Precalculus · F.IF.B.6

Rates at an instant from shrinking intervals

A speedometer shows a speed at one instant, but speed is distance over time and an instant has no time in it. How can a rate belong to a single moment?

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The difference quotientPrecalculus

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Concavity: when the rate of change itself changesPrecalculus

Where you are 4 in Average rate of change

Functions13

What a function is5

Graphs of functions5

Features of a graph10

Average rate of change4

Average rate of change from a formula Average rate of change from a table or graph

Rates at an instant from shrinking intervals

The difference quotient

Why this exists

An average rate of change needs two inputs: change in output over change in input. A speedometer says how fast the car is going now, at one input, where there is no change at all and the fraction would be \displaystyle \frac{0}{0}.

The way out is a whole family of intervals around the instant, narrower and narrower. Each has an ordinary average rate. If those averages settle on one number as the width shrinks, from both sides, that number is the rate at the instant. The difference quotient already showed averages crowding toward such a number; this lesson says what it is, how to estimate it well, how to read it on the graph, and when there is none.

Definition

Fix an input a. For a small width h \ne 0, the average rate of f between a and a + h is the difference quotient

\frac{f(a + h) - f(a)}{h}.

With h > 0 the interval [a, a + h] lies to the right of a; with h < 0 the interval [a + h, a] lies to the left. Call these the right estimates and the left estimates.

The rate of change of f at a, also called the instantaneous rate, is the number that the right estimates and the left estimates both approach as h shrinks toward 0. If the two sides approach different numbers, or either side approaches no number, f has no rate of change at a.

A third kind of interval sits evenly around a, from a - h to a + h, with width 2h. Its average rate is the centered average:

\frac{f(a + h) - f(a - h)}{2h}, \qquad h > 0.

The rate has the units of an average rate, such as meters per second.

Intuition

Mark the point P = (a, f(a)) on the graph. Every estimate is the slope of a secant through P and one other point of the graph. As h shrinks, that point slides toward P and the secant pivots about P, homing in, for a smooth curve, on the line that touches the curve at P and runs in its direction. That line is the tangent line at P, and the rate of change at a is its slope.

Figure 1The maglev train’s position in Examples 1 to 3. The right secant has slope 11.5, the left one 5.5; as the width shrinks both pivot toward the dashed tangent line of slope 8.

Equally, zoom in on a smooth curve around P and it looks straighter and straighter, until it cannot be told apart from its tangent line. A sharp corner never straightens: y = |x| at the origin is the same V at every scale, slope -1 on the left and 1 on the right. That is why a corner has no rate.

Results

The rate is the slope of the tangent line. If f has rate r at a, the tangent line at (a, f(a)) is

y = f(a) + r(x - a),

and near a it estimates f itself well (Tangent lines and normal lines).

The centered average is the mean of a left and a right estimate of the same width:

\begin{aligned} &\frac{f(a + h) - f(a - h)}{2h} \\ &\quad = \frac{1}{2}\left(\frac{f(a + h) - f(a)}{h} + \frac{f(a) - f(a - h)}{h}\right). \end{aligned}

For a quadratic, the centered average is exact. If f(x) = px^2 + qx + c, the centered average at a equals 2pa + q for every h > 0, and that is the rate at a. For a line every estimate equals its slope.

Bracketing. If, as the width shrinks, the left estimates rise and the right estimates fall toward the same number, that number lies between every left estimate and every right estimate.

Why it is true

The mean formula is algebra. Put the two fractions over their common denominator 2h: the sum \big(f(a + h) - f(a)\big) + \big(f(a) - f(a - h)\big) telescopes to f(a + h) - f(a - h).

For the quadratic, expand: f(a + h) - f(a - h) is p\big((a + h)^2 - (a - h)^2\big) + q\big((a + h) - (a - h)\big), which is 4pah + 2qh, because the a^2, h^2 and constant terms cancel in pairs. Dividing by 2h gives 2pa + q, with no h left. Meanwhile the one-sided estimate is 2pa + q + ph, which approaches 2pa + q from both sides as h shrinks, so 2pa + q is the rate.

For bracketing, a rising list of numbers that approaches L never passes L, so every left estimate is at most L; likewise every right estimate is at least L. The tangent-line statement is the definition read on the graph: the secant slopes through P approach r.

Worked examples

Example 1

The maglev train from both sides

Mihaela is running a maglev train test. In the first seconds of the run the train’s distance from the start is s(t) = 0.5t^3 + 2t meters, t seconds after it starts. Estimate its speed at t = 2 from intervals on both sides.

Solution

First simplify the difference quotient at t = 2, so that each estimate is quick. Here s(2) = 4 + 4 = 8, and expanding (2 + h)^3 = 8 + 12h + 6h^2 + h^3,

\begin{aligned} s(2 + h) - s(2) &= 0.5(12h + 6h^2 + h^3) + 2h \\ &= 8h + 3h^2 + 0.5h^3. \end{aligned}

Dividing by h \ne 0, the average rate between 2 and 2 + h is 8 + 3h + 0.5h^2 meters per second.

h	interval	average rate
1	[2, 3]	11.5
0.1	[2, 2.1]	8.305
0.01	[2, 2.01]	8.03005
0.001	[2, 2.001]	8.0030005
-0.001	[1.999, 2]	7.9970005
-0.01	[1.99, 2]	7.97005
-0.1	[1.9, 2]	7.705
-1	[1, 2]	5.5

The right estimates fall toward 8 and the left ones rise toward 8; the excess h(3 + 0.5h) has the sign of h for every width used. By bracketing the speed lies between 7.9970005 and 8.0030005: the train is moving at 8 meters per second at t = 2.

Check the first row directly: s(3) = 13.5 + 6 = 19.5, and \displaystyle \frac{19.5 - 8}{1} = 11.5.

Example 2

Centered intervals

Use centered intervals around t = 2 for the same train, and compare their accuracy with the one-sided estimates of Example 1.

Solution

Example 1 gives s(2 + h) = 8 + 8h + 3h^2 + 0.5h^3, and putting -h for h gives s(2 - h) = 8 - 8h + 3h^2 - 0.5h^3. Subtracting, the 8 and the 3h^2 cancel:

\frac{s(2 + h) - s(2 - h)}{2h} = \frac{16h + h^3}{2h} = 8 + 0.5h^2.

With h = 1 this is 8.5, the mean of 11.5 and 5.5, as the Results promise. With h = 0.1 it is 8.005, and with h = 0.01 it is 8.00005.

The one-sided estimates miss 8 by about 3h, the centered ones by 0.5h^2: at h = 0.1, 0.3 against 0.005. Left and right errors are nearly equal and opposite; the mean cancels most of them.

Example 3

The tangent line

Batbayar, on the trackside crew, wants a quick forecast of where the train will be shortly after t = 2. Find the tangent line to s(t) = 0.5t^3 + 2t at t = 2 and use it to estimate s(2.1).

Solution

The point is (2, 8) and the rate there is 8 (Example 1), so the tangent line is

y = 8 + 8(t - 2) = 8t - 8.

At t = 2.1 it gives 8(2.1) - 8 = 8.8 meters. The true position is s(2.1) = 0.5(9.261) + 4.2 = 8.8305 meters, so the forecast is short by 0.0305 meters: the curve bends upward away from the line, as in the figure, because the train is speeding up.

Example 4

From a table of measurements

Chidinma logs the train’s position as it brakes, every half second.

t (seconds)	x (meters)
10	850.0
10.5	856.4
11	861.6
11.5	865.4
12	867.6

Estimate the train’s speed at t = 11.

Solution

With no formula, the width cannot shrink below the table’s spacing (average rates from a table).

Right, h = 0.5: \displaystyle \frac{865.4 - 861.6}{0.5} = 7.6. Left, h = 0.5: \displaystyle \frac{861.6 - 856.4}{0.5} = 10.4.

They disagree by almost 3 meters per second: the train is slowing quickly. The centered average with h = 0.5 uses the rows on either side:

\frac{865.4 - 856.4}{1} = 9.0,

which is the mean of 7.6 and 10.4. With h = 1 the centered average is \displaystyle \frac{867.6 - 850.0}{2} = 8.8. The two centered values agree much more closely than the one-sided ones, so the best estimate from this log is about 9 meters per second.

Example 5

When the estimates settle

Estimate the rate of change of f(x) = 2^x at x = 3 to three decimal places.

Solution

Here 2^{3 + h} - 8 = 8(2^h - 1) has no factor of h to cancel, so compute numerically. Centered averages, rounded:

h	centered average
0.1	5.54962
0.01	5.54522
0.001	5.54518

From h = 0.01 to h = 0.001 the first four digits, 5.545, no longer change, so the rate at x = 3 is 5.545 to three decimal places. The one-sided estimates at h = 0.001 are 5.54710 on the right and 5.54326 on the left; they bracket 5.545 but already differ in the third decimal place, the same story as Example 2.

Settling is judged from the numbers: shrink the width until the digits you need stop changing, and confirm with both sides. So the graph of the exponential function 2^x climbs about 5.545 units per unit at x = 3.

Example 6

A corner

Find the left and right estimates and the centered averages of g(x) = |x^2 - 1| at x = 1. Does g have a rate of change at 1?

Solution

Here g(1) = 0. For h > 0, (1 + h)^2 > 1, so the absolute value changes nothing and

\frac{g(1 + h) - g(1)}{h} = \frac{2h + h^2}{h} = 2 + h.

For h < 0 near 0, (1 + h)^2 < 1, so the absolute value flips the sign: g(1 + h) = -2h - h^2, and the quotient is -2 - h.

The right estimates approach 2 and the left estimates approach -2. They do not approach the same number, so g has no rate of change at x = 1; the graph has a corner there, as in the figure.

Figure 2The corner in Example 6. The right secant has slope 3 and the left one slope -1; shrinking the width sends them toward 2 and -2, never toward one line.

The centered average hides this. For h > 0, g(1 - h) = 2h - h^2, so

\frac{g(1 + h) - g(1 - h)}{2h} = \frac{2h^2}{2h} = h,

which approaches 0, the mean of 2 and -2 and a slope neither side has. A centered average is trustworthy only once both sides are known to agree.

Boundaries

The two sides must agree. A rate at a exists only when the left and right estimates approach one number. At a corner, such as |x| at 0, they approach different numbers, and at a point like the origin on \displaystyle y = \sqrt[3]{x} the estimates grow without bound and approach none (see Differentiability: corners, cusps and piecewise functions).
An estimate is not the rate. Every estimate belongs to an interval of positive width. Putting h = 0 into \displaystyle \frac{f(a + h) - f(a)}{h} gives \displaystyle \frac{0}{0}, not the rate.
Measured data limit the width. A table cannot give a width below its spacing, and rounding makes tiny widths useless: the difference of two readings rounded to the nearest 0.1 meter can be off by up to 0.1 meter, which over 0.01 second is 10 meters per second of error.
The function must be defined around a. At the end of a domain, such as x = 0 for \displaystyle \sqrt{x}, only one side exists, and the rate as defined here does not.

Where this goes next

In Example 3 the curve bent up away from its tangent line because the train was speeding up; in Chidinma’s log the rate fell. A rate that itself changes is the subject of Concavity: when the rate of change itself changes. In calculus these estimates become exact: the rate at every input at once is the derivative, found from its definition.

Exercises

Shrinking intervals from a formula

For f(x) = x^2 + 3x, find the average rates of change on [1, 1.1], [0.9, 1], [1, 1.01] and [0.99, 1]. Estimate the rate of change at x = 1.
For g(x) = x^3, simplify \displaystyle \frac{g(1 + h) - g(1)}{h} and use it for h = 0.1, -0.1, 0.01 and -0.01. Estimate the rate at x = 1.
For \displaystyle k(x) = \frac{1}{x}, show that \displaystyle \frac{k(2 + h) - k(2)}{h} = \frac{-1}{2(2 + h)}, find it for h = \pm 0.1 and h = \pm 0.01 to four decimal places, and estimate the rate at x = 2.
Use centered averages with h = 0.1 and h = 0.01 to estimate the rate of change of \displaystyle \sqrt{x} at x = 9. What fraction does it appear to be?

Centered intervals

Find the centered averages of f(x) = x^3 - 2x at x = 2 with h = 0.5 and h = 0.1, and estimate the rate there.
Show that the centered average of q(x) = 3x^2 - 5x at x = 4 is the same for h = 1, 0.5 and 0.1. What is the rate at 4?
A left estimate and a right estimate of the same width at some point are 6.2 and 6.8. What is the centered average of that width?

From a table

On a different run Chidinma logs the train as it speeds up.

t (seconds)	x (meters)
4	40.0
4.5	49.5
5	60.0
5.5	71.5
6	84.0

Estimate the speed at t = 5 with a right estimate, a left estimate and two centered averages (h = 0.5 and h = 1).
Estimate the speed at t = 4.5 and at t = 5.5 with centered averages. By about how much does the speed rise over that one second?
Why can the table not give a centered average at t = 6? Give the best estimate it does allow.

Tangent lines

Using your rate from exercise 1, find the tangent line to f(x) = x^2 + 3x at x = 1, and use it to estimate f(1.2). Compare with the true value.
Using your rate from exercise 3, find the tangent line to \displaystyle y = \frac{1}{x} at x = 2. Where does it cross the x-axis?
The tangent line to the graph of f at x = 3 is y = 2x - 1. Find f(3) and the rate of change of f at 3.

Corners and other failures

For f(x) = |x - 1| + x, find the right and left estimates at x = 1 with h = \pm 0.1 and h = \pm 0.01. Does f have a rate of change at 1?
For m(x) = |x^2 - 9| at x = -3, find the right and left estimates with h = \pm 0.1 and the centered average with h = 0.1. What do they say?
For \displaystyle y = \sqrt[3]{x}, find the centered averages at x = 0 with h = 0.001 and h = 0.000001. Do they settle?

Batbayar’s run

On a second test Batbayar’s train is r(t) = 0.25t^3 + 3t meters from the start after t seconds.

Estimate the speed at t = 2 from centered averages with h = 0.1 and h = 0.01.
Find the tangent line at t = 2, use it to forecast r(2.2), and compare with the true value.

Reasoning

Mihaela’s right estimates of a rate, for h = 0.1, 0.01 and 0.001, are 4.31, 4.0301 and 4.003001; her left estimates are 3.71, 3.9701 and 3.997001. Explain what she may conclude about the rate, and how sure she can be.
Explain why a linear function has the same rate of change at every input, and why every estimate of it, of any width and on either side, gives that rate exactly.
Explain why the centered average at a corner can settle on a number that is not a rate, using Example 6.
Show that for f(x) = x^3 the centered average at a is 3a^2 + h^2. Use this to explain why, in Example 2, the centered estimates missed 8 by exactly 0.5h^2.

Answers

5.1, 4.9, 5.01 and 4.99; the rate is 5. The quotient at 1 is \displaystyle \frac{(1 + h)^2 + 3(1 + h) - 4}{h} = 5 + h.
3 + 3h + h^2, giving 3.31, 2.71, 3.0301 and 2.9701; the rate is 3. (1 + h)^3 - 1 is 3h + 3h^2 + h^3.
\displaystyle \frac{1}{2 + h} - \frac{1}{2} = \frac{-h}{2(2 + h)}; divide by h. The values are -0.2381, -0.2632, -0.2488 and -0.2513, so the rate is -0.25.
0.166669 and 0.1666667, rounded; the rate appears to be \displaystyle \frac{1}{6}.
10.25 and 10.01; the rate is 10. With h = 0.5: f(2.5) = 10.625, f(1.5) = 0.375, and \displaystyle \frac{10.25}{1} = 10.25. In general the centered average is 10 + h^2.
All three are 19, the rate: by the quadratic result, 2(3)(4) - 5 = 19 for every h.
6.5, the mean of 6.2 and 6.8.
Right 23, left 21, centered 22 for both widths. \displaystyle \frac{71.5 - 60.0}{0.5} = 23, \displaystyle \frac{60.0 - 49.5}{0.5} = 21, \displaystyle \frac{71.5 - 49.5}{1} = 22, \displaystyle \frac{84.0 - 40.0}{2} = 22. About 22 meters per second.
20 at t = 4.5 and 24 at t = 5.5, so about 4 meters per second faster. \displaystyle \frac{60.0 - 40.0}{1} = 20 and \displaystyle \frac{84.0 - 60.0}{1} = 24.
A centered interval needs a reading after t = 6. The best left estimate is \displaystyle \frac{84.0 - 71.5}{0.5} = 25 meters per second, likely a little low since the speed is rising.
y = 5x - 1, from y = 4 + 5(x - 1); it gives 5, below f(1.2) = 1.44 + 3.6 = 5.04 by 0.04.
\displaystyle y = 1 - \frac{1}{4}x, crossing at x = 4. The line is \displaystyle y = \frac{1}{2} - \frac{1}{4}(x - 2).
f(3) = 5 and the rate is 2. The tangent line passes through (3, f(3)), and 2(3) - 1 = 5; its slope is the rate.
Every right estimate is 2 and every left estimate is 0, so there is no rate at 1. f(1) = 1 and f(1 + h) = |h| + 1 + h, so the quotient is \displaystyle \frac{|h| + h}{h}.
Right 5.9, left -6.1, centered -0.1. For h > 0 the quotient is 6 - h, since m(-3 + h) = 6h - h^2; for h < 0 it is -6 + h. The sides head to 6 and -6: a corner, no rate. The centered average is -h, heading to 0, the mean of the sides.
100 and 10{,}000; they do not settle. \displaystyle \frac{\sqrt[3]{h} - \sqrt[3]{-h}}{2h} = \frac{2\sqrt[3]{h}}{2h}; with h = 0.001 that is \displaystyle \frac{0.2}{0.002} = 100. The graph has a vertical tangent at the origin.
6.0025 and 6.000025; the speed is 6 meters per second. The centered average is 0.25(12 + h^2) + 3 = 6 + 0.25h^2.
y = 6t - 4, forecasting 9.2 meters; the true value is r(2.2) = 2.662 + 6.6 = 9.262 meters. The line is y = 8 + 6(t - 2), since r(2) = 8.
The rate lies between 3.997001 and 4.003001, so it is 4.00 to two decimal places, and the digits suggest exactly 4. The right estimates fall and the left ones rise toward one number, so by bracketing it lies between any left and any right estimate.
Between any two points of a line the slope is its one slope m: \displaystyle \frac{(mx_2 + c) - (mx_1 + c)}{x_2 - x_1} = m. Every estimate is such a slope, so all equal m, at every input.
The centered average is the mean of a right estimate near 2 and a left estimate near -2, so it approaches 0. A rate needs each side alone to approach one number; here they approach 2 and -2. Averaging hides the disagreement.
(a + h)^3 - (a - h)^3 = 6a^2h + 2h^3, since the a^3 and 3ah^2 terms cancel; dividing by 2h gives 3a^2 + h^2. For s(t) the cube part gives 0.5(12 + h^2) and the linear part exactly 2, so the centered average is 8 + 0.5h^2.