Finding The Arc Length Of A Curve

Let \(f(x)\) be a smooth function defined over \([a,b]\). We want to calculate the length of the curve from the point \((a,f(a))\) to the point \((b,f(b))\). If we want to find the area under the curve, we would divide the area into \(n\) rectangles and add the area of those rectangles:

Image from openstax.org (Calculus Volume 1)

The estimate of the area under the curve would get better and better as \(n \to \infty\). Can we do a similar thing with the arc length (i.e. divide the arc into \(n\) horizontal intervals and add the width of those intervals)? The problem with this is that for any curve, the estimated length will always be \(\lim_{n \to \infty} \Delta x = \int_a^b dx = b-a\). This is clearly wrong. Instead, for \(i=1,2,…,n\), we can construct a line segment from the point \((x_{i-1},f(x_{i-1}))\) to the point \((x_i,f(x_i))\):

Image from openstax.org (Calculus Volume 1)

To find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval:

Image from openstax.org (Calculus Volume 1)

By the Pythagorean theorem the length of the line segment \(s\) is:

\[\begin{align} s &= \sqrt{(\Delta x)^2 + (\Delta y_i)^2} \\ &= \Delta x \sqrt{1 + \frac{\Delta y_i}{\Delta x}^2}\end{align}\]

According to the Mean Value Theorem, there should be a value \(x_i^*\) such that \(f'(x_i^*) = \frac{\Delta y_i}{\Delta x}\).:

\[s = \sqrt{1+[f'(x_i^*)]^2} \Delta x \]

Adding up all the segments we get:

\[ \text{Arc Length} \approx \sum^n_{i=1} \sqrt{1+[f'(x_i^*)]^2} \Delta x \]

Taking the limit as \(n→∞\):

\[ \text{Arc Length} = \int^b_a \sqrt{1+ [f'(x)]^2} \ dx \]

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