Arc length = .

We find the length of a curved segment.

example 1 Find the length of the curve from to .

Since the graph of this function is a line, calculus is not required to find the length of the indicated segment. Instead, we can use the distance formula, with the points and to get We will also use the formula from the above theorem as a verification that the theorem is correct. Since , we have and the length of the segment is as expected.

Since the graph of this function is a line, calculus is not required to find the length of the indicated segment. Instead, we can use the distance formula, with the points and to get We will also use the formula from the above theorem as a verification that the theorem is correct. Since , we have and the length of the segment is as expected.

(problem 1b) Find the length of the curve from to using the distance
formula.

The endpoints of the line segment are and

Arc length = .

example 2 Find the length of the graph of from to .

First, and so . Then the arc length is To compute this integral, we will use a -substitution with and . Thus, the anti-derivative is,

First, and so . Then the arc length is To compute this integral, we will use a -substitution with and . Thus, the anti-derivative is,

Returning to our computation of arc length,

example 3 Find the length of the graph of from to .

In this example, computing a simplified form of the integrand is laborious, so we will proceed slowly. First, , so and Next, we add one: and the key to this problem is that this expression is a perfect square: Now, we can compute the square root: Finally, we can integrate to get the arc length: \begin{align*} L &= \int _1^2 \sqrt{1+f'(x)^2} \; dx \\ &= \int _1^2 \left (\frac{x^2}{4} + \frac{1}{x^2} \right ) \; dx \\ &= \int _1^2 \left (\frac{x^2}{4} + x^{-2} \right ) \; dx \\ &= \left (\frac{x^3}{12} + \frac{x^{-1}}{-1} \right ) \bigg |_1^2 \\ &= \left (\frac{x^3}{12} - \frac{1}{x} \right ) \bigg |_1^2 \\ &= \left (\frac 23 - \frac 12\right ) - \left (\frac{1}{12} - 1\right ) \\ &= \frac{13}{12}. \end{align*}

In this example, computing a simplified form of the integrand is laborious, so we will proceed slowly. First, , so and Next, we add one: and the key to this problem is that this expression is a perfect square: Now, we can compute the square root: Finally, we can integrate to get the arc length: \begin{align*} L &= \int _1^2 \sqrt{1+f'(x)^2} \; dx \\ &= \int _1^2 \left (\frac{x^2}{4} + \frac{1}{x^2} \right ) \; dx \\ &= \int _1^2 \left (\frac{x^2}{4} + x^{-2} \right ) \; dx \\ &= \left (\frac{x^3}{12} + \frac{x^{-1}}{-1} \right ) \bigg |_1^2 \\ &= \left (\frac{x^3}{12} - \frac{1}{x} \right ) \bigg |_1^2 \\ &= \left (\frac 23 - \frac 12\right ) - \left (\frac{1}{12} - 1\right ) \\ &= \frac{13}{12}. \end{align*}

Here is a detailed, lecture style video on arc length:

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In this section, we derive the arc length formula. Suppose is a differentiable function on the open interval and continuous on the closed interval . Let and let for . Note that and . Next, to approximate the arc length, connect the adjacent points and with straight line segments for From the distance formula, the length of the segment is

Now, we apply the Mean Value Theorem to on the interval to conclude that for some number in the interval . Thus,

Returning to the arc length, , of on , we have Taking the limit as , we define the arc length as

2024-09-27 13:57:56