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## Arc Length

Suppose $$f$$ is a continuously differentiable function on $$[a,b]$$, i.e., $$f'$$ is continuous on $$[a,b]$$. Then the length $$L$$ of the curve $$y=f(x)$$ on $$[a,b]$$ is $L=\int_a^b \sqrt{1+\left[ f'(x) \right]^2} \;dx=\int_a^b \sqrt{1+\left( \frac{dy}{dx} \right)^2} \;dx.$

Break $$[a,b]$$ into $$n$$ subintervals $$[x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]$$ where $$x_i=a+i \Delta x$$ and $$\Delta x=(b-a)/n$$. Consider the $$n+1$$ points on the curve $$y=f(x)$$: $P_0(x_0,y_0),P_1(x_1,y_1),\ldots,P_n(x_n,y_n),$ where $$y_i=f(x_i)$$ for $$i=0,1,2,\ldots,n$$. Note that $$L\approx \sum_{i=1}^n |P_{i-1}P_i|$$ where $|P_{i-1}P_i|=\sqrt{(x_i-x_{i-1})^2+(y_i-y_{i-1})^2}=\sqrt{(\Delta x)^2+(f(x_i)-f(x_{i-1}))^2}.$ By the Mean Value Theorem on $$f$$ on $$[x_{i-1},x_i]$$, we get $f(x_i)-f(x_{i-1})=f'(x_i^*)(x_i-x_{i-1})=f'(x_i^*)\Delta x,$ for some $$x_i^*$$ in $$(x_{i-1},x_i)$$. Then $|P_{i-1}P_i|=\sqrt{(\Delta x)^2+(f(x_i)-f(x_{i-1}))^2}=\sqrt{(\Delta x)^2+(f'(x_i^*)\Delta x)^2}= \sqrt{1+\left[ f'(x_i^*) \right]^2} \Delta x.$ Therefore the length $$L$$ of the curve $$y=f(x)$$ on $$[a,b]$$ is $L\approx \sum_{i=1}^n |P_{i-1}P_i|=\sum_{i=1}^n \sqrt{1+\left[ f'(x_i^*) \right]^2} \Delta x.$ This approximation of $$L$$ gets better as $$n\to \infty$$. Thus $L=\lim_{n\to \infty} \sum_{i=1}^n \sqrt{1+\left[ f'(x_i^*) \right]^2} \Delta x =\int_a^b \sqrt{1+\left[ f'(x) \right]^2} \;dx.$

Example. Prove that the circumference of a circle of radius $$r$$ is $$2\pi r$$.

Solution. The circumference of a circle of radius $$r$$ is twice the arc length $$L$$ of the semicircle $$y=\sqrt{r^2-x^2}$$, $$-r\leq x\leq r$$. \begin{align*} L=\;&\int_{-r}^r \sqrt{1+\left( \frac{dy}{dx} \right)^2} \;dx\\ =\; &\int_{-r}^r \sqrt{1+\left( \frac{-x}{\sqrt{r^2-x^2}} \right)^2} \;dx\\ =\; &\int_{-r}^r \sqrt{1+\frac{x^2}{r^2-x^2}} \;dx\\ =\; &\int_{-r}^r \frac{r}{\sqrt{r^2-x^2}} \;dx\\ =\; & \left. r\sin^{-1}\left( \frac{x}{r} \right) \right\vert_{-r}^r\\ =\; & r\sin^{-1}(1)-r\sin^{-1}(-1)\\ =\; & 2r\sin^{-1}(1)\\ =\; & 2r\frac{\pi}{2}\\ =\; & \pi r. \end{align*} Therefore the circumference of a circle of radius $$r$$ is $$2L=2 \pi r$$.

Example. Find the arc length of the curve $$y=\ln(\sec x)$$ on $$\left[0,\frac{\pi}{3}\right]$$.

Solution. By the chain rule, $\frac{dy}{dx}=\frac{1}{\sec x}\frac{d}{dx}\left( \sec x \right)=\frac{1}{\sec x} \sec x \tan x=\tan x.$ The arc length of the curve $$y=\ln(\sec x)$$ on $$\left[0,\frac{\pi}{3}\right]$$ is \begin{align*} &\int_0^{\frac{\pi}{3}} \sqrt{1+\left( \frac{dy}{dx} \right)^2} \;dx\\ =\; &\int_0^{\frac{\pi}{3}} \sqrt{1+\tan^2 x} \;dx\\ =\; &\int_0^{\frac{\pi}{3}} \sqrt{\sec^2 x} \;dx\\ =\; &\int_0^{\frac{\pi}{3}} \sec x \;dx\\ =\; & \left. \ln|\sec x +\tan x| \right\vert_0^{\frac{\pi}{3}}\\ =\; & \ln\left\vert \sec\left( \frac{\pi}{3}\right) +\tan \left( \frac{\pi}{3}\right) \right\vert - \ln|\sec 0 +\tan 0|\\ =\; & \ln(2 +\sqrt{3}). \end{align*}

The roles of $$x$$ and $$y$$ in the formula of arc length are switched when the graph is given by $$x=f(y)$$ from $$y=c$$ to $$y=d$$:
$L=\int_c^d \sqrt{1+\left[ \frac{df}{dy} \right]^2} \;dy=\int_c^d \sqrt{1+\left( \frac{dx}{dy} \right)^2} \;dy.$

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