Rates of Change

Objectives

1. Compute the average rate of change of a function over an interval (slope of a secant line), including dimension and units.
2. Compute the average rate of change over multiple small intervals to estimate the instantaneous rate of change at a given input.
3. Describe the derivative at a given input as the instantaneous rate of change and as the slope of a tangent line.
4. Given a function and a scenario, interpret the meaning of the derivative at a given input, including units.
5. Write and understand Lagrange (“prime”) and Leibniz (“fraction”) derivative notation.
6. Given a function and input value, give the equation of the tangent line at that input.

1. Average Rate of Change

Objective. Compute the average rate of change of a function over an interval (slope of a secant line), including dimension and units.

A single input has no average rate of change. Over $[2, 2]$ the change in inputs is $2 - 2 = 0$, and the quotient $\bigl(f(2) - f(2)\bigr)/0$ is undefined; the definition applies only to an interval $[a, b]$ with $a < b$.

The line $y = 0$ meets the graph of $f(x) = x^2$ only at $(0, 0)$. It is not a secant line, because the definition requires two points on the graph.

The average rate of change of $f(x)$ over $[a, b]$ is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$.

Units and dimension. The units of an average rate of change are output units per input unit, as in miles per hour. The dimension is $[\text{output}]/[\text{input}]$.

When the output is itself a rate, the units of its rate of change have a second “per input unit.” A velocity in meters per second, with time in seconds, has a rate of change in meters per second per second, written m/s$^2$. Its dimension follows from the dimension arithmetic of the Lesson 12 Notes, section 3:

Example 1

Compute the average rate of change of $f$ over $[0, 2]$ and over $[1, 2]$:

$$f(x) = x^3 - 5x.$$

ShowHide solution

By the definition of average rate of change, the average rate of change of $f(x)$ over $[a, b]$ is

$$\frac{f(b) - f(a)}{b - a}.$$

Over $[0, 2]$, evaluate $f$ at the endpoints $x = 0$ and $x = 2$:

$$f(0) = 0^3 - 5(0) = 0, \qquad f(2) = 2^3 - 5(2) = 8 - 10 = -2.$$

Substitute $a = 0$ and $b = 2$ into the formula for the average rate of change:

$$\frac{f(2) - f(0)}{2 - 0} = \frac{-2 - 0}{2 - 0} = \frac{-2}{2} = -1.$$

Over $[1, 2]$, evaluate $f$ at the endpoints $x = 1$ and $x = 2$. The output $f(2) = -2$ comes from the first interval:

$$f(1) = 1^3 - 5(1) = 1 - 5 = -4.$$

Substitute $a = 1$ and $b = 2$ into the formula for the average rate of change:

$$\frac{f(2) - f(1)}{2 - 1} = \frac{-2 - (-4)}{2 - 1} = \frac{2}{1} = 2.$$

The average rate of change is $-1$ over $[0, 2]$ and $2$ over $[1, 2]$. The signs differ because $f(2) = -2$ lies below $f(0) = 0$ and above $f(1) = -4$. By the definition of a secant line, the line through the graph points $(0, 0)$ and $(2, -2)$ is a secant line; it falls with slope $-1$, and the secant line through $(1, -4)$ and $(2, -2)$ rises with slope 2.

f = makeFun(x^3 - 5*x ~ x)
f_change = makeFun((f(b) - f(a))/(b - a) ~ a & b)
f_change(0, 2)
[1] -1
f_change(1, 2)
[1] 2

Parentheses. The line f(2) - f(0)/(2 - 0) returns $-2$, because R divides $f(0)$ by 2 before subtracting. Put the numerator and the denominator each in their own parentheses: (f(2) - f(0))/(2 - 0).

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2. Estimating the Instantaneous Rate of Change

Objective. Compute the average rate of change over multiple small intervals to estimate the instantaneous rate of change at a given input.

A car’s average speed over a trip is an average rate of change. The speedometer reading at one moment is an instantaneous rate of change.

The function $f(x) = |x|$ has no instantaneous rate of change when $x = 0$. For every $h > 0$, its average rate of change over $[0, h]$ is $(|h| - 0)/h = 1$, and over $[-h, 0]$ it is $(0 - |{-h}|)/h = -1$. The average rates from the two sides stay apart, and the definition requires them to approach one unique, fixed value.

Intervals. An interval containing $x = a$ may be symmetric, as in $[a - 0.01, a + 0.01]$. It may also be one-sided, as in $[a, a + 0.01]$ or $[a - 0.01, a]$. When the instantaneous rate of change exists, the average rates over symmetric and one-sided intervals approach the same value.

Stopping test. To estimate the instantaneous rate of change when $x = a$, follow these steps:

1. Compute average rates of change over intervals containing $x = a$, shrinking $h$, the distance from $a$ to the far end of each interval, by a factor of 10 at each step. Early steps may use symmetric intervals; every step tested for stopping uses both one-sided intervals, $[a, a + h]$ and $[a - h, a]$.
2. Stop at a step where, to three decimals, the two one-sided rates agree with each other and with the rate or rates on the step before. A rounded $-0.000$ counts as $0.000$.

f = makeFun(x^3 - 5*x ~ x)
f_change = makeFun((f(b) - f(a))/(b - a) ~ a & b)
f_change(-1, 1)
[1] -4
f_change(-0.1, 0.1)
[1] -4.99
f_change(-0.01, 0.01)
[1] -4.9999
f_change(0, 0.001)
[1] -4.999999
f_change(-0.001, 0)
[1] -4.999999

Data tables. At the first or last input of a table, use the one-sided interval between that input and its neighboring input. At an interior input, use the interval between the neighboring inputs on each side. For 2005 in the table of Example 2, that interval is $[2004, 2006]$, and the estimate is

thousand dollars per year.

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3. Derivative

Objective. Describe the derivative at a given input as the instantaneous rate of change and as the slope of a tangent line.

The figure shows secant lines of $f(x) = x^3 - 5x$ over $[0, 2]$, $[0, 1]$, and $[0, 0.5]$. The slope over $[0, 2]$ is $-1$, from Example 1. The other two slopes are

As the interval shrinks toward $x = 0$, the slopes approach $-5$, the instantaneous rate of change estimated in Example 4, and the secant lines approach the dashed line through $(0, 0)$ with slope $-5$. That line is the tangent line, defined next.

The vertical line $x = 0$ meets the graph of $f(x) = x^3 - 5x$ only at $(0, 0)$, and it is not the tangent line there, because a vertical line has no slope. The definition requires a slope equal to the instantaneous rate of change, here about $-5$.

The function $f(x) = |x|$ has no derivative when $x = 0$, because its instantaneous rate of change there is undefined: its average rates are 1 from the right and $-1$ from the left, as shown after the definition of instantaneous rate of change.

Height and slope. The output $f(1) = -4$ is the height of the graph of $f(x) = x^3 - 5x$ at $x = 1$. The derivative $f'(1)$ is the slope of the tangent line there.

Example 6

The graph shows four lines near the point $(1, -4)$ on the graph of

$$f(x) = x^3 - 5x.$$

The derivative at that point is $f'(1) = -2$. Which line is the tangent line to $f(x)$ when $x = 1$?

ShowHide solution

By the definition of a tangent line, the tangent line to $f(x)$ when $x = 1$ passes through $(1, -4)$ with slope $f'(1) = -2$. Line C does both, and it is the tangent line. Line A is not the tangent line. It passes through the graph points $(1, -4)$ and $(2, -2)$, and its slope is $\bigl(-2 - (-4)\bigr)/(2 - 1) = 2$ instead of $f'(1) = -2$; it is the secant line over $[1, 2]$. Line B is not the tangent line, because it is horizontal, with slope 0 instead of $-2$. Line D is not the tangent line: it passes through $(0, 0)$ and $(1, -2)$, with slope $-2$, and misses the point $(1, -4)$.

Example 7

The graph shows $g(x) = \sqrt{x}$ and its tangent line at $(4, 2)$. The tangent line also passes through $(0, 1)$. Find $g'(4)$, and describe it as a slope and as a rate of change.

ShowHide solution

By the definition of a tangent line, the slope of the tangent line at $(4, 2)$ is the instantaneous rate of change of $g(x)$ when $x = 4$, which the definition of the derivative names $g'(4)$. Compute the slope from the two marked points on the tangent line:

$$g'(4) = \frac{2 - 1}{4 - 0} = \frac{1}{4} = 0.25.$$

The derivative $g'(4) = 0.25$ is the slope of the tangent line at $(4, 2)$. It is also the instantaneous rate of change: when $x = 4$, $g(x)$ is increasing at 0.25 output units per input unit. As a check, the average rate of change over $[4, 4.01]$ is $\bigl(\sqrt{4.01} - \sqrt{4}\bigr)/(4.01 - 4) \approx (2.002498 - 2)/0.01 \approx 0.2498$, close to 0.25.

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4. Interpreting the Derivative

Objective. Given a function and a scenario, interpret the meaning of the derivative at a given input, including units.

The height $H$ in meters of a ball thrown straight up, $t$ seconds after release, is

The graphs show $H$ with its tangent line at $t = 0.5$, 1, 2, and 3. The derivatives are estimated from one-sided average rates on each side of each input. Used R.

H = makeFun(19.6*t - 4.9*t^2 ~ t)
H_change = makeFun((H(b) - H(a))/(b - a) ~ a & b)
H_change(0.5, 0.5001)
[1] 14.69951
H_change(0.4999, 0.5)
[1] 14.70049
H_change(0.5, 0.50001)
[1] 14.69995
H_change(0.49999, 0.5)
[1] 14.70005
H_change(1, 1.0001)
[1] 9.79951
H_change(0.9999, 1)
[1] 9.80049
H_change(1, 1.00001)
[1] 9.799951
H_change(0.99999, 1)
[1] 9.800049
H_change(2, 2.0001)
[1] -0.00049
H_change(1.9999, 2)
[1] 0.00049
H_change(2, 2.00001)
[1] -4.899974e-05
H_change(1.99999, 2)
[1] 4.900009e-05
H_change(3, 3.0001)
[1] -9.80049
H_change(2.9999, 3)
[1] -9.79951
H_change(3, 3.00001)
[1] -9.800049
H_change(2.99999, 3)
[1] -9.799951

At each input, the two one-sided rates over intervals of width 0.0001 and of width 0.00001 all round to the same value to three decimals, with $-0.000$ counted as $0.000$. The stopping test is met: $H'(0.5) \approx 14.700$, $H'(1) \approx 9.800$, $H'(2) \approx 0.000$, and $H'(3) \approx -9.800$. The graphs round these to one decimal.

Positive derivative

The derivative $H'(1) \approx 9.8$ is positive, and the tangent line rises. One second after release, the ball’s height is increasing at about 9.8 meters per second.

Negative derivative

The derivative $H'(3) \approx -9.8$ is negative, and the tangent line falls. Three seconds after release, the ball’s height is decreasing at about 9.8 meters per second.

Zero derivative

The derivative $H'(2)$ is approximately 0, and the tangent line is horizontal. Two seconds after release, the ball’s height is neither increasing nor decreasing: the ball is at the top of its flight.

Size of the derivative

The derivative $H'(0.5) \approx 14.7$ is larger than $H'(1) \approx 9.8$, and the orange tangent line at $t = 0.5$ is steeper than the red one at $t = 1$. Half a second after release, the ball’s height is increasing at about 14.7 meters per second, a rate of greater size than the 9.8 meters per second at one second.

Interpretation. An interpretation of $f'(a)$ is one sentence that states

1. the input value, with its units,
2. whether the output is increasing or decreasing, and
3. the size of the rate of change, $|f'(a)|$, in output units per input unit.

For a zero derivative at a peak or a valley of the graph, the output is neither increasing nor decreasing. A zero derivative can also occur where the output keeps increasing: $f(x) = x^3$ has $f'(0) = 0$ and is increasing through $x = 0$.

Weight = makeFun(781.9*exp(0.1216*(Week - 25)) ~ Week)
Weight_change = makeFun((Weight(b) - Weight(a))/(b - a) ~ a & b)
Weight_change(26, 26.0001)
[1] 107.3736
Weight_change(25.9999, 26)
[1] 107.3723
Weight_change(26, 26.00001)
[1] 107.373
Weight_change(25.99999, 26)
[1] 107.3729
Weight_change(26, 26.000001)
[1] 107.373
Weight_change(25.999999, 26)
[1] 107.373

Temp = makeFun(60 - 0.5*(Hour - 14)^2 ~ Hour)
Temp_change = makeFun((Temp(b) - Temp(a))/(b - a) ~ a & b)
Temp_change(10, 10.0001)
[1] 3.99995
Temp_change(9.9999, 10)
[1] 4.00005
Temp_change(10, 10.00001)
[1] 3.999995
Temp_change(9.99999, 10)
[1] 4.000005
Temp_change(12, 12.0001)
[1] 1.99995
Temp_change(11.9999, 12)
[1] 2.00005
Temp_change(12, 12.00001)
[1] 1.999995
Temp_change(11.99999, 12)
[1] 2.000005
Temp_change(14, 14.0001)
[1] -4.999997e-05
Temp_change(13.9999, 14)
[1] 4.999997e-05
Temp_change(14, 14.00001)
[1] -5.000089e-06
Temp_change(13.99999, 14)
[1] 5.000089e-06

Lesson map

5. Derivative Notation

Objective. Write and understand Lagrange (“prime”) and Leibniz (“fraction”) derivative notation.

The Leibniz fraction is written like an average rate of change, the change in outputs divided by the change in inputs. The derivative is the value that these average rates approach as the intervals shrink. The bar with $x = a$ at its lower right marks the input where the derivative is taken. The fraction gives the dimension of the derivative as $[df/dx] = [f]/[x]$.

Inverse notation. The symbols $f^{-1}(a)$ and $f'(a)$ name different objects. The superscript $-1$ in $f^{-1}(a)$ marks the inverse function evaluated at $a$, and the prime in $f'(a)$ marks the derivative of $f(x)$ when $x = a$.

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6. Equation of the Tangent Line

Objective. Given a function and input value, give the equation of the tangent line at that input.

1. Calculate $f(a)$.
2. Calculate or approximate $f'(a)$.
3. Substitute these values into the equation of the tangent line.

A line through the point $(a, f(a))$ with slope $m$ has the equation $y - f(a) = m(x - a)$, its point-slope form. The equation of the tangent line is that form with $m = f'(a)$, solved for $y$. Distributing $m$ and collecting the constants gives the slope-intercept form: the slope times $x$ plus the vertical intercept, as in $y = 2x + 3$.

A horizontal line $y = f(a)$ passes through $(a, f(a))$, and it is not the tangent line when $f'(a) \neq 0$, because its slope 0 differs from $f'(a)$.

Example 12

Find the equation of the tangent line to $f$ when $x = 2$, with $f'(2)$ estimated in R:

$$f(x) = x^3 - 5x.$$

ShowHide solution

The equation of the tangent line to $f(x)$ at $(a, f(a))$ is

$$y = f(a) + f'(a)(x - a),$$

here with $a = 2$. To write the equation, find the output $f(2)$, the height of the point of tangency, and the slope $f'(2)$. Calculate the output by hand:

$$f(2) = 2^3 - 5(2) = 8 - 10 = -2.$$

Estimate $f'(2)$ as in Example 4, with one-sided intervals on each side of $x = 2$. Used R.

f = makeFun(x^3 - 5*x ~ x)
f_change = makeFun((f(b) - f(a))/(b - a) ~ a & b)
f_change(2, 2.001)
[1] 7.006001
f_change(1.999, 2)
[1] 6.994001
f_change(2, 2.0001)
[1] 7.0006
f_change(1.9999, 2)
[1] 6.9994
f_change(2, 2.00001)
[1] 7.00006
f_change(1.99999, 2)
[1] 6.99994
f_change(2, 2.000001)
[1] 7.000006
f_change(1.999999, 2)
[1] 6.999994

The rates round to 7.006 and 6.994 on the first step, to 7.001 and 6.999 on the second, and to 7.000 on both sides on the last two. The stopping test is met:

$$f'(2) \approx 7.000.$$

Substitute $f(2) = -2$ and $f'(2) \approx 7$ into the equation of the tangent line, then distribute and collect the constants:

$$y = f(2) + f'(2)(x - 2) \approx -2 + 7(x - 2) = -2 + 7x - 14 = 7x - 16.$$

As a check, the line passes through the point of tangency: $7(2) - 16 = -2 = f(2)$. On the graph, the tangent line touches $f$ at $(a, f(a)) = (2, -2)$ and climbs with slope $f'(2) \approx 7$.

Example 13

The function Weight models the average weight in grams of a male fetus, with input Week the gestational age in weeks:

$$\text{Weight} = 781.9e^{0.1216(\text{Week} - 25)}.$$

Find the equation of the tangent line at week 26 in slope-intercept form, and plot it with the model in R.

ShowHide solution

The slope is $m = \text{Weight}'(26) \approx 107.373$, from Example 8. Distribute $m$ in the equation of the tangent line to reach slope-intercept form:

$$y = \text{Weight}(26) + m(\text{Week} - 26) = m\,\text{Week} + \bigl(\text{Weight}(26) - 26m\bigr).$$

Compute the vertical intercept $\text{Weight}(26) - 26m$ in R twice, with $m$ first from the right-side rate and then from the left-side rate, to confirm that the two sides give the same intercept. Keep $m$ at full precision, because multiplying $m$ by 26 multiplies its rounding error by 26. Used R.

Weight = makeFun(781.9*exp(0.1216*(Week - 25)) ~ Week)
Weight_change = makeFun((Weight(b) - Weight(a))/(b - a) ~ a & b)
m = Weight_change(26, 26.000001)
Weight(26) - 26*m
[1] -1908.696
m = Weight_change(25.999999, 26)
Weight(26) - 26*m
[1] -1908.696

Both sides give $-1908.696$. By hand, with $\text{Weight}(26)$ and $m$ rounded to three decimals, the computation is

$$\begin{align*} \text{Weight}(26) &= 781.9e^{0.1216(26 - 25)} = 781.9e^{0.1216} \approx 883.001,\\ \text{Weight}(26) - 26m &\approx 883.001 - 26(107.373) = 883.001 - 2791.698 = -1908.697. \end{align*}$$

The hand result differs from R’s in the last decimal, the rounding error that R avoids by keeping $\text{Weight}(26)$ and $m$ at full precision. The tangent line is

$$y \approx 107.373\,\text{Week} - 1908.696.$$

Plot the model and the tangent line together. Used R.

tangent = makeFun(Weight(26) + m*(Week - 26) ~ Week)
plotFun(Weight(Week) ~ Week, xlim = range(20, 35))
plotFun(tangent(Week) ~ Week, add = TRUE, col = "red")

R’s plot shows the model curve and, with col = "red", the tangent line in red, as in the figure. The red line is $y \approx 107.373\,\text{Week} - 1908.696$, and it touches the model at $(26, \text{Weight}(26)) \approx (26, 883.001)$. At week 27 the tangent line’s value is $\text{Weight}(26) + m(27 - 26) \approx 883.001 + 107.373 = 990.374$ grams, below the model’s $\text{Weight}(27) = 781.9e^{0.1216(27 - 25)} \approx 997.176$ grams, because the model curves upward.

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