1 The length of a curve

To find the length of a curve in the $xy$ plane we first divide the curve into a large number of pieces. We measure (or, at least, approximate) the length of each piece and then by an obvious summation process obtain an estimate for the length of the curve. Theoretically, we allow the number of pieces to increase without bound, implying that the length of each piece will tend to zero. In this limit the summation process becomes an integration process.

Figure 11

Figure 11 shows the portion of the curve $y\left(x\right)$ between $x=a$ and $x=b$ . A small piece of this curve has been selected and can be considered as the hypotenuse of a triangle with base $\delta x$ and height $\delta y$ . (Here $\delta x$ and $\delta y$ are intended to be ‘small’ so that the curved segment can be regarded as a straight segment .)

Using Pythagoras’ theorem, the length of the hypotenuse is: $\sqrt{\delta x^2+\delta y^2}=\sqrt{1+{\left(\frac{\delta y}{\delta x}\right)}^2}\delta x$

By summing all such contributions between $x=a$ and $x=b$ , and letting $\delta x\to 0$ we obtain an expression for the total length of the curve:

$\underset{\delta x\to 0}{lim}{\sum }_{x=a}^{x=b}\sqrt{1+{\left(\frac{\delta y}{\delta x}\right)}^2}\delta x$

But we already know how to write such an expression in terms of an integral. We obtain the following result:

Because of the complicated form of the integrand, and in particular the square root, integrals of this type are often difficult to calculate. In practice, approximate numerical methods rather than exact methods are normally needed to perform the integration. We shall first illustrate the application of the formula in Key Point 7 by a problem which could be calculated in a much simpler way, before looking at some harder problems.

Example 5

Find the length of the curve $y=3x+2$ between $x=1$ and $x=5$ .

Solution

In this Example, the curve is in fact a straight line segment, and its length could be obtained using Pythagoras’ theorem without the need for integration.

Notice from the formula in Key Point 7 that it is necessary to find $\frac{dy}{dx}$ , which in this case is 3.

Applying the formula we find

Thus the length of the curve $y=3x+2$ between the points where $x=1$ and $x=5$ is 12.65 units.

Task!

Find the length of the curve $y=coshx$ between $x=0$ and $x=2$ shown in the diagram.

First write down $\frac{dy}{dx}$ :

Answer

Hence write down the required integral:

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This integral can be evaluated by making use of the hyperbolic identity ${cosh}^{2}x-{sinh}^{2}x\equiv 1.$

Write down the integral which results after applying this identity:

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Perform the integration to find the required length:

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The next Task is more complicated still and requires the use of a hyperbolic substitution and knowledge of the hyperbolic identities.

Task!

Find the length of the curve $y=x^2$ between $x=0$ and $x=3$ .

Given $y=x^2$ then $\frac{dy}{dx}=2x$ . Use this result and apply the formula in Key Point 7 to obtain the integral required:

Answer

Make the substitution $x=\frac{1}{2}sinhu$ , giving $\frac{dx}{du}=\frac{1}{2}coshu$ , to obtain an integral in terms of $u$ :

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Use the hyperbolic identity ${cosh}^{2}u-{sinh}^{2}u\equiv 1$ to eliminate ${sinh}^{2}u$ :

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Use the hyperbolic identity ${cosh}^{2}u\equiv \frac{1}{2}\left(cosh2u+1\right)$ to rewrite the integrand in terms of $cosh2u$ :

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Finally, perform the integration to complete the calculation:

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Exercises

1. Find the length of the line $y=2x+7$ between $x=1$ and $x=3$ using the technique of this Section. Verify your result from your knowledge of the straight line.
2. Find the length of $y=x^{3∕2}$ between $x=0$ and $x=5$ .
3. Calculate the length of the curve $y^2=4x^3$ between $x=0$ and $x=2$ , in the first quadrant.

Answer