Module 37 of 47Lesson 1 of 8 · AI 5.1 — Limits and the derivative as a rate of change

Course map · Topic 5: Calculus

5.1 Limits and the derivative as a rate of change

SL & HL11 worked questions

The derivative tells you how fast something is changing at one instant. We find it by taking the gradient of a chord and letting the chord become shorter and shorter — the gradient tends to a limit, and that limit is the gradient of the tangent.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ + 🖩 · Paper 1[4 marks]

The point

P(2,\,4)

lies on the curve

y=x^{2}

. The point

Q

also lies on the curve and has

x

-coordinate

2+h

.
Copy and complete the table for the gradient of the chord

PQ

, and hence write down the gradient of the tangent at

P

$h$	1	0.1	0.01	0.001
gradient of $PQ$	5

Show the full step-by-step solution

Step 1🧠 ThinkWhat is a chord gradient?

Q

is a second point on the curve, a little to the right of

P

. The gradient of the straight line

PQ

is “rise ÷ run”. As

h

gets smaller,

Q

slides towards

P

and the chord turns into the tangent.

Step 2✍️ By handOne gradient by hand (h = 0.1)

When

h=0.1

Q

has

x=2.1

and

y=2.1^{2}=4.41

m=\dfrac{4.41-4}{2.1-2}=\dfrac{0.41}{0.1}=4.1

In general:

m=\dfrac{(2+h)^{2}-4}{h}

Step 3🖩 On the GDCLet the spreadsheet do the whole table

ctrl+doc💻 Ctrl+I›=((2+h)^2-4)/h›enter💻 Enter

Add a Lists & Spreadsheet page. Name column A h and type the four values. In the grey formula cell of column B type the chord formula.

chord gradients 5, 4.1, 4.01, 4.001

Step 4✍️ By handSee the pattern

5,\ 4.1,\ 4.01,\ 4.001\ \to\ 4

\displaystyle\lim_{h\to0}m_{PQ}=4

so the tangent at

P

has gradient

4

Final answer

4.1,\ 4.01,\ 4.001

; the gradient of the tangent at

P

4

⚠️ Common mistake Trying

h=0

: you get

\dfrac{0}{0}

, which has no value. The limit is what the values approach.

Question 2●○○ Warm-up✍️ By hand · Paper 1[2 marks]

Water is pumped into a reservoir near Gaborone. The volume of water is

V

megalitres,

t

hours after pumping starts. When

t=10

\dfrac{dV}{dt}=3.5

.
Interpret this value in context, including its units.

Show the full step-by-step solution

Step 1🧠 ThinkRead the notation aloud

\dfrac{dV}{dt}

is read “the rate of change of

V

with respect to

t

”. It is how fast the volume is changing. Units: units of $V$ per unit of $t$ = megalitres per hour.

Step 2✍️ By handWrite the interpretation

t=10

hours, the volume of water is increasingR1

at a rate of

3.5

megalitres per hour.A1

💡 Remember A full interpretation answers four things: what (volume of water), which way (increasing, because 3.5 is positive), how fast (3.5 megalitres per hour), when (10 hours after starting).

Final answer 10 hours after pumping starts, the volume of water is increasing at 3.5 megalitres per hour.

⚠️ Common mistake Saying “the reservoir holds 3.5 megalitres”.

\dfrac{dV}{dt}

is a rate, not an amount.

Question 3●○○ Warm-up✍️ + 🖩 · Paper 1[3 marks]

Let

g(x)=\dfrac{x^{2}-4}{x-2}

x\ne2

.
(a) Find

g(1.99)

and

g(2.01)

.
(b) Hence write down

\displaystyle\lim_{x\to2}g(x)

Show the full step-by-step solution

Step 1🧠 ThinkWhy can't we just put x = 2?

x=2

the bottom is

0

, and we can never divide by 0. But we can look at what happens when

x

is very close to 2, from both sides.

Step 2🖩 On the GDCEvaluate close to 2 on a Calculator page

define g(x)=(x^2-4)/(x-2)›enter💻 Enter›g(1.99)›enter💻 Enter›g(2.01)›enter💻 Enter

Define the function once, then use it.

values either side of 2

Step 3✍️ By handConclude

g(1.99)=3.99,\quad g(2.01)=4.01

From both sides the values approach 4, so

\displaystyle\lim_{x\to2}g(x)=4

✅ Check it Factorise:

\dfrac{(x-2)(x+2)}{x-2}=x+2

for

x\ne2

, and

2+2=4

. ✔

Final answer (a)

3.99

and

4.01

(b)

4

Question 4●●○ Standard🖩 GDC · Paper 1[2 marks]

Let

f(x)=x^{3}-2x

. Use your graphic display calculator to find the gradient of the curve

y=f(x)

at the point where

x=1.5

Show the full step-by-step solution

Step 1🧠 ThinkGradient of a curve = gradient of its tangent

A curve has a different steepness at every point. “Gradient at

x=1.5

” means the gradient of the tangent there, which is

f'(1.5)

. The GDC can measure it for us.

Step 2🖩 On the GDCRoute 1: on the graph

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx›1.5›enter💻 Enter

Graph

f_1(x)=x^3-2x

. Choose dy/dx, click on the curve, type 1.5 and press enter. The calculator draws the point and writes the gradient next to it.

dy/dx at x = 1.5 is 4.75

Step 3🖩 On the GDCRoute 2: on a Calculator page

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Choose Numerical Derivative at a Point. In the box set Variable = x, Value = 1.5, 1st Derivative, then OK. Type the function into the template.

both forms give 4.75

Final answer

f'(1.5)=4.75

Question 5●●○ Standard✍️ + 🖩 · Paper 1[5 marks]

A stone is dropped from a bridge. After

t

seconds it has fallen

s=4.9t^{2}

metres.
(a) Find the average speed of the stone between

t=2

and

t=3

.
(b) Find the average speed between

t=2

and

t=2+h

for

h=0.1,\ 0.01

and

0.001

.
(c) Hence estimate the speed of the stone at the instant

t=2

, giving units.

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkAverage speed = distance ÷ time

Average speed is the gradient of the chord on the distance–time graph: change in

s

divided by change in

t

Step 2✍️ By handChord from t = 2 to t = 3

s(2)=4.9\times4=19.6,\quad s(3)=4.9\times9=44.1

\dfrac{44.1-19.6}{3-2}=24.5\ \text{m s}^{-1}

Part (b)

Step 1🖩 On the GDCShrink the time interval

Use the formula

\dfrac{4.9(2+h)^{2}-19.6}{h}

in a spreadsheet (or type each one).

20.09, 19.649, 19.6049

Step 2✍️ By handRecord the results

h=0.1:\ 20.09\qquad h=0.01:\ 19.649\qquad h=0.001:\ 19.6049

Part (c)

Step 1✍️ By handWhat are the values approaching?

h\to0

, the average speed

\to19.6

Speed at

t=2

19.6\ \text{m s}^{-1}

✅ Check it On the GDC,

\text{nDeriv}(4.9x^2,x=2)

gives

19.6

. ✔

Final answer (a)

24.5\ \text{m s}^{-1}

(b)

20.09,\ 19.649,\ 19.6049\ \text{m s}^{-1}

(c)

19.6\ \text{m s}^{-1}

Question 6●●○ Standard✍️ By hand · Paper 1[4 marks]

Mpho makes a cup of tea and measures its temperature

T\,^{\circ}\mathrm{C}

t

minutes later. The graph shows

T

against

t

, and the tangent to the curve at the point

P

where

t=5

. The tangent passes through

(0,\,85)

and

(20,\,10)

.
(a) Find the gradient of the tangent.
(b) Interpret your answer to (a) in context.

Show the full step-by-step solution

Step 1🧠 ThinkThe tangent's gradient is the rate at that moment

The dashed line touches the curve at

P

. Its gradient is the rate of change of temperature exactly at

t=5

. It is a straight line, so we use two points on the line (not on the curve).

Part (a)

Step 1✍️ By handRise over run

m=\dfrac{10-85}{20-0}

m=\dfrac{-75}{20}=-3.75

Part (b)

Step 1✍️ By handSay what it means

\dfrac{dT}{dt}=-3.75

: 5 minutes after it was made, the tea is coolingR1

3.75\,^{\circ}\mathrm{C}

per minute.A1

Final answer (a)

-3.75

(b) At

t=5

the temperature is decreasing at

3.75\,^{\circ}\mathrm{C}

per minute.

⚠️ Common mistake Using two points on the curve (that gives a chord, i.e. an average rate). Use points on the tangent line.

Question 7●●○ Standard🖩 GDC · Paper 2[3 marks]

The number of bacteria in a culture is modelled by

N(t)=500\times1.08^{t}

, where

t

is the time in hours.
(a) Find the number of bacteria after 5 hours.
(b) Find

N'(5)

and interpret it in context.

Show the full step-by-step solution

Step 1🧠 ThinkTwo different questions

(a) asks how many bacteria: substitute

t=5

. (b) asks how fast the number is growing at that moment: the derivative. We do not need a rule for differentiating this — the GDC finds

N'(5)

numerically.

Step 2🖩 On the GDCParts (a) and (b) on one Calculator page

500·1.08^5›enter💻 Enter›nDeriv(500·1.08^x,x=5)›enter💻 Enter

📍 Where to find it on the TI-Nspire

nDeriv(is inmenu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

N(5) ≈ 735, N′(5) ≈ 56.5

Step 3✍️ By handAnswer with units

(a)

N(5)=734.66\ldots\approx735

bacteriaA1

(b)

N'(5)=56.5

(3 s.f.)A1

After 5 hours the population is increasing at about

56.5

bacteria per hour.R1

Final answer (a)

735

bacteria (b)

N'(5)\approx56.5

: after 5 hours the number of bacteria is growing at about 56.5 bacteria per hour.

Question 8●●● Exam-level✍️ + 🖩 · Paper 2[8 marks]

Kagiso throws a ball straight up. Its height above the ground after

t

seconds is

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

metres.

(a) Find the average velocity of the ball between $t=1$ and $t=1.5$ .[2]
(b) Show that the gradient of the chord from $t=1$ to $t=1+k$ is $10-5k$ .[3]
(c) Hence write down the velocity of the ball at $t=1$ .[1]
(d) Use your GDC to find $h'(3)$ , and explain what the sign of your answer tells you.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handAverage velocity = chord gradient

h(1)=20-5=15,\qquad h(1.5)=30-11.25=18.75

\dfrac{18.75-15}{1.5-1}

=\dfrac{3.75}{0.5}=7.5\ \text{m s}^{-1}

Part (b)

Step 1🧠 ThinkUse a letter for the gap

Instead of 0.5, the gap is

k

. We find

h(1+k)

, subtract

h(1)

, and divide by

k

. This is a “show that”, so every line must be written.

Step 2✍️ By handExpand and simplify

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

\dfrac{h(1+k)-h(1)}{k}=\dfrac{10k-5k^{2}}{k}

=10-5k

Part (c)

Step 1✍️ By handLet k shrink to 0

k\to0

10-5k\to10

, so the velocity at

t=1

10\ \text{m s}^{-1}

Part (d)

Step 1🖩 On the GDCdy/dx on the graph at t = 1 and t = 3

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx›3›enter💻 Enter

Graph

f_1(x)=20x-5x^2

(the GDC uses

x

for

t

). Use dy/dx twice: type 1, then 3.

h′(1) = 10 and h′(3) = −10

Step 2✍️ By handInterpret the sign

h'(3)=-10

Negative: the height is decreasing — the ball is falling (at

10\ \text{m s}^{-1}

).R1

Final answer (a)

7.5\ \text{m s}^{-1}

(b) shown (c)

10\ \text{m s}^{-1}

(d)

h'(3)=-10

; the ball is coming down.

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[6 marks]

Lesedi's bakery in Francistown makes

x

loaves of bread a day. The daily cost, in pula, is

C(x)=0.02x^{2}+15x+400

(a) Find the average rate of change of $C$ between $x=50$ and $x=60$ .[2]
(b) Use your GDC to find $C'(50)$ .[2]
(c) Interpret your answer to (b), including units.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handAverage rate = chord gradient

C(50)=50+750+400=1200,\quad C(60)=72+900+400=1372

\dfrac{1372-1200}{60-50}

=17.2

pula per loafA1

Part (b)

Step 1🖩 On the GDCDerivative at a point

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

C′(50) = 17

Step 2✍️ By handRecord it

C'(50)=17

Part (c)

Step 1✍️ By handMeaning

When 50 loaves are made, the daily cost is increasing at P 17 per loafR1

— making one more loaf adds about P 17 to the cost.A1

💡 Remember Economists call

C'(x)

the marginal cost. The chord (17.2) and the tangent (17) are close, but they answer different questions.

Final answer (a) P 17.20 per loaf (b)

17

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

A water tank in Maun is being drained. The volume of water in the tank

t

minutes after draining begins is

V(t)=2000-40t+0.2t^{2}

litres,

0\le t\le100

(a) Write down the volume of water in the tank when draining begins.[1]
(b) Find the value of $\dfrac{dV}{dt}$ when $t=20$ .[2]
(c) Interpret your answer to (b) in context.[2]
(d) Find the time when the water is draining at 20 litres per minute.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handDraining begins at t = 0

V(0)=2000

litresA1

Part (b)

Step 1🖩 On the GDCNumerical derivative at t = 20

dV/dt = −32 at t = 20

Step 2✍️ By handRecord

\dfrac{dV}{dt}=-32

Part (c)

Step 1✍️ By handNegative means decreasing

20 minutes after draining starts, the volume is decreasingR1

at 32 litres per minute.A1

Part (d)

Step 1🧠 ThinkDraining at 20 L/min means dV/dt = −20

“Draining” already tells us the volume goes down, so the rate is

-20

. Graph the derivative itself: on the GDC,

f_2(x)=\text{nDeriv}(f_1(x),x)

draws the gradient function. Then intersect it with

y=-20

Step 2🖩 On the GDCGraph the gradient function and intersect

f2(x)=nDeriv(f1(x),x)›enter💻 Enter›f3(x)=-20›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

📍 Where to find it on the TI-Nspire

nDeriv(is inmenu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

dV/dt = −20 when t = 50

Step 3✍️ By handAnswer

t=50

minutesA1

Final answer (a) 2000 L (b)

-32

t=50

minutes.

⚠️ Common mistake Solving

\dfrac{dV}{dt}=+20

. The volume is going down, so the rate is negative.

Question 11●●○ Standard✍️ + 🖩 · Paper 1[3 marks]

The number of members of a running club in Kasane,

t

years after it starts, is modelled by

M(t)=1200-\dfrac{800}{t+1}

.
(a) Find

M(0)

M(10)

and

M(100)

.
(b) Hence write down

\displaystyle\lim_{t\to\infty}M(t)

and interpret it.

Show the full step-by-step solution

Step 1🧠 ThinkA limit as t gets very large

“

t\to\infty

” means: what happens when

t

gets bigger and bigger without stopping? As

t

grows,

\dfrac{800}{t+1}

becomes tiny.

Part (a)

Step 1🖩 On the GDCEvaluate on a Calculator page

press ctrl+enter for decimals

Step 2✍️ By handRecord

M(0)=400,\ M(10)\approx1127,\ M(100)\approx1192

Part (b)

Step 1✍️ By handFind the limit

\dfrac{800}{t+1}\to0

\displaystyle\lim_{t\to\infty}M(t)=1200

The club's membership gets closer and closer to 1200 but never reaches it.R1

Final answer (a)

400,\ 1127,\ 1192

(b)

1200

: in the long run membership approaches 1200 members.

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 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.

🎯 What you must know

Chord: a straight line joining two points on a curve. Its gradient $=\dfrac{\text{change in } y}{\text{change in } x}$ is the average rate of change between the two points.
Tangent: a line that just touches the curve at one point. Its gradient is the instantaneous rate of change at that point.
Move the second point closer and closer ( $h \to 0$ ): the chord gradient gets closer and closer to a fixed number — the limit. That limit is the gradient of the tangent.
Notation: $\displaystyle\lim_{h\to0}$ means “the value the expression approaches as $h$ gets closer to 0”. We never put $h=0$ itself (we would divide by 0).
The derivative is written $\dfrac{dy}{dx}$ or $f'(x)$ . It is the gradient function: put in an $x$ -value, get out the gradient of the curve there.
$\dfrac{dV}{dt}$ means “rate of change of $V$ with respect to $t$ ”. Units = units of $V$ per unit of $t$ , e.g. litres per minute. Positive → increasing; negative → decreasing.
On the GDC: Graphs → Analyze Graph → dy/dx gives the gradient at a point on a graph; on a Calculator page, Numerical Derivative at a Point (or nDeriv() gives $f'(a)$ directly.

Gradient of a chord (average rate of change)

m=\dfrac{y_2-y_1}{x_2-x_1}

📘 In the formula booklet

Derivative as a limit

f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Gradient at a point on a graph (Graphs page): choose dy/dx, then click the curve and type the x-value

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx›1.5›enter💻 Enter

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 5: dy/dx.
Press click on the curve, type the x-value and press enter. The gradient at that point is shown.
Type 1.5.
Press enter.

The menu path

dy/dx at x = 1.5 is 4.75

Worked example: Question 4 — Route 1: on the graph

Graph

f_1(x)=x^3-2x

. Choose dy/dx, click on the curve, type 1.5 and press enter. The calculator draws the point and writes the gradient next to it.

Derivative at a point on a Calculator page (fills in the d/dx template)

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 4: Calculus.
Choose 1: Numerical Derivative at a Point….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

The menu path

both forms give 4.75

Worked example: Question 4 — Route 2: on a Calculator page

Choose Numerical Derivative at a Point. In the box set Variable = x, Value = 1.5, 1st Derivative, then OK. Type the function into the template.

Or type the command directly

nDeriv(x^3-2x,x=1.5)›enter💻 Enter

📍 Where to find it on the TI-Nspire

nDeriv(is inmenu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Type nDeriv(x^3-2x,x=1.5).
Press enter.

both forms give 4.75

Worked example: Question 4 — Route 2: on a Calculator page

Choose Numerical Derivative at a Point. In the box set Variable = x, Value = 1.5, 1st Derivative, then OK. Type the function into the template.

Get a decimal instead of an exact value

ctrl+enter💻 Ctrl+Enter

Press ctrl then enter to get the answer as a decimal.

values either side of 2

Worked example: Question 3 — Evaluate close to 2 on a Calculator page

Define the function once, then use it.

⚠️ Where students lose marks

Giving the average rate (chord) when the question asks for the rate at an instant (tangent) — or the other way round.
Leaving out units, or writing the wrong ones. $\dfrac{dV}{dt}$ with $V$ in litres and $t$ in minutes has units litres per minute.
Ignoring the sign: a negative derivative means the quantity is decreasing. Say so in your interpretation.
Putting $h=0$ into the chord-gradient formula. You must look at what the values approach, not substitute 0.
Interpreting in vague words (“it changes by 3.5”). Always say what changes, how fast, in which direction, and at what moment.

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL AI 5.1 — Limits and the derivative as a rate of change

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

5.1 Limits and the derivative as a rate of change

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks