Link Between Distance Function and Derivative of Position Vector

THROW YOUR RULER, an integral will be all you need to measure distance. In fact, the integral can measure distances your ruler cannot.

You’re an engineer tasked with measuring the distance traveled by a rollercoaster on a parabolic path. Let the function $y = x^2$ on the interval $[-1, 1]$ represent that parabola. Keep in mind that the rollercoaster doesn’t exist yet, so we cannot directly go and measure it. We only know that it will follow the shape of the above quadratic.

The $x^2$ function curves too much for a straight line to measure it accurately. Since a parabola is symmetric, if we can measure the distance from 0 to 1, doubling our result will give us the distance on the full interval $[-1, 1]$.

[FIGURE 1: The parabola $y = x^2$ on the interval $[0, 1]$ that we want to measure]

Building the Approximation

You know how to find the distance between two points on a curve using the distance formula and the Pythagorean theorem. For two points $P_i = (x_i, f(x_i))$ and $P_{i-1} = (x_{i-1}, f(x_{i-1}))$ on our curve, the distance between them is:

$$d(P_{i-1}, P_i) = \sqrt{(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2}$$

[FIGURE 2: Two consecutive points on the parabola illustrating the distance calculation with the horizontal and vertical components $\Delta x$ and $\Delta y$]

By segmenting our interval into $n$ different points, we can approximate the arc length of the function, though initially quite crudely. We partition the interval $[0, 1]$ into $n$ subintervals, each of width $\Delta x = \frac{1}{n}$.

The total arc length is approximately:

$$L \approx \sum_{i=1}^{n} d(P_{i-1}, P_i) = \sum_{i=1}^{n} \sqrt{(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2}$$

[FIGURE 3: Piecewise linear approximation with $n=10$ segments showing how we connect consecutive points with straight lines]

[FIGURE 4: Comparison showing approximations with $n = 5, 10, 20, 50$ segments demonstrating how the approximation improves as $n$ increases]

The Limit Process and the Integral

If you know calculus, you can predict the technique we’re going to use to refine our approximation. The limit that this sum approaches as $n \to \infty$ is given by a definite integral.

Let’s expand our distance function more carefully. For each segment:

$$d(P_{i-1}, P_i) = \sqrt{(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2}$$

Let $\Delta x = x_i - x_{i-1}$ and $\Delta y = f(x_i) - f(x_{i-1})$. Then:

$$d(P_{i-1}, P_i) = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

Factoring out $(\Delta x)^2$:

$$d(P_{i-1}, P_i) = \sqrt{(\Delta x)^2\left(1 + \frac{(\Delta y)^2}{(\Delta x)^2}\right)}$$$$= |\Delta x| \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}$$$$= \Delta x \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}$$

(assuming $\Delta x > 0$, which is true as we move from left to right along the curve)

Invoking the Mean Value Theorem

By the Mean Value Theorem, for a differentiable function $f$ on the interval $[x_{i-1}, x_i]$, there exists a point $c_i \in (x_{i-1}, x_i)$ such that:

$$f'(c_i) = \frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}} = \frac{\Delta y}{\Delta x}$$

This is a crucial step! The Mean Value Theorem guarantees that somewhere in each subinterval, the instantaneous rate of change (the derivative) equals the average rate of change. This allows us to replace the ratio $\frac{\Delta y}{\Delta x}$ with the derivative at some point $c_i$.

Therefore:

$$d(P_{i-1}, P_i) = \Delta x \sqrt{1 + [f'(c_i)]^2}$$

The total arc length approximation becomes:

$$L \approx \sum_{i=1}^{n} \Delta x \sqrt{1 + [f'(c_i)]^2}$$

This is a Riemann sum! As $n \to \infty$ (and $\Delta x \to 0$), this Riemann sum converges to the definite integral:

$$L = \int_{0}^{1} \sqrt{1 + [f'(x)]^2} \, dx$$

We call the integrand the differential arc length element:

$$ds = \sqrt{1 + [f'(x)]^2} \, dx$$

Computing the Arc Length for $y = x^2$

For our specific function $f(x) = x^2$, we have $f'(x) = 2x$. Therefore, the arc length from $x = 0$ to $x = 1$ is:

$$L = \int_{0}^{1} \sqrt{1 + (2x)^2} \, dx = \int_{0}^{1} \sqrt{1 + 4x^2} \, dx$$

To evaluate this integral, we use the trigonometric substitution:

$$2x = \tan\theta \quad \Rightarrow \quad dx = \frac{1}{2}\sec^2\theta \, d\theta$$

When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \arctan(2)$.

Substituting:

$$L = \int_{0}^{\arctan(2)} \sqrt{1 + \tan^2\theta} \cdot \frac{1}{2}\sec^2\theta \, d\theta$$$$= \int_{0}^{\arctan(2)} \sec\theta \cdot \frac{1}{2}\sec^2\theta \, d\theta$$$$= \frac{1}{2}\int_{0}^{\arctan(2)} \sec^3\theta \, d\theta$$

Using the standard integral $\int \sec^3\theta \, d\theta = \frac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|) + C$, we get:

$$L = \frac{1}{2}\left[\frac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|)\right]_{0}^{\arctan(2)}$$$$= \frac{1}{4}\left[(\sqrt{5} \cdot 2 + \ln|\sqrt{5} + 2|) - (1 \cdot 0 + \ln|1|)\right]$$$$= \frac{1}{4}\left[2\sqrt{5} + \ln(\sqrt{5} + 2)\right]$$$$\approx \frac{1}{4}(4.472 + 1.444) \approx 1.479$$

Alternatively, using the hyperbolic substitution formula:

$$L = \left[\frac{x}{2}\sqrt{1 + 4x^2} + \frac{1}{4}\sinh^{-1}(2x)\right]_{0}^{1} = \frac{\sqrt{5}}{2} + \frac{1}{4}\sinh^{-1}(2) \approx 1.479$$

The total arc length of the parabola $y = x^2$ on the full interval $[-1, 1]$ is therefore:

$$L_{total} = 2L \approx 2(1.479) = 2.958$$

In Summary

We’ve seen how the integral emerges naturally from the problem of measuring curved distances. By:

1. Approximating the curve with straight line segments
2. Expressing the length of each segment using the Pythagorean theorem
3. Applying the Mean Value Theorem to connect the discrete approximation to the continuous derivative
4. Taking the limit as the number of segments approaches infinity

We arrived at the arc length formula:

$$\boxed{L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx}$$