Wemazi

Topic 5.1

Apply the theorem to a real situation

Builds on using a graph or a table to identify a guarantee. 7 minutes.

A cart on a ramp

The top graph shows the distance in meters that a cart rolls down a ramp in the first 4 seconds. The bottom graph shows its speed at each moment, in meters per second.

The solid straight line on the top graph is the secant line from the start to the end. The cart rolls 20 m in 4 s, so the slope of the secant line is the average speed, 5 m/s.

Drag the point. The slope of the orange tangent is the speed at that moment, and the orange bar below shows the same number.

The speed rises from 1 m/s at the start to 13 m/s at the end, so it passes 5 m/s. At that moment the bar reaches the dashed line at 5 m/s, and the tangent lies along the dashed copy of the secant line.

A rate at a moment and on average

The curve now shows any quantity that changes with time. The slope of the secant line from \(a\) to \(b\) is the average rate. The slope of the tangent is the rate at one moment.

Drag the point. At least one moment inside the interval has the two rates equal, and there the tangent is parallel to the secant line.

Notation

  1. To find the average rate of change, divide the total change by the elapsed time.

    \[\cy{m} = \dfrac{f(b) - f(a)}{b - a}\]

  2. The quantity must be continuous on the interval and differentiable inside it. The problem states this, or the setting makes it clear.

  3. Then at some time \(c\) inside the interval, the instantaneous rate, the rate at that one moment, equals the average rate.

    \[f'(c) = \cy{m}\]

  4. Write the answer in words, with units: at some moment between 0 s and 4 s, the speed was 5 m/s.

When it fails

The first button moves the point to the middle, \(t = 2\) s. The speed there is 4 m/s, not 5 m/s, so the tangent is not parallel to the secant line.

The speed is 5 m/s at about 2.31 s. The theorem does not give the time, so do not name one unless you solve for it.

Example 1: a tank

A tank holds 200 L at 8:00 and 380 L at 8:06. The volume is a differentiable function of time. Show that at some moment the tank was filling at 30 L/min.

  1. Let \(V(t)\) be the volume in liters, \(t\) minutes after 8:00. The volume is differentiable, so the conditions are met on \([0, 6]\).

  2. The slope of the secant line between the readings is the average rate of change.

    \[\cy{m} = \dfrac{380 - 200}{6 - 0} = \cy{30} \text{ L/min}\]

  3. By the theorem, at some time between 8:00 and 8:06 the volume was increasing at 30 L/min.

    \[V'(c) = 30\]

Example 2: a speed limit

A car passes gate A at 8:00 and gate B at 9:00. The gates are 110 km apart, and the limit is 100 km/h. The distance from gate A is \(s(t) = 100t + 10t^3\) km after \(t\) hours. Show that the car broke the limit at some moment.

  1. The car covers 110 km in 1 h, so its average speed is 110 km/h, the dashed line on the bottom graph.

    \[\begin{aligned} \cy{m} &= \dfrac{s(1) - s(0)}{1 - 0} \\ &= \dfrac{110 - 0}{1 - 0} = 110 \end{aligned}\]

  2. A tempting wrong move: The moment is the middle of the hour, so the speed was 110 km/h at 8:30.

    \[s'(\tfrac12) = 110\]

    At 8:30 the speed is \(s'(\tfrac12) = 100 + 30 \cdot \tfrac14 = 107.5\) km/h, not 110. The bar stops below the dashed line, and the tangent is not parallel to the secant line.

  3. The theorem guarantees a time when the speed was 110 km/h, but it does not name the time. Since 110 is above 100, the car broke the limit.

  4. As a check, the time is about 0.58 h after 8:00, which is about 8:35.

    \[\begin{aligned} 100 + 30c^2 &= 110 \\ c &= \dfrac{1}{\sqrt3} \end{aligned}\]

Practice

A cyclist's odometer reads 18 km at 1:00 pm and 63 km at 3:30 pm. Her distance is a differentiable function of time. State a speed that she had at some moment.

The average speed is the slope of the line between the two readings, with the time in hours.

Solution

  1. The secant line joins the readings at 1:00 pm and 3:30 pm, 2.5 h apart.

  2. The average speed is the change in distance divided by the elapsed time.

    \[\cy{m} = \dfrac{63 - 18}{2.5} = 18 \text{ km/h}\]

  3. By the theorem, at some time between 1:00 pm and 3:30 pm her speed was 18 km/h.

Questions

The graph shows the distance in kilometers that a bus travels in the first \(t\) hours after it leaves, a differentiable function of time.

  1. What speed does the bus have at some moment between \(t = 0\) and \(t = 4\)?

    • 40 km/h
    • 160 km/h
    • 70 km/h
    Answer

    40 km/h

  2. What speed does the theorem guarantee at some moment between \(t = 1\) and \(t = 3\)?

    • 35 km/h
    • 70 km/h
    • 40 km/h
    • \(\tfrac{110}{3}\) km/h
    Answer

    35 km/h

  3. A runner finishes a 42 km race in 2.5 h, and her distance is a differentiable function of time. Which statement does the theorem support?

    • At some moment her speed was 16.8 km/h.
    • At 1.25 h her speed was 16.8 km/h.
    • Her speed was 16.8 km/h at every moment.
    • At some moment her speed was 105 km/h.
    Answer

    At some moment her speed was 16.8 km/h.

Summary

You divide the total change by the elapsed time and report that the rate at some moment had this value, with units.

This works because the rate at one moment is the slope of a tangent, and some tangent is parallel to the secant line.

You use it when a problem gives two readings and asks about the rate at some moment.

This is one of five skills in topic 5.1 of Which & Why: AP Calculus.