Wemazi

Topic 9.3 · BC

Set up and evaluate length

Builds on choosing the parameter interval that traces a path once. 8 minutes.

Adding up short straight segments

A point moves along \(x = t^3\) and \(y = t^2\) for \(0 \le t \le 1\), from the origin to \((1, 1)\). How long is the path?

One straight segment from the first point to the last point has length \(\sqrt2 \approx 1.414\). The path bends away from this segment, so the path is longer.

Use 4 segments. They form a polygon with corners at \(t = 0\), 0.25, 0.5, 0.75 and 1, and its sides add up to about 1.4375.

Use more segments. The total grows and approaches 1.4397, and this limit is the length of the path.

Each segment has length about the speed times the time step \(\Delta t\), so the polygon length is a sum of these products.

The area under the speed graph

Remove the numbers. The upper figure shows the path from \(t = a\) to \(t = b\), and the lower graph plots the speed against \(t\).

Each teal strip below has height equal to the speed and width \(\Delta t\), so its area is about the length of one segment.

Use more steps. The strips become thin, and their total area approaches the area under the speed graph.

The length of the path is therefore the area under the speed graph, which is the integral of the speed.

Notation

  1. Over a short time step \(\Delta t\), the orange segment \(\cy{\Delta s}\) is the hypotenuse of a right triangle with legs \(\Delta x\) and \(\Delta y\). A leg is green when its change is positive and red when it is negative. Because \(\Delta x \approx \frac{dx}{dt}\Delta t\) and \(\Delta y \approx \frac{dy}{dt}\Delta t\), the Pythagorean theorem gives the speed times \(\Delta t\).

    \[\Delta s \approx \cy{\sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}}\,\Delta t\]

  2. Add the steps. As \(\Delta t\) shrinks, the sum becomes an integral over the parameter interval.

    \[L = \int_a^b \cy{\sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}}\,dt\]

  3. The integral needs \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to be continuous on \([a, b]\). The path must be traced once on that interval, because the integral counts a retraced part twice.

  4. The integral of \(\frac{dx}{dt}\) alone is a different quantity: the net change in \(x\), the final \(x\) minus the starting \(x\). The arc length formula for a graph \(y = f(x)\) is the special case \(t = x\).

    \[\int_a^b \dfrac{dx}{dt}\,dt = x(b) - x(a)\]

When it fails

Take the ellipse \(x = 4\cos t\), \(y = 3\sin t\) for \(0 \le t \le 2\pi\). The speed is \(\sqrt{16\sin^2 t + 9\cos^2 t}\), and the speed graph is no longer level.

This speed has no antiderivative made of the usual functions, so you evaluate the definite integral with a calculator. The length is about 22.10.

Example 1: An arc of a circle

A point moves with \(x = 6\cos t\) and \(y = 6\sin t\). Find the length of the path for \(0 \le t \le \frac{\pi}{3}\).

  1. Differentiate each coordinate.

    \[\dfrac{dx}{dt} = -6\sin t,\qquad \dfrac{dy}{dt} = 6\cos t\]

  2. Square and add, and use \(\sin^2 t + \cos^2 t = 1\). The speed is constant, which is why the graph below is level.

    \[\sqrt{36\sin^2 t + 36\cos^2 t} = \sqrt{36} = 6\]

  3. Integrate the speed over the interval.

    \[L = \int_0^{\pi/3} 6\,dt = 2\pi\]

    As a check, the arc is one sixth of a circle with circumference \(12\pi\), and \(\frac{12\pi}{6} = 2\pi\).

Example 2: A curve that speeds up

A point moves with \(x = 6t\) and \(y = 2t^{3/2}\), in meters, for \(0 \le t \le 5\). Find the length of the path.

  1. Find the two rates.

    \[\dfrac{dx}{dt} = 6,\qquad \dfrac{dy}{dt} = 3t^{1/2}\]

  2. A tempting wrong move: Use the arc length formula for a graph, with the slope \(\frac{dy}{dx} = \frac{\sqrt t}{2}\), and integrate with respect to \(t\).

    \[L = \int_0^5 \sqrt{1 + \dfrac{t}{4}}\,dt = \cr{\dfrac{19}{3}}\]

    In the figure, the faint segment from the first point to the last point is about 37.4 m long. A path is never shorter than this segment, but the red bar is only \(\frac{19}{3} \approx 6.3\) long. The graph formula integrates with respect to \(x\), and here \(dx = 6\,dt\), so this integral misses a factor of 6.

  3. Use the speed instead, with both rates under the square root.

    \[\sqrt{6^2 + 9t} = 3\sqrt{4 + t}\]

  4. Integrate. The antiderivative of \(3\sqrt{4 + t}\) is \(2(4 + t)^{3/2}\).

    \[\begin{aligned} L &= \int_0^5 3\sqrt{4 + t}\,dt \\ &= 2(4 + t)^{3/2}\Big|_0^5 \\ &= 2(27 - 8) = 38 \end{aligned}\]

    The length 38 m is more than the 37.4 m segment, as it must be.

Practice

A point moves with \(x = e^t\cos t\) and \(y = e^t\sin t\) for \(0 \le t \le 1\). Find the length of the path.

On the figure, the lower graph shows the speed against t. The length is the area under that graph.

Solution

  1. Differentiate each coordinate with the product rule.

    \[\begin{aligned} \dfrac{dx}{dt} &= e^t(\cos t - \sin t) \\ \dfrac{dy}{dt} &= e^t(\sin t + \cos t) \end{aligned}\]

  2. Square and add. The cross terms cancel, and \(\sin^2 t + \cos^2 t = 1\).

    \[\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 = 2e^{2t}\]

  3. The speed is \(\sqrt2\,e^t\). Integrate it.

    \[L = \int_0^1 \sqrt2\,e^t\,dt = \cy{\sqrt2\,(e - 1)}\]

    As a check, \(\sqrt2\,(e - 1) \approx 2.43\).

Questions

Questions 1 and 2 use the figure. A point moves along \(x = \frac{t^3}{3}\) and \(y = t^2\) for \(0 \le t \le 2\). The small triangle at \(t = 1\) is for a time step of 0.5, and the orange polygon has 2 segments.

  1. In the figure, the legs of the small triangle at \(t = 1\) are 0.5 and 1. About how long is the orange segment of the triangle?

    • 1.12
    • 1.5
    • 1.25
    Answer

    1.12

  2. The polygon with 2 segments has length about 4.85. How does the length of the path compare?

    • A little more than 4.85
    • Equal to 4.85
    • Less than 4.85
    Answer

    A little more than 4.85

  3. A point moves with \(x = 3t^2\) and \(y = 2t^3\) for \(0 \le t \le 1\). What is the length of the path?

    • \(4\sqrt2 - 2\)
    • 5
    • \(\sqrt{13}\)
    • \(\frac{96}{5}\)
    Answer

    \(4\sqrt2 - 2\)

Summary

You find the length of a parametric path by integrating the speed, \(\sqrt{(x')^2 + (y')^2}\), over the parameter interval.

This works because a short step has length about the speed times \(\Delta t\), and the integral adds all the steps.

You use it when the path is traced once on the interval and \(x'\) and \(y'\) are continuous.

This is one of four skills in topic 9.3 of Which & Why: AP Calculus.