqrt3 \end{aligned}

Answer. $L=\sqrt3+\tfrac12\ln(2+\sqrt3)$ .

Check. The endpoint is $(\sqrt3,3/2)$ , so the chord is $\sqrt{21}/2=2.291\ldots$ . The result $2.390529\ldots$ exceeds this and is below the horizontal-plus-vertical distance $\sqrt3+3/2$ .

Do not replace $\sqrt{1+(y')^2}$ by $1+y'$ : Pythagoras combines squared perpendicular changes before taking the root.

Problem-solving

Simplify the expression under the square root first. A square, a trigonometric identity or a hyperbolic substitution can turn the length integral into familiar calculus.

Use the same curve over a fresh interval.

Parametric arc length

A parameter can describe both coordinate changes at once. If $x=x(t)$ and $y=y(t)$ , the length element comes from $\mathrm{d}x=x'(t)\,\mathrm{d}t$ and $\mathrm{d}y=y'(t)\,\mathrm{d}t$ .

Parametric arc length

For a smooth parametrised curve traced once over $\alpha\le t\le\beta$ ,

L=\int_\alpha^\beta\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+ \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t.

Use increasing parameter limits; split the interval if necessary. The integral counts every traversal, so choose an interval that does not retrace the required arc.

Not in the formula booklet: learn it.

This formula also works where $\mathrm{d}x/\mathrm{d}t=0$ ; there is no division by that derivative. A zero velocity at an isolated point does not itself invalidate the length integral.

Worked example 3: a square hidden in the speed

Question. Find the exact length of $x=e^t-t$ , $y=4e^{t/2}$ , for $0\le t\le\ln4$ .

Plan. Calculate both derivatives and simplify their squared sum before integrating.

\begin{aligned} x'&=e^t-1 \\ y'&=2e^{t/2} \\ (x')^2+(y')^2&=(e^t-1)^2+4e^t=(e^t+1)^2 \end{aligned}

Since $e^t+1>0$ , its positive square root is $e^t+1$ . Also $y$ strictly increases, so the curve is not retraced. Therefore

L=[e^t+t]_0^{\ln4}=3+\ln4.

Answer. $L=3+\ln4$ .

Check. The endpoints are $(1,4)$ and $(4-\ln4,8)$ . Their straight-line distance is $\sqrt{(3-\ln4)^2+16}=4.313\ldots$ , below $L=4.386\ldots$ . Since both coordinates increase, their total changes give the upper bound $7-\ln4=5.614\ldots$ , also consistent.

Change only the parameter, using $u=e^{t/2}$ . A different speed must give the same geometric length over the matching interval.

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2.4 - Lengths of curves

Cartesian arc length

Parametric arc length

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