Tangent Lines and Rates of Change

Main idea

A curve doesn't have a single slope, but you can measure the slope between two points on it. Slide the second point toward the first and those slopes settle on one number: the slope right at that point. The same move turns average speed into the speed at one instant.

{\htmlData{sym=1}{m_{\tan}}} = \lim_{x\to a}{\htmlData{sym=0}{\frac{f(x) - f(a)}{x - a}}}

What the symbols mean

$\frac{f(x) - f(a)}{x - a}$
Secant slope
Rise over run between two points on the curve: the average rate of change.
$m_{\tan}$
Tangent slope
The instantaneous rate of change at x = a.

Explanation

A tangent line is simply the straight line that just grazes a curve at one point, heading the same direction the curve is heading right there. (The name comes from the Latin tangere, "to touch.") Its slope tells you how steep the curve is at that spot, which is the same as how fast something is changing at that instant. The formula $\dfrac{f(x) - f(a)}{x - a}$ looks busy, but it's just rise over run between two points: $f(x) - f(a)$ (how much the height changes) divided by $x - a$ (how far over you moved). The $\lim_{x\to a}$ in front just means "now slide the second point in close to the first." That's all there is to it, so bear with the notation.

You already know how to find the slope of a line: pick two points, rise over run. The only problem with a curve is that rise over run needs two points, and "right here" is only one.

Here's the trick. Use two points anyway, and then push them together. Take $f(x) = x^2$ and the point $(1, 1)$ . The line through $(1, 1)$ and a second point $(x, x^2)$ is called a secant line, and its slope is $\dfrac{x^2 - 1}{x - 1}$ . Now slide the second point in:

$x$	$2$	$1.5$	$1.1$	$1.01$	$1.001$
slope	$3$	$2.5$	$2.1$	$2.01$	$2.001$

Notice the slopes closing in on $2$ . We never plug in $x = 1$ itself (that would be $\frac00$ ), but we don't need to. The number the slopes are heading toward is the slope of the tangent line, the line that just grazes the curve at the point. That's what we'll call the slope of the curve there.

Rates of change. The same idea shows up the moment something moves. If $s(t)$ is position, the secant slope $\dfrac{s(t) - s(a)}{t - a}$ is distance over time, which is just the average velocity over $[a, t]$ . Make the time interval shorter and shorter, and the averages settle on the instantaneous velocity at $t = a$ . That's the number a speedometer shows.

It works for any quantity you'd ask "how fast is it changing right now?" about: temperature, cost, population. The one fuzzy part is "the value the slopes are heading toward." The next few lessons make that precise. It's called a limit, and it's the idea the rest of calculus is built on.

See it

Watch the second point slide in toward the first. The secant line pivots and settles onto the tangent line, and its slope settles on a single number: the slope of the curve at that point.

A drone's altitude is h(t) = 40t − 5t² meters, the same drone as in the practice problems. The line through t = 1 and a second time b has slope 35 − 5b, the average velocity between them. With b = 4 it's 15 m/s, far too low; at b = 2 it's 25, and at b = 1.01 it's 29.95. As the second point closes in, the line pivots onto the tangent and its slope settles on 30 m/s, the speedometer reading at t = 1.

Worked examples

Example 1

Find the tangent line to $f(x) = x^2$ at $x = 1$ .

1
The table above shows the secant slopes heading to $2$ , so the tangent slope is $m = 2$ .

Example 2

A ball's height is $s(t) = 20t - 4.9t^2$ meters. Find its average velocity on $[1, 2]$ .

1
Find the height at each end: $s(2) = 40 - 19.6 = 20.4$ and $s(1) = 20 - 4.9 = 15.1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the average rate of change of $f(x) = x^3$ on $[1, 3]$ .

Problem 2+5 XP

Find the slope of the secant line to $f(x) = x^2$ through $x = 3$ and $x = 3.1$ .

Problem 3+5 XP

Estimate the slope of the tangent line to $f(x) = x^2$ at $x = 3$ .

Problem 4+5 XP

Write the tangent line to $f(x) = x^2$ at $x = 3$ .

Problem 5+5 XP

A cart's position is $s(t) = t^2 + 2t$ . Find its average velocity on $[2, 4]$ .

Problem 6+5 XP

A drone's altitude is $h(t) = 40t - 5t^2$ meters after $t$ seconds. Compute its average velocity over $[1, 1.1]$ and over $[1, 1.01]$ , and use them to estimate its velocity at exactly $t = 1$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the line tangent to $f(x) = \frac1x$ at $x = 2$ by taking the limit of secant slopes.

Where it's used

A trip's average speed is a secant slope; the speedometer reading is the tangent slope. Economists call the tangent slope of a cost curve the marginal cost: the cost of one more unit.

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