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Suppose that y=f(x), is a smooth curve on the interval [a,b]. Define and find a formula for the arc length L of the curve y=f(x) over the interval [a,b].

  Arc Length y= f(x) x=a to x=b Arc Length y= f(x) x=a to x=b n strips

The basic idea for defining arc length is to break up the curve into small segments, approximating the curve segments by line segments, add the lengths of the line segments to form a Riemann sum that approximates the arc length L, and take the limit of the Riemann sums to obtain an integral for L.

To obtain this idea, divide the interval [a,b] into n subintervals by inserting points x1,x2,x3,...,xn-1 between a and b. As shown in the figure above, let P0, P1, P2,...,Pn be the points on the curve with x-coordinates a,x1,x2,...,xn-1,b and join these points with straight line segments. These line segments form a polygonal path that we can regard as approximation to the curve y=f(x). As suggested by the Figure below, the Length Lk of the kth line segment in the polygonal path is Length Lk of the kth line segment

  Arc Length Polygonal Path

If we now add the lengths of these line segments, we obtain the following approximation to the length L of the curve


Length of all line segments

To put this in the form of a Riemann sum we will apply the Mean-Value Theorem. The theorem implies that there is a point any point in the kth subinterval between xk-1 and xk such that


Length of all segments Mean Value Theorem i


and hence we can rewrite as


Length of all segments Mean Value Theorem ii

Thus, taking the limit as n increases and the widths of the subintervals approach zero yields the following integral that defines the arc length L:


Length of all segments Mean Value Theorem i


EXAMPLE

Find the arc length of the curve y=x3/2  from x=1 to x=2

 Arc Length y=x^1.5 x=1 to x=2.828

Solution:

  Arc Length Solution i
Arc Length Solution ii
Arc Length Solution iii

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