Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

Lessons

Senior 4 · Mathematics

Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

5 lessons

Lesson 3 of 5

Tangents, normals, rates of change and motion

25 min

What a speedometer measures

A car's odometer counts the distance travelled; its speedometer shows how fast that distance is changing, right now. If $s (t)$ is the distance at time $t$ , the speedometer shows $s^{'} (t)$ . In the same way, the derivative gives the slope of the tangent to a curve, the rate at which a balloon's volume grows as its radius increases, or the extra cost of producing one more item. This lesson turns the rules of Lessons 1 and 2 into equations of tangents and normals, and into rates of change and motion.

Definition

Tangent to $y = f (x)$ at $A (a, f (a))$ : the line through $A$ with slope $f^{'} (a)$ :

y - f (a) = f^{'} (a) (x - a) .

Normal at $A$ : the line through $A$ perpendicular to the tangent. If $f'(a) \ne 0$ its slope is $-\frac{1}{f'(a)}$ :

y - f(a) = -\frac{1}{f'(a)}(x - a) ;

if $f^{'} (a) = 0$ the tangent is horizontal and the normal is the vertical line $x = a$ .

Worked example 1: tangent and normal at a point

Find the tangent and the normal to $y = x^{3} - 2 x + 1$ at the point where $x = 1$ .

Solution $y (1) = 0$ , so the point is $(1, 0)$ . $y^{'} = 3 x^{2} - 2$ , so the slope is $y^{'} (1) = 1$ .

Tangent: $y - 0 = 1 (x - 1)$ , that is $y = x - 1$ . Normal: slope $- 1$ : $y = - (x - 1) = - x + 1$ .

At $(1, 0)$ the tangent $y = x - 1$ (red) touches the curve, and the normal $y = - x + 1$ (green) is perpendicular to it. The slopes $1$ and $- 1$ multiply to $- 1$ .

Worked example 2: a tangent with a given slope

Find the tangent to $y = x^{2} - 4 x + 5$ that is parallel to the line $y = 2 x + 1$ .

Solution Parallel means slope 2: $y^{'} = 2 x - 4 = 2$ , so $x = 3$ and $y = 9 - 12 + 5 = 2$ . Tangent: $y - 2 = 2 (x - 3)$ , that is $y = 2 x - 4$ .

Worked example 3: tangents from a point outside the curve

Find the tangents to $y = x^{2}$ that pass through $P (1, - 3)$ .

Solution $P$ is not on the curve ( $1^2 \ne -3$ ), so let the point of contact be $(a, a^{2})$ . The tangent there is $y - a^{2} = 2 a (x - a)$ . It passes through $(1, - 3)$ when

-3 - a^2 = 2a(1 - a) \iff a^2 - 2a - 3 = 0 \iff a = 3 \ \text{or}\ a = -1 .

Tangents: at $(3, 9)$ , $y = 6 x - 9$ ; at $(- 1, 1)$ , $y = - 2 x - 1$ . Check: both give $y = - 3$ at $x = 1$ ✓.

Definition

Rate of change of $y$ with respect to $x$ : the average rate over $[a, a + h]$ is $\frac{f(a + h) - f(a)}{h}$ ; the instantaneous rate at $a$ is $f^{'} (a)$ , in units of $y$ per unit of $x$ .

Example: a spherical balloon of radius $r$ cm has volume $V = \frac{4}{3}\pi r^3$ , so $\frac{dV}{dr} = 4\pi r^2$ . At $r = 10$ , $\frac{dV}{dr} = 400\pi \approx 1257\,\mathrm{cm^3}$ per cm: near that size, each extra centimetre of radius adds about 1 257 cm³.

Important

Motion in a straight line. If $s (t)$ is the position (displacement from a fixed point) at time $t$ , then

v(t) = \frac{ds}{dt} \ \text{(velocity)}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} \ \text{(acceleration)} .

$v \gt 0$ : moving in the positive direction; $v \lt 0$ : the negative direction; $v = 0$ : momentarily at rest.
Speed is $∣ v ∣$ . The object speeds up when $v$ and $a$ have the same sign, and slows down when they have opposite signs.

Worked example 4: a particle moving back and forth

A particle moves along a line with $s (t) = t^{3} - 6 t^{2} + 9 t$ metres, $t \ge 0$ in seconds. (a) Find $v$ and $a$ . (b) When is it at rest, and when does it move backwards? (c) Find the total distance travelled in the first 4 s.

Solution

$v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3) \,\mathrm{m\,s^{-1}}$ ; $a = 6t - 12 \,\mathrm{m\,s^{-2}}$ .
At rest at $t = 1$ and $t = 3$ . $v \lt 0$ between the roots: it moves backwards for $1 \lt t \lt 3$ .
$s (0) = 0$ , $s (1) = 4$ , $s (3) = 0$ , $s (4) = 64 - 96 + 36 = 4$ . It goes 4 m forward, 4 m back, then 4 m forward: total distance 12 m (although its final displacement is only 4 m).

Position–time graph of Worked example 4. The slope of the curve is the velocity: zero at the turning points $t = 1$ and $t = 3$ , negative between them.

Worked example 5: a ball thrown upwards

A ball is thrown up from a 2 m high balcony; its height after $t$ seconds is $h = 2 + 20 t - 5 t^{2}$ metres. Find its greatest height, and its velocity when it hits the ground.

Solution $v = 20 - 10 t$ , which is 0 at $t = 2$ : the greatest height is $h(2) = 2 + 40 - 20 = 22\,\mathrm{m}$ . It hits the ground when $5 t^{2} - 20 t - 2 = 0$ : $t = \frac{20 + \sqrt{440}}{10} \approx 4.10\,\mathrm{s}$ (the other root is negative). Then $v \approx 20 - 41.0 = -21.0\,\mathrm{m\,s^{-1}}$ : about $21\,\mathrm{m\,s^{-1}}$ , downwards.

Warning

Distance travelled is not the same as displacement. When the velocity changes sign, split the time interval at the moments of rest and add the distances of each piece, as in Worked example 4(c).

Tip

In motion problems, write down $s$ , $v$ and $a$ with their units first, then mark the times where $v = 0$ on a time line with the sign of $v$ . Almost every question (direction, rest, speeding up, distance) can be read off that line.

Patrick JMT: Position, velocity and acceleration using derivatives (about 9 min). Watch how velocity and acceleration are found by differentiating, and how the sign of the velocity shows the direction of motion.

✦

Practice

Give exact answers unless told otherwise.

Find the tangent and the normal to $y = 2 x^{2} - 3 x$ at $(2, 2)$ .
Find the tangent to $y = \frac{2x + 1}{x - 3}$ at $x = 4$ .
Find the points on $y = x^{3} - 3 x^{2}$ where the tangent is parallel to $y = 9 x$ .
Find the tangents to $y = x^{2} + 1$ that pass through the origin.
The side of a square grows. Find the rate of change of its area with respect to its side when the side is 5 cm.
The cost of making $x$ items is $C(x) = 20\,000 + 50x - 0.01x^2$ Frw (illustrative, for $x \le 2000$ ). Find the marginal cost $C^{'} (x)$ at $x = 100$ and say what it means.
A particle has $s (t) = t^{2} - 8 t + 7$ metres. When is it at rest? Find its velocity and speed at $t = 2$ .
For $s (t) = 2 t^{3} - 9 t^{2} + 12 t$ , $t \ge 0$ , find when the particle is at rest and the total distance travelled from $t = 0$ to $t = 3$ .

Show answersHide answers

$y^{'} = 4 x - 3 = 5$ at $x = 2$ . Tangent $y = 5 x - 8$ ; normal $y - 2 = -\frac{1}{5}(x - 2)$ , that is $y = -\frac{1}{5}x + \frac{12}{5}$ .
$y (4) = 9$ , $y' = -\frac{7}{(x - 3)^2} = -7$ : $y = - 7 x + 37$ .
$3x^2 - 6x = 9 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow x = 3$ or $- 1$ : points $(3, 0)$ and $(- 1, - 4)$ .
Contact $(a, a^{2} + 1)$ , tangent $y - a^{2} - 1 = 2 a (x - a)$ through $(0, 0)$ : $a^{2} = 1$ . Tangents $y = 2 x$ and $y = - 2 x$ .
$A = x^{2}$ , $\frac{dA}{dx} = 2x = 10\,\mathrm{cm^2}$ per cm.
$C^{'} (x) = 50 - 0.02 x$ , $C^{'} (100) = 48$ : the 101st item costs about 48 Frw extra.
$v = 2 t - 8 = 0$ at $t = 4$ . At $t = 2$ : $v = -4\,\mathrm{m\,s^{-1}}$ , speed $4\,\mathrm{m\,s^{-1}}$ .
$v = 6 t^{2} - 18 t + 12 = 6 (t - 1) (t - 2)$ : at rest at $t = 1, 2$ . $s (0) = 0$ , $s (1) = 5$ , $s (2) = 4$ , $s (3) = 9$ : distance $5 + 1 + 5 = 11\,\mathrm{m}$ .

Quick check

The tangent to a curve at a point has slope 4. What is the slope of the normal there?

Key points

Tangent at $a$ : $y - f (a) = f^{'} (a) (x - a)$ ; normal: slope $-\frac{1}{f'(a)}$ , or $x = a$ if $f^{'} (a) = 0$ .
For a tangent through an outside point, call the point of contact $(a, f (a))$ and solve for $a$ .
$f^{'} (a)$ is the instantaneous rate of change, in units of $y$ per unit of $x$ .
$v = s^{'}$ , $a = v^{'} = s^{''}$ ; split at $v = 0$ to find the distance travelled.

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples (the cost figures are illustrative), the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Photo: Speedometer in driving car, MikeRun, Wikimedia Commons, CC BY-SA 4.0.
Video: Position, Velocity, Acceleration using Derivatives, Patrick JMT, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1What a speedometer measures

What a speedometer measures