Wemazi

Topic 9.3 · BC

Account for retracing

Builds on setting up and evaluating the length of a parametric path. 7 minutes.

Out and back on a rail

A cart rolls on a straight rail. Its position is \(x = 3(1 - \cos t)\) and \(y = 4(1 - \cos t)\) for \(0 \le t \le 2\pi\), in meters, and it starts at the origin.

Set \(t = \pi\). The cart is at \((6, 8)\), the far end of the 10 m rail, and the distance traveled is 10 m.

Set \(t = 2\pi\). The cart returns to the origin along the same rail, which turns red where the cart passes over it a second time.

The orange bar, the distance traveled, grows to 20 m, but the faint bar, the rail covered, stays at 10 m.

The distance traveled counts every pass. The length of the path counts each point of the rail once.

Every pass, or each point once

Remove the numbers. The orange bar shows the distance traveled so far, and the faint bar shows the length of the path covered so far.

The speed is never negative, so the distance traveled only grows, even along a stretch that the point passed before.

The path is the set of points that the moving point visits, and its length counts each point once.

Move the time. On a new part of the circle, both bars grow. On a part passed a second time, which turns red, only the orange bar grows.

The two bars are equal only while the point passes along each part once.

Notation

  1. The distance traveled from \(t = a\) to \(t = b\) is the integral of the speed, and it adds every pass because the speed is never negative.

    \[\begin{aligned} \text{distance} &= \int_a^b \cy{\text{speed}}\,dt \\ \cy{\text{speed}} &= \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} \end{aligned}\]

  2. To find the length of the path, find where the point turns back or begins to repeat, and split the interval there. Then count each part of the path once.

    \[\begin{aligned} \text{length} &= \text{distance} \\ &\quad - \text{repeated passes} \end{aligned}\]

  3. When you simplify the root of a square, keep the absolute value, because a square root is never negative.

    \[\sqrt{(2t - 4)^2} = |2t - 4|\]

When it fails

The length formula assumes that the path is traced once. Extend the interval to \(0 \le t \le 4\pi\), and the point goes around the circle twice.

The integral of the speed is now twice the length of the circle. It gives the distance traveled, not the length of the path.

Example 1: Going around a circle

A point moves with \(x = 4\cos\pi t\) and \(y = 4\sin\pi t\) for \(0 \le t \le 3\), in meters. Find the distance traveled and the length of the path traced.

  1. The derivatives are \(\dfrac{dx}{dt} = -4\pi\sin\pi t\) and \(\dfrac{dy}{dt} = 4\pi\cos\pi t\), so the speed is constant.

    \[\begin{aligned} &\sqrt{16\pi^2\sin^2\pi t + 16\pi^2\cos^2\pi t} \\ &= 4\pi \end{aligned}\]

  2. Integrate the speed over the whole interval.

    \[\text{distance} = \int_0^3 4\pi\,dt = 12\pi\]

    As a check, \(12\pi \approx 37.7\) m.

  3. The functions \(\cos\pi t\) and \(\sin\pi t\) repeat when \(t\) increases by 2, so \(0 \le t \le 2\) traces the circle once. The path is the circle.

    \[\text{length} = \int_0^2 4\pi\,dt = 8\pi\]

  4. From \(t = 2\) to \(t = 3\), the point goes over the upper half again, the red arc. The difference is the length of that half circle.

    \[12\pi - 8\pi = 4\pi\]

Example 2: A bug that turns around

A bug moves with \(x = 4(t^2 - 4t)\) and \(y = 3(t^2 - 4t)\) for \(0 \le t \le 5\), in meters. Find the distance traveled and the length of the path.

  1. Because \(y = \tfrac{3}{4}x\), the bug stays on a line through the origin. The factor \(t^2 - 4t\) has derivative \(2t - 4\).

    \[\begin{aligned} \dfrac{dx}{dt} &= 4(2t - 4) \\ \dfrac{dy}{dt} &= 3(2t - 4) \end{aligned}\]

  2. A tempting wrong move: The speed is \(\sqrt{25(2t - 4)^2}\). Simplify the root and integrate.

    \[\int_0^5 5(2t - 4)\,dt = \cr{25}\]

    The root of a square is never negative, but \(5(2t - 4)\) is negative for \(t < 2\). The integral counts the 20 m that the bug travels in that time as \(-20\). The value 25 is only the distance from the first point to the last point, the teal arrow.

  3. The speed is \(5|2t - 4|\). The bug turns around at \(t = 2\), so split the integral there. Remove the absolute value on each part.

    \[\begin{aligned} \int_0^2 5(4 - 2t)\,dt &= 20 \\ \int_2^5 5(2t - 4)\,dt &= 45 \end{aligned}\]

  4. The distance traveled is 65 m. The bug covers the line from \((-16, -12)\) to \((20, 15)\), which is 45 m long, so it travels the stretch from the origin to \((-16, -12)\) twice.

    \[\text{distance} = 65,\qquad \text{length} = 45\]

Practice

A bug moves with \(x = 12(4t - t^2)\) and \(y = 5(4t - t^2)\) for \(0 \le t \le 5\), in meters. Find the distance traveled and the length of the path.

Move the slider to find the time when the bug turns around. The distance adds both passes, but the length counts the red stretch once.

Solution

  1. The speed \(13|4 - 2t|\) is 0 at \(t = 2\), where the bug turns around.

    \[\begin{aligned} &\sqrt{144(4 - 2t)^2 + 25(4 - 2t)^2} \\ &= 13|4 - 2t| \end{aligned}\]

  2. Integrate on each side of \(t = 2\). The distance traveled is \(52 + 117 = 169\) m.

    \[\begin{aligned} \int_0^2 13(4 - 2t)\,dt &= 52 \\ \int_2^5 13(2t - 4)\,dt &= 117 \end{aligned}\]

  3. Each point is \(4t - t^2\) times \((12, 5)\), a vector of length 13. The factor takes every value from \(-5\) at \(t = 5\) to 4 at \(t = 2\), a range of 9, so the path is \(13 \cdot 9 = 117\) m long.

    \[\text{length} = 117\]

Questions

Questions 1 and 2 use the figure. A point moves along \(x = 5\cos t\) and \(y = 5\sin t\) for \(0 \le t \le 3\pi\). The red part is traced a second time.

  1. Which part of the circle is traced twice?

    • The upper half
    • The lower half
    • The whole circle
    Answer

    The upper half

  2. What is the distance traveled for \(0 \le t \le 3\pi\)?

    • \(15\pi\)
    • \(10\pi\)
    • \(5\pi\)
    Answer

    \(15\pi\)

  3. A point moves with \(x = 3(t^3 - 3t)\) and \(y = 4(t^3 - 3t)\) for \(0 \le t \le 2\). What is the length of the path, with each part counted once?

    • 20
    • 30
    • 10
    Answer

    20

Summary

The distance traveled is the integral of the speed, and it counts every pass because a speed is never negative.

The length of the path counts each point once, so it is shorter when the point turns back or repeats a lap.

Before you call the integral of the speed a length, check that the interval traces the path once.

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