# Arc length

## Definition of arc length

• Recall how we defined the area under the graph of $$f(x)\ge0$$ for $$a\le x\le b$$:
• For a positive integer $$n$$, we split the interval $$[a,b]$$ to $$n$$ pieces of length $$\Delta x=\frac{b-a}{n}$$.
• In each subinterval $$[x_i,x_{i+1}]=[a+i\Delta x,a+(i+1)\Delta x]$$, we select a test point $$x_i\le x_i^*\le x_{i+1}$$.
• Adding up the areas of the rectangles $$[x_i,x_{i+1}]\times[0,f(x_i^*)]$$ gives the Riemann partial sum $$\sum_{i=0}^{n-1} f(x_i^*)\Delta x$$.
• The area is defined as the limit of this as $$n\to\infty$$: $\int_a^bf(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_i^*)\Delta x.$
• We will now define the length $$L$$ of the graph $$y=f(x)$$ for $$a\le x\le b$$ using a similar method.
• Again, for a positive integer $$n$$, we split the interval $$[a,b]$$ to $$n$$ pieces of length $$\Delta x=\frac{b-a}{n}$$.
• For each subinterval $$[x_i,x_{i+1}]=[a+i\Delta x,a+(i+1)\Delta x]$$, we will be adding the length of the line segment $|(x_i,f(x_i))(x_{i+1},f(x_{i+1}))|=\sqrt{(x_{i+1}-x_{i})^2+(f(x_{i+1})-f(x_{i}))^2}$
• Let us write $$P_i=(x_i,f(x_i))$$ and $$f(x_{i+1})-f(x_i)=\Delta y_i$$ so that the line segment length is written $$|P_iP_{i+1}|=\sqrt{(\Delta x)^2+(\Delta y_i)^2}$$.
• The arc length of $$y=f(x)$$, $$a\le x\le b$$ is defined as the limit $L=\lim_{n\to\infty}\sum_{i=0}^{n-1}\sqrt{(\Delta x)^2+(\Delta y_i)^2}.$
• Theorem. Let $$C$$ be the curve $$y=f(x)$$, $$a\le x\le b$$.
• Suppose that $$C$$ is smooth, that is for $$a<x<b$$, the derivative $$f'(x)$$ exists, and it is continuous.
• Then we have the following formula for the arc length. $L=\int_a^b\sqrt{1+(f'(x))^2}\,\mathrm dx.$
• Remark. If we use the Leibniz notation: $L=\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx,$ then it looks as if we have pulled the $$\mathrm dx$$ out from $$\sqrt{(\mathrm dx)^2+(\mathrm dy)^2}$$. Of course, this argument is entirely symbolic as $$\mathrm dx$$ is not a number.
• Example. Let's calculate this way the arc length of the parabola $$y=x^2$$, $$1\le x\le 4$$.
• We have $L=\int_1^4\sqrt{1+4x^2}\,\mathrm dx\stackrel{x=\frac12\tan\theta}{=}\int_{\tan^{-1}2}^{\tan^{-1}8}\frac12\sec^3\theta\,\mathrm d\theta.$
• Recall that we have $\int\sec\theta\,\mathrm d\theta=\int\frac{\sec\theta\cdot\sec\theta+\sec\theta\cdot\tan\theta}{\tan\theta+\sec\theta}\,\mathrm d\theta\stackrel{u=\sec\theta+\tan\theta}{=}\int u^{-1}\,\mathrm du=\ln|\sec\theta+\tan\theta|+C,$ and thus $\int\sec^3\theta\,\mathrm d\theta\stackrel{u=\sec\theta,\,\mathrm dv=\sec^2\theta\,\mathrm dx}{=}\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\,\mathrm d\theta=\sec\theta\tan\theta-\int\sec^3\theta\,\mathrm d\theta+\int\sec\theta\,\mathrm d\theta=\frac12\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right)+C.$
• Therefore, we need to find $$\sec\tan^{-1}2$$ and $$\sec\tan^{-1}8$$.
• Let $$\bigtriangleup ABC$$ be a right triangle with side lengths $$|AB|=1$$ and $$|AC|=2$$ so that $$\tan\theta=\frac{|AC|}{|AB|}=2$$. Then its hypotenuse will have length $$|BC|=\sqrt5$$, and thus we get $$\sec\tan^{-1}2=\sec\theta=\frac{|BC|}{|AB|}=\sqrt5$$.
• Similarly, we can get $$\sec\tan^{-1}8=\sqrt17$$.
• We can finish: $\int_{\tan^{-1}2}^{\tan^{-1}8}\frac12\sec^3\theta\,\mathrm d\theta=\frac14\left(8\sqrt{17}-2\sqrt5+\ln(8+\sqrt{17})-\ln(2+\sqrt 5)\right).$

## The arc length function

• Let $$C$$ be the curve $$y=f(x)$$, $$a\le x\le b$$.
• Its arc length function is $s(x)=\int_a^x\sqrt{1+(f'(t))^2}\,\mathrm dt.$
• It measures the path length along $$C$$ from $$(a,f(x=a))$$ to $$(x,f(x))$$.
• Example. Let $$C$$ be the top half semicircle with centre the origin and radius 1.
• Then we have $$C:y=\sqrt{1-x^2},\,-1\le x\le 1$$.
• The arc length function is: $s(x)=\int_{-1}^x\sqrt{1+\frac{t^2}{1-t^2}}\,\mathrm dt=\int_{-1}^x\sqrt{\frac{1}{1-t^2}}\,\mathrm dt\stackrel{t=\sin\theta}{=}\int_{-\pi/2}^{\sin^{-1}x}\frac{\cos\theta}{\cos\theta}\,\mathrm d\theta=\sin^{-1}x+\frac{\pi}{2}.$
• Note that we get $$s(x=1)=\pi$$.
• Exercises. 8.1: 1, 9, 13, 19, 35, 39