What if the derivative is undefined at a point in the interval?

If the derivative $f'(x)$ approaches infinity, express the curve as a function of $y$, $x = g(y)$, and integrate with respect to $y$ over the corresponding interval.

Can arc length be negative?

No. The integrand $\sqrt{1 + [f'(x)]^2}$ is always strictly positive, and integrating from $a$ to $b$ where $b > a$ yields a positive result.

What if the integral cannot be evaluated exactly?

Many arc length integrals cannot be solved using elementary functions. In these cases, use numerical integration techniques such as Simpson's rule or a calculator to approximate the definite integral.

How to find arc length

The arc length of a curve is found by integrating the differential of arc length over a specified interval. This method applies to smooth curves where the derivative of the function is continuous on the interval of integration.

The setup

For a curve defined by $y = f(x)$ from $x = a$ to $x = b$ , the arc length $L$ is given by $L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$ . If the curve is instead defined by $x = g(y)$ from $y = c$ to $y = d$ , use $L = \int_c^d \sqrt{1 + [g'(y)]^2} dy$ .

The steps

Identify the function and the interval of integration. 2. Compute the first derivative of the function. 3. Square the derivative and add 1. 4. Set up the integral by taking the square root of the result from step 3. 5. Evaluate the definite integral.

Checking the result

Compare the calculated arc length to the straight-line distance between the endpoints, which is $\sqrt{(b-a)^2 + (f(b)-f(a))^2}$ . The arc length must be greater than or equal to this straight-line distance.

Common errors

A frequent error is forgetting to square the derivative before adding 1. Another common mistake is incorrectly expanding squares or failing to recognize when an integral requires numerical approximation because the square root expression is algebraically intractable.

Worked example

Find the arc length of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$ .

Identify $f(x) = \frac{2}{3}x^{3/2}$ on $[0, 3]$ . 2. Compute the derivative: $f'(x) = x^{1/2}$ . 3. Square and add one: $1 + [f'(x)]^2 = 1 + x$ . 4. Set up the integral: $L = \int_0^3 \sqrt{1 + x} dx$ . 5. Evaluate the integral using substitution $u = 1 + x$ , $du = dx$ . The limits change to $u(0)=1$ and $u(3)=4$ . $L = \int_1^4 u^{1/2} du = [\frac{2}{3}u^{3/2}]_1^4 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{14}{3}$ .

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References: Calculus: Early Transcendentals by James Stewart · OpenStax Calculus Volume 2