Instantaneous Rate of Change

A car’s speedometer doesn’t show your average speed for the trip. It shows how fast you’re going right now. That’s an instantaneous rate of change: the rate at one single moment. You can’t compute it with one point, because a rate needs two. Instead, you estimate it with average rates of change over smaller and smaller intervals.

Key ideas

Instantaneous rate and the tangent

The instantaneous rate of change of $f$ at $x = a$ is the slope of the tangent line at $\big(a, f(a)\big)$: the line that just touches the curve there and points in the same direction as the curve.

Average rate of change	Instantaneous rate of change
over an interval, from $a$ to $b$	at a single point, $x = a$
slope of a secant (two points)	slope of a tangent (one point)
example: $240$ km in $3$ h is $80$ km/h on average	example: the speedometer reads $95$ km/h

Estimating with secants

Pick a second point close to $a$, at $x = a + h$, and find the secant slope:

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

As $h$ gets smaller ($1$, $0.1$, $0.01$, …), the secant swings closer and closer to the tangent, and its slope settles toward the instantaneous rate. Use small negative values of $h$ too, to approach from the left. If the slopes from both sides close in on the same number, that number is your estimate.

As the second point slides toward $P(2, 40)$, the secant slopes $0$, $5$, $7.5$ approach the tangent’s slope, $10$ m/s.

The centred interval

A quicker and usually more accurate estimate uses points equally spaced on both sides of $a$:

$$\text{instantaneous rate at } a \approx \frac{f(a + h) - f(a - h)}{2h}$$

For quadratic functions this centred estimate is exactly right for any $h$. For other functions it’s very close when $h$ is small, like $0.01$ or $0.001$.

Reading the sign

Instantaneous rate	Tangent	The function is …
positive	slopes up	increasing at that point
zero	horizontal	momentarily flat, often at a maximum or minimum
negative	slopes down	decreasing at that point

The units are the same as for an average rate: $y$-units per $x$-unit.

From a table of data

With data instead of an equation, use the closest values on each side. The centred estimate from the values just before and just after usually beats a one-sided estimate.

Looking ahead. In calculus, the idea of letting $h$ shrink toward $0$ is made exact using limits, and the instantaneous rate of change becomes the derivative. Everything on this page is the groundwork for it.

Worked examples

Example 1: Secants over shrinking intervals

A ball’s height is $h(t) = -5t^2 + 30t$ metres after $t$ seconds. Estimate how fast it’s rising at $t = 2$ s by finding secant slopes on both sides of $t = 2$.

Solution. $h(2) = -20 + 60 = 40$. For each second time, compute $\dfrac{h(t) - h(2)}{t - 2}$:

Interval	$h(t)$	Secant slope (m/s)
$1 \le t \le 2$	$h(1) = 25$	$\frac{40 - 25}{1} = 15$
$1.9 \le t \le 2$	$h(1.9) = 38.95$	$\frac{40 - 38.95}{0.1} = 10.5$
$1.99 \le t \le 2$	$h(1.99) = 39.8995$	$\frac{40 - 39.8995}{0.01} = 10.05$
$2 \le t \le 2.01$	$h(2.01) = 40.0995$	$\frac{40.0995 - 40}{0.01} = 9.95$
$2 \le t \le 2.1$	$h(2.1) = 40.95$	$\frac{40.95 - 40}{0.1} = 9.5$
$2 \le t \le 3$	$h(3) = 45$	$\frac{45 - 40}{1} = 5$

From the left, the slopes go $15$, $10.5$, $10.05$; from the right, $5$, $9.5$, $9.95$. Both close in on $10$.

The ball is rising at about $10$ m/s at $t = 2$ s.

Example 2: From a table of data

A sprinter’s distance from the start line is recorded each second.

Time (s)	$0$	$1$	$2$	$3$	$4$	$5$	$6$
Distance (m)	$0$	$3$	$10$	$18$	$27$	$36.5$	$46$

Estimate her speed at $t = 4$ s.

Solution. The one-sided estimates are

$$\frac{27 - 18}{4 - 3} = 9 \text{ m/s (from the left)}, \qquad \frac{36.5 - 27}{5 - 4} = 9.5 \text{ m/s (from the right)}$$

The centred estimate uses $t = 3$ and $t = 5$:

$$\frac{36.5 - 18}{5 - 3} = \frac{18.5}{2} = 9.25 \text{ m/s}$$

Her speed at $4$ s is about $9.25$ m/s. (With equal steps, the centred estimate is always exactly the average of the two one-sided estimates. Since she’s still speeding up, one of those is too low and the other too high, so their average is a better estimate than either.)

Example 3: A centred estimate for an exponential model

The caffeine in your body $t$ hours after a large coffee is $A(t) = 200(0.5)^{t/5}$ mg. Estimate the instantaneous rate of change at $t = 5$ h using a centred interval with $h = 0.1$, then with $h = 0.01$.

Solution. With $h = 0.1$:

$$\frac{A(5.1) - A(4.9)}{0.2} \approx \frac{98.6233 - 101.3959}{0.2} \approx -13.863 \text{ mg/h}$$

With $h = 0.01$:

$$\frac{A(5.01) - A(4.99)}{0.02} \approx \frac{99.86147 - 100.13873}{0.02} \approx -13.863 \text{ mg/h}$$

(rounded to three decimal places). The two estimates agree, so at $t = 5$ h the caffeine level is dropping at about $13.9$ mg per hour. Compare this with Example 3 on the average rate of change page: the average rates on either side were $-20$ and $-10$ mg/h, and the instantaneous rate sits between them.

Example 4: Interpreting positive, zero, and negative rates

The depth of water in a harbour $t$ hours after high tide is $d(t) = 3\cos\left(\dfrac{\pi}{6}t\right) + 5$ metres (angles in radians). Estimate the instantaneous rate of change at $t = 3$, $6$, and $9$ using centred intervals with $h = 0.01$, and interpret each.

Solution.

$$\begin{aligned} t = 3\text{: } \quad \frac{d(3.01) - d(2.99)}{0.02} &\approx -1.571 \text{ m/h} \\ t = 6\text{: } \quad \frac{d(6.01) - d(5.99)}{0.02} &= 0 \text{ m/h} \\ t = 9\text{: } \quad \frac{d(9.01) - d(8.99)}{0.02} &\approx 1.571 \text{ m/h} \end{aligned}$$

(rounded to three decimal places). At $t = 3$ h, halfway between high and low tide, the water is falling fastest, at about $1.57$ m/h. At $t = 6$ h it’s low tide ($d = 2$ m): the depth is momentarily not changing, and the tangent is horizontal. At $t = 9$ h the tide is coming in, and the water is rising at about $1.57$ m/h.

Common mistakes

Using one big interval. The secant slope over $[2, 3]$ in Example 1 is $5$ m/s, half the true rate. Shrink the interval until the slopes settle down.

Only approaching from one side. One-sided slopes can all be too big or all too small. Check both sides, or use a centred interval.

Dividing by the wrong width in a centred estimate. The interval from $a - h$ to $a + h$ has width $2h$, not $h$. In Example 2 the times $3$ and $5$ are $2$ seconds apart, so divide by $2$.

Rounding too early. With $h = 0.01$, the change in $y$ is tiny, so rounding the function values to one or two decimal places can wreck the estimate. Keep several decimal places until the final division.

Mixing up “rate is zero” with “value is zero”. A rate of $0$ means the quantity isn’t changing at that moment, like the water at low tide in Example 4. The depth itself is $2$ m, not $0$.

Practice

1. (Warm-up) Average or instantaneous rate of change?

• (a) A car’s speedometer reads $80$ km/h.
• (b) A family drives $240$ km to Ottawa in $3$ hours.
• (c) A radar gun measures a hockey shot at $130$ km/h as it leaves the stick.
• (d) A town grew by about $1200$ people per year from 2010 to 2020.

Solution

(a) Instantaneous: the speed at that moment.

(b) Average: $\frac{240}{3} = 80$ km/h over the whole trip.

(c) Instantaneous: the speed at one moment.

(d) Average: over a ten-year interval.

2. (Warm-up) The height of a ball is changing at $+7$ m/s at $t = 1.5$ s, $0$ m/s at $t = 3$ s, and $-10$ m/s at $t = 4$ s. Describe what the ball is doing at each time.

Solution

At $1.5$ s it’s going up at $7$ m/s. At $3$ s it’s at its highest point, momentarily neither rising nor falling. At $4$ s it’s coming down at $10$ m/s.

3. (Core) For $f(x) = x^2$, find the secant slopes from $x = 3$ to $x = 3 + h$ for $h = 1$, $0.1$, $0.01$, and $-0.1$. Estimate the instantaneous rate of change at $x = 3$.

Solution

$f(3) = 9$.

• $h = 1$: $\frac{16 - 9}{1} = 7$
• $h = 0.1$: $\frac{9.61 - 9}{0.1} = 6.1$
• $h = 0.01$: $\frac{9.0601 - 9}{0.01} = 6.01$
• $h = -0.1$: $\frac{8.41 - 9}{-0.1} = 5.9$

The slopes close in on $6$ from both sides, so the instantaneous rate of change at $x = 3$ is about $6$.

4. (Core) A stone dropped from a bridge falls $d(t) = 4.9t^2$ metres in $t$ seconds. Use a centred interval with $h = 0.1$ to estimate its speed at $t = 2$ s.

Solution

$$\frac{d(2.1) - d(1.9)}{0.2} = \frac{4.9(4.41) - 4.9(3.61)}{0.2} = \frac{21.609 - 17.689}{0.2} = \frac{3.92}{0.2} = 19.6 \text{ m/s}$$

Because $d$ is quadratic, the centred estimate is exact: the stone is falling at $19.6$ m/s.

5. (Core) Water drains from a tank. The volume left is recorded every $2$ minutes.

Time (min)	$0$	$2$	$4$	$6$	$8$	$10$
Volume (L)	$600$	$470$	$360$	$270$	$200$	$150$

Estimate the instantaneous rate of change of the volume at $t = 4$ min and at $t = 8$ min. What do your answers tell you?

Solution

Use centred intervals:

$$t = 4\text{: } \frac{270 - 470}{6 - 2} = \frac{-200}{4} = -50 \text{ L/min}, \qquad t = 8\text{: } \frac{150 - 270}{10 - 6} = \frac{-120}{4} = -30 \text{ L/min}$$

The water is draining at about $50$ L/min at $4$ minutes but only about $30$ L/min at $8$ minutes. The tank drains more slowly as it empties.

6. (Core) $1000 is invested at $5\%$ per year, compounded annually, so the value after $t$ years is $V(t) = 1000(1.05)^t$. Estimate how fast the value is growing at $t = 10$ years, using a centred interval with $h = 0.01$. Round to the nearest cent.

Solution

$$\frac{V(10.01) - V(9.99)}{0.02} \approx \frac{1629.68956 - 1628.10008}{0.02} \approx 79.47$$

At $10$ years, the investment is growing at about $79.47 per year. (Keep plenty of decimal places in the two values before you subtract.)

7. (Core) A Ferris wheel rider’s height is $h(t) = 12 - 10\cos\left(\dfrac{\pi}{10}t\right)$ metres after $t$ seconds (angles in radians). Estimate the instantaneous rate of change of height at $t = 5$ s and at $t = 10$ s with centred intervals, $h = 0.01$. Interpret each.

Solution

$$t = 5\text{: } \frac{h(5.01) - h(4.99)}{0.02} \approx 3.142 \text{ m/s}, \qquad t = 10\text{: } \frac{h(10.01) - h(9.99)}{0.02} = 0 \text{ m/s}$$

At $5$ s the rider is level with the centre of the wheel ($h = 12$ m) and rising fastest, at about $3.14$ m/s. At $10$ s the rider is at the top ($h = 22$ m), where the height is momentarily not changing.

8. (Challenge) Let $f(x) = \dfrac{1}{x}$.

• (a) Find the secant slopes from $x = 2$ to $x = 2.1$ and from $x = 2$ to $x = 2.01$, and a centred estimate with $h = 0.01$. Round to five decimal places.
• (b) Use your estimate to write an equation of the tangent line at $\left(2, \frac{1}{2}\right)$.

Solution

(a)

$$\frac{\frac{1}{2.1} - 0.5}{0.1} \approx -0.23810, \qquad \frac{\frac{1}{2.01} - 0.5}{0.01} \approx -0.24876$$$$\frac{\frac{1}{2.01} - \frac{1}{1.99}}{0.02} \approx -0.25001$$

The instantaneous rate of change is about $-0.25$, or $-\frac{1}{4}$.

(b) The tangent has slope $-\frac{1}{4}$ and passes through $\left(2, \frac{1}{2}\right)$:

$$y - \frac{1}{2} = -\frac{1}{4}(x - 2) \quad\Rightarrow\quad y = -\frac{1}{4}x + 1$$

9. (Challenge) Again let $h(t) = -5t^2 + 30t$.

• (a) Show that the centred secant slope over $[3 - k, 3 + k]$ is $0$ for every $k \gt 0$. What does that tell you about $t = 3$?
• (b) Example 1 found that the rate at $t = 2$ s is $10$ m/s. Use the symmetry of the parabola to find when the rate is $-10$ m/s.

Solution

(a)

$$h(3 + k) = -5(9 + 6k + k^2) + 90 + 30k = 45 - 5k^2$$$$h(3 - k) = -5(9 - 6k + k^2) + 90 - 30k = 45 - 5k^2$$

The two heights are equal, so the secant slope $\dfrac{h(3 + k) - h(3 - k)}{2k} = 0$ for every $k$. The tangent at $t = 3$ is horizontal: the instantaneous rate is $0$, and the ball is at its maximum height, $45$ m.

(b) The parabola is symmetric about $t = 3$. The point at $t = 2$ is $1$ s before the peak, so its mirror image is at $t = 4$, $1$ s after. There the tangent has the opposite slope. The rate is $-10$ m/s at $t = 4$ s.

Check with a centred interval: $\dfrac{h(4.01) - h(3.99)}{0.02} = \dfrac{39.8995 - 40.0995}{0.02} = -10$ ✓