Differentiation 1 · Rates of change & the gradient of a curve

Module 33 of 46Lesson 1 of 12 · Differentiation 1 · Rates of change & the gradient of a curve

Differentiation, stationary points & optimisation · Essential understanding 1

Rates of change & the gradient of a curve

IB AA SL 5.1with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: A straight line has one gradient. A curve has a different gradient at every point. The gradient of a curve at a point is its instantaneous rate of change, and we find it by squeezing a chord down until it becomes a tangent.

By the end of this lesson you can:

find an average rate of change between two points
explain how a chord turns into a tangent as the two points come together
say in plain words what a limit is
understand the gradient of a curve at a point as the derivative f′(a) and write it as f′(a) or $\dfrac{dy}{dx}$
interpret a rate of change with units in a real situation

💡 AA Paper 1 has no calculator

In AA, Paper 1 is calculator-free, so the by-hand working is what earns the marks there. Use the TI-Nspire steps for Paper 2, and to check your hand answers.

🧑🏾 Imagine this…

Lesedi drives from Gaborone to Francistown, about 430 km, in 5 hours. Her average speed is

\dfrac{430}{5} = 86

km/h. But near Palapye the speedometer reads 110 km/h, and in the queue at a roadblock it reads 0.
So there are two different questions: how fast on average over the whole trip? and how fast right now, at this exact instant? Differentiation is the maths of the second question.

Average rate of change

the change in y divided by the change in x between two points:

\dfrac{\text{rise}}{\text{run}}

. It is the gradient of the straight line joining the two points.

Chord

a straight line joining two points on a curve.

Tangent

a straight line that just touches a curve at one point and has the same direction as the curve there.

Instantaneous rate of change

how fast y is changing with x at one exact value of x. It equals the gradient of the tangent.

Limit

the value a number gets closer and closer to. Here: what the chord's gradient settles down to when the two points are squeezed together.

Derivative

the gradient of a curve at a point. We write f′(a) (say “f prime of a”) or

\dfrac{dy}{dx}

(say “dy by dx”).

★ Average rate of change and the chord

Between $x = a$ and $x = b$ : average rate $= \dfrac{f(b) - f(a)}{b - a}$ (this is the gradient of the chord).
Starting at $x = a$ and moving a small step $h$ : chord gradient $= \dfrac{f(a + h) - f(a)}{h}$ .
Make $h$ smaller and smaller. If the chord gradients settle on one number, that number is the gradient of the curve at $x = a$ , written $f'(a)$ .
The units of a rate are (units of y) per (unit of x), for example metres per second.

⚠️ Mistakes that lose marks

You can never put $h = 0$ straight into the chord gradient: that gives $\dfrac{0}{0}$ . Simplify first, then let $h$ get close to 0.
Always give the units and the sign. A rate of −10 m/s is not “10 m/s”: the minus sign means the quantity is going down.
The gradient of a curve at a point is the gradient of the tangent, not of the chord. They are only close when the points are very close together.

Prof. Kitso says

Zoom in on any smooth curve far enough and it looks like a straight line. The gradient of that tiny straight piece is the gradient of the curve. The lab below lets you zoom in without any maths first.

🧪 Try it yourselfA ball thrown upwards

The height of a ball is

h = 20t - 5t^2

metres after t seconds. Drag the tangent along the curve and watch the gradient. It is +10 at t = 1 (going up at 10 m/s), 0 at the top, and −10 at t = 3 (coming down). Check the first and last against Example 3.

Worked examples

Example 1.1Average rate of change from a table

The table shows how far Lesedi has driven t hours after leaving Gaborone.

t (hours)	0	1	2	3	4
d (km)	0	45	110	175	230

(a) Find her average speed over the whole 4 hours.
(b) Find her average speed between t = 1 and t = 3.
(c) In which hour was her average speed the lowest?

🧠 Think it through

Average speed is the average rate of change of distance: $\dfrac{\text{change in } d}{\text{change in } t}$ .
Use the two end values of the interval only. What happens in between does not matter.
For (c), work out the average speed in each hour and compare.

✍️ By hand

(a)

\dfrac{230 - 0}{4 - 0} = 57.5

km/h

(b)

\dfrac{175 - 45}{3 - 1} = \dfrac{130}{2} = 65

km/h

🔢 On your TI-Nspire CX II

On a Calculator page type each division.

✅ Answer (a) 57.5 km/h. (b) 65 km/h. (c) The first hour (45 km/h).

Example 1.2Squeezing a chord: the gradient of y = x² at x = 3

Find the gradient of the curve

y = x^2

at the point where

x = 3

, by looking at chords from

x = 3

x = 3 + h

🧠 Think it through

The point is (3, 9). The other end of the chord is $(3 + h,\ (3 + h)^2)$ .
Try some small values of h and look for a pattern.
Then do it with algebra, simplify, and only then let h shrink.

✍️ By hand

Chord gradient

= \dfrac{(3 + h)^2 - 9}{h} = \dfrac{9 + 6h + h^2 - 9}{h} = \dfrac{6h + h^2}{h} = 6 + h

h

gets closer to 0,

6 + h

gets closer to 6.

🔢 On your TI-Nspire CX II

Type the chord gradient for h = 1, 0.1, 0.01 and 0.001. The answers close in on 6.

✅ Answer The gradient at x = 3 is 6.

Example 1.3Instantaneous rate of change in context

The height of a ball thrown upwards is

h(t) = 20t - 5t^2

metres, t seconds after it is thrown.

(a) Find the gradient of the chord from $t = 1$ to $t = 1 + h$ , and so find the rate of change of height at $t = 1$ .
(b) Do the same at $t = 3$ .
(c) Describe what the ball is doing at each time.

🧠 Think it through

Find h(1) = 15 and expand h(1 + h) carefully.
Divide by h, simplify, then let h → 0.
A positive rate means the height is increasing. A negative rate means it is decreasing.

✍️ By hand

(a)

h(1 + h) = 20(1 + h) - 5(1 + h)^2 = 15 + 10h - 5h^2

, and

h(1) = 15

Chord gradient

= \dfrac{10h - 5h^2}{h} = 10 - 5h \to 10

. Rate

= 10

m/s.

(b)

h(3) = 15

and

h(3 + h) = 15 - 10h - 5h^2

. Chord gradient

= -10 - 5h \to -10

. Rate

= -10

m/s.

🔢 On your TI-Nspire CX II

Check with the lab, or on a Graphs page draw 20x-5x^2 and add a tangent at x = 1 and at x = 3 (menu💻 click menu on keypad ▸ Geometry ▸ Tangent). The tangent gradients are 10 and −10.

✅ Answer (a) 10 m/s. (b) −10 m/s. (c) Rising at t = 1, falling at t = 3.

Both heights are 15 m, but the rates have opposite signs. Same height, opposite direction.

Example 1.4Estimating a gradient from a tangent

A tangent is drawn to the curve

y = \dfrac{1}{2}x^2

at the point where

x = 3

. It passes through the points

(2,\ 1.5)

and

(4,\ 7.5)

. Find the gradient of the curve at

x = 3

, and write your answer using

f'

notation.

🧠 Think it through

The gradient of the curve at x = 3 is the gradient of the tangent there.
Use the two points on the tangent line: $\dfrac{\text{rise}}{\text{run}}$ .

✍️ By hand

Gradient

= \dfrac{7.5 - 1.5}{4 - 2} = \dfrac{6}{2} = 3

f'(3) = 3

🔢 On your TI-Nspire CX II

Graph 0.5x^2 and the line through (2, 1.5) and (4, 7.5), which is 3x-4.5. It touches the curve at (3, 4.5).

✅ Answer f′(3) = 3.

Example 1.5Notation and meaning

Write each statement using derivative notation, or explain it in words.

(a) The gradient of $y = x^2$ at $x = 3$ is 6.
(b) A tank holds V litres of water at time t minutes. When t = 4, $\dfrac{dV}{dt} = -12$ .
(c) N(t) is the number of bacteria in a dish after t hours. N′(5) = 240.

🧠 Think it through

f′(a) and $\dfrac{dy}{dx}$ both mean “the rate of change at x = a”. The letters change with the situation: V and t, N and t.
In words, always say: what quantity, how fast, up or down, with which units, and at what time.

✍️ By hand

(a)

f'(3) = 6

, or

\dfrac{dy}{dx} = 6

when

x = 3

(b) At 4 minutes the volume is decreasing at 12 litres per minute (the tank is draining).

✅ Answer (a) f′(3) = 6. (b) The tank is losing 12 L/min at t = 4. (c) The bacteria are growing at 240 per hour at t = 5.

N′(5) = 240 does not mean there are 240 bacteria. It is a rate: 240 more every hour, at that moment.

Challenge questions

⭐ Challenge 1.6Average rate of changePaper 1 style

Find the average rate of change of

f(x) = x^2 - 2x

between

x = 1

and

x = 4

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Work out f(1) and f(4).
Divide the change in f by the change in x.

✍️ By hand

f(1) = 1 - 2 = -1

f(4) = 16 - 8 = 8

\dfrac{8 - (-1)}{4 - 1} = \dfrac{9}{3} = 3

🔢 On your TI-Nspire CX II

Type it directly.

✅ Answer 3

⭐ Challenge 1.7From chord to tangentPaper 2 style

A runner's distance is

s(t) = t^2 + t

metres after t seconds.

(a) Find $s(2)$ and show that $s(2 + h) = 6 + 5h + h^2$ .
(b) Hence write down the average speed between t = 2 and t = 2 + h, in its simplest form.
(c) Find the average speed for h = 0.1 and for h = 0.01.
(d) Write down the speed of the runner at t = 2 seconds, and say what the units are.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Expand (2 + h)² and add (2 + h).
Subtract s(2) and divide by h.
Put in the two small values of h, then decide what number the speeds are heading towards.

✍️ By hand

(a)

s(2) = 4 + 2 = 6

s(2 + h) = (2 + h)^2 + (2 + h) = 4 + 4h + h^2 + 2 + h = 6 + 5h + h^2

✓

(b)

\dfrac{s(2 + h) - s(2)}{h} = \dfrac{5h + h^2}{h} = 5 + h

(d) The values are heading to 5, so the speed at t = 2 is 5 metres per second.

🔢 On your TI-Nspire CX II

Check (c) on the Calculator page.

✅ Answer (a) shown. (b) 5 + h. (c) 5.1 and 5.01 m/s. (d) 5 m/s.

“Hence” means you must use your answer to (a) in (b). Do not start again from scratch.

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Differentiation 1 · Rates of change & the gradient of a curve

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Rates of change & the gradient of a curve

Worked examples

Challenge questions