Velocity and Tangent, calculus lesson

Your car’s speedometer shows one number at one moment, yet speed is distance divided by time, and at a single moment no time passes. The slope of a curve has the same puzzle. A slope needs two points, but the tangent line touches the curve at one. This first lesson sets up both problems side by side and shows that they are really one problem, which the rest of the course solves with limits.

Key terms

Slope

Definition The slope of the line through two points $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 \neq x_2$ is

m = \frac{y_2 - y_1}{x_2 - x_1}.

A nonvertical line can be written as $y = mx + b$ , and its slope $m$ is the same for every pair of distinct points on it.

In plain words Slope is rise over run. It tells you how much $y$ changes for each unit step in $x$ .

Example Through the points $(1, 2)$ and $(4, 11)$ the slope is

m = \frac{11 - 2}{4 - 1} = 3,

so $y$ goes up by 3 for every step of 1 in $x$ .

Average velocity

Definition If $s(t)$ is the position of an object at time $t$ , its average velocity on the time interval $[a, b]$ with $a < b$ is

v_{\text{avg}} = \frac{s(b) - s(a)}{b - a}.

In plain words Distance travelled divided by the time it took, where moving backwards counts as negative distance.

Example A ball dropped from a tower has fallen $s(t) = 4.9t^2$ meters after $t$ seconds. From $t = 1$ to $t = 2$ its average velocity is

\frac{s(2) - s(1)}{2 - 1} = \frac{4.9 \cdot 4 - 4.9 \cdot 1}{2 - 1} = 14.7 \text{ meters per second.}

Difference quotient

Definition For a function $f$ and numbers $a$ and $h \neq 0$ with $a$ and $a + h$ in the domain of $f$ , the difference quotient is

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

In plain words It is the slope of the line through the points $(a, f(a))$ and $(a + h, f(a + h))$ on the graph. Average velocity is the difference quotient of the position function.

Example For $f(x) = x^2$ and $a = 3$ ,

\frac{(3 + h)^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h, \qquad h \neq 0.

Secant line

Definition A secant line of the graph of $f$ is the line through two distinct points $(a, f(a))$ and $(b, f(b))$ of the graph, with $a \neq b$ . Its slope is the difference quotient

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

In plain words Pick two points on the curve and draw the straight line through them.

Example For $f(x) = x^2$ the secant line through $(1, 1)$ and $(3, 9)$ has slope

\frac{9 - 1}{3 - 1} = 4

and equation $y = 4x - 3$ .

Instantaneous velocity as a limiting idea

Definition The instantaneous velocity at time $a$ is the number that the average velocities over $[a, a + h]$ (or $[a + h, a]$ when $h < 0$ ) approach as $h$ gets closer and closer to $0$ , if they approach a single number. Written as a limit,

v(a) = \lim_{h \to 0} \frac{s(a + h) - s(a)}{h}.

The exact meaning of $\lim$ comes in the next lessons. For now it means the value the quotient settles on as $h$ shrinks.

In plain words Measure the average velocity over shorter and shorter time intervals starting at $a$ and see what number the results close in on.

Example For the falling ball, $s(t) = 4.9t^2$ and $a = 1$ .

$h$	Average velocity on $[1, 1 + h]$
$1$	$14.7$
$0.1$	$10.29$
$0.01$	$9.849$
$0.001$	$9.8049$

Algebra confirms the pattern. For $h \neq 0$

\frac{4.9(1 + h)^2 - 4.9}{h} = \frac{9.8h + 4.9h^2}{h} = 9.8 + 4.9h.

As $h$ shrinks this approaches $9.8$ , so the ball moves at $9.8$ meters per second at $t = 1$ .

Tangent line

Definition Let $f$ be defined on an open interval containing $a$ . If the slopes of the secant lines through $(a, f(a))$ and $(a + h, f(a + h))$ approach a single number $m$ as $h \to 0$ , that is

m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

then the tangent line to the graph at $(a, f(a))$ is the line through that point with slope $m$ ,

y = f(a) + m(x - a).

In plain words Slide the second point of a secant line toward the first. The line those secants settle into is the tangent line.

Example For $f(x) = x^2$ at $a = 3$ the secant slopes are $6 + h$ , which approach $m = 6$ . The tangent line is $y = 9 + 6(x - 3)$ , or $y = 6x - 9$ . You can try this yourself in the tangent line explorer.

Normal line

Definition The normal line to a curve at a point is the line through that point that is perpendicular to the tangent line there. If the tangent slope $m$ is not zero, the normal line at $(a, f(a))$ has slope $-1/m$ . If $m = 0$ , the normal line is the vertical line $x = a$ , and if the tangent line is vertical, the normal line is the horizontal line $y = f(a)$ .

In plain words The tangent line runs along the curve and the normal line sticks straight out of it at a right angle.

Example For $f(x) = x^2$ at $(3, 9)$ the tangent slope is $6$ , so the normal line has slope $-1/6$ and equation

y = 9 - \frac{1}{6}(x - 3).

Vertical tangents

Definition Let $f$ be continuous at $a$ . The graph of $f$ has a vertical tangent line $x = a$ at $(a, f(a))$ when the secant slopes grow without bound in the same direction from both sides, that is

\lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = \infty \quad \text{or} \quad \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = -\infty.

In plain words The curve stands straight up for an instant. The tangent exists as a line, but its slope is not a number.

Example For $f(x) = \sqrt[3]{x}$ at $a = 0$ the difference quotient is

\frac{\sqrt[3]{h} - 0}{h} = \frac{1}{\sqrt[3]{h^2}}, \qquad h \neq 0,

which is positive and grows without bound as $h \to 0$ from either side. So the graph has the vertical tangent $x = 0$ at the origin.

Check yourself

A car’s position is $s(t) = t^2 + 2t$ kilometers after $t$ hours. What is its average velocity between $t = 1$ and $t = 3$ ?

Show the answer

The average velocity is

\frac{s(3) - s(1)}{3 - 1} = \frac{15 - 3}{2} = 6 \text{ kilometers per hour.}

Find the slope of the tangent line to $f(x) = x^2$ at $x = -2$ by simplifying the difference quotient.

Show the answer

For $h \neq 0$ the difference quotient simplifies to

\frac{(-2 + h)^2 - 4}{h} = \frac{-4h + h^2}{h} = -4 + h,

which approaches $-4$ as $h \to 0$ . The tangent slope is $-4$ .

Velocity and Tangent

Theory, part 1

Practice, part 2