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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 \(y=d\):
\[L=\int_c^d \sqrt{1+\left[ \frac{df}{dy} \right]^2} \;dy=\int_c^d \sqrt{1+\left( \frac{dx}{dy} \right)^2} \;dy. \]

Arc Length