Examples

Arc length of ln x with trig sub

Find the length of the curve $y=\ln x$ for $1\le x\le \sqrt{3}$:

Solution

(1) Set up arc length formula:

First note that $dy/dx=1/x$. Then:

$$\begin{array}{c}L={\int }_a^b\sqrt{1+(f^{\prime }{)}^2}dx\gg \gg {\int }_1^{\sqrt{3}}\sqrt{1+{\left(\frac{1}{x}\right)}^2}dx\\ \\ \gg \gg {\int }_1^{\sqrt{3}}\sqrt{\frac{x^2+1}{x^2}}dx\gg \gg {\int }_1^{\sqrt{3}}\frac{\sqrt{x^2+1}}{x}dx\end{array}$$

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(2) Integrate using trig substitution:

Observe $\sqrt{x^2}+1$. Choose $x=\tan \theta$ and therefore $dx={\sec }^2\theta d\theta$.

$$\begin{array}{c}{\int }_1^{\sqrt{3}}\frac{\sqrt{x^2+1}}{x}dx\gg \gg {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\sqrt{{\tan }^2\theta +1}}{\tan \theta }({\sec }^2\theta )d\theta \\ \\ \gg \gg {\int }_{\pi /4}^{\pi /3}\frac{{\sec }^3\theta }{\tan \theta }d\theta \gg \gg {\int }_{\pi /4}^{\pi /3}\frac{\sec \theta }{\tan \theta }(1+{\tan }^2\theta )d\theta \\ \\ \gg \gg {\int }_{\pi /4}^{\pi /3}\csc \theta +\sec \theta \tan \theta d\theta \gg \gg \ln |\csc \theta -\cot \theta |+\sec \theta {|}_{\pi /4}^{\pi /3}\\ \\ \gg \gg \ln |\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}|+2-\ln |\sqrt{2}-1|-\sqrt{2}\\ \\ \gg \gg \ln \left(\frac{\sqrt{3}(\sqrt{2}+1)}{3}\right)+2-\sqrt{2}\end{array}$$

Arc length of chain, via position

A hanging chain describes a catenary shape. (‘Catenary’ is to hyperbolic trig as ‘sinusoid’ is to normal trig.) The graph of the hyperbolic cosine is a catenary:

$$y=f(x)=\cosh x$$

Let us compute the arc length of this catenary on the portion from $x=0$ to $x=a$.

Solution

(1) Arc-length function:

$$s(a)={\int }_0^a\sqrt{1+(f^{\prime }{)}^2}dx$$

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(2) Compute $f^{\prime }(x)$:

$$\frac{d}{dx}\cosh (x)=\sinh (x)$$

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(3) Plug into arclength formula:

$$s(a)={\int }_0^a\sqrt{1+{\sinh }^2(x)}dx$$

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(4) Hyperbolic trig identity:

$$\begin{array}{c}{\cosh }^{2}x-{\sinh }^{2}x=1\\ \\ 1+{\sinh }^{2}x={\cosh }^{2}x\end{array}$$

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(5) Simplify integrand and integrate:

$$\begin{array}{c}{\int }_0^a\sqrt{1+{\sinh }^2(x)}dx\gg \gg {\int }_0^a\sqrt{{\cosh }^{2}x}dx\\ \\ \gg \gg {\int }_0^a\cosh xdx\gg \gg \sinh a\end{array}$$

Coincidence?

The arc length of a catenary curve matches the area under the catenary curve!