Mini-Lesson

Calculus (AI SL)

3 missions. Guess, learn, then check. Your place is saved.

Mission 1A car's distance graph curves. How fast is it at exactly 3 s?
Mission 2A tap's flow rate keeps changing. How much filled the bath?
Mission 3What shape of can uses the least metal for 330 ml?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A car's distance graph curves. How fast is it at exactly 3 s?

Make a guess: A car's distance graph curves. How fast is it at exactly 3 s?

Calculus · differentiation

The gradient of a curve as a limit

A curve has no single gradient, but a chord between two points does.
Fix the point at $x = 3$ x = 3 on $y = x^{2}$ y = x² and move a second point closer. The chord gradients approach a limit:

Gap $h$ h	1	0.5	0.1	0.01
Chord gradient	7	6.5	6.1	6.01

That limit, 6, is the gradient of the tangent at $x = 3$ x = 3.
The gradient function is the derivative, written $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x). It is also a rate of change: $\frac{d s}{d t}$ ds/dt is how fast $s$ s changes with $t$ t.

Calculate

?Find the gradient of the chord joining the points on

y = x^{3}

where

x = 2

and

x = 2.1

. Give your answer to 2 decimal places.

Calculus · differentiation

Differentiating $a x^{n}$ axⁿ

For integer powers: if $f (x) = a x^{n}$ f(x) = axⁿ, then $f^{'} (x) = an x^{n - 1}$ f'(x) = anxⁿ⁻¹. Multiply by the power, then reduce the power by 1.
Differentiate a sum term by term. A constant term differentiates to 0, and $a x$ ax differentiates to $a$ a.
Rewrite fractions as negative powers first: $\frac{4}{x} = 4 x^{- 1}$ 4/x = 4x⁻¹ and $\frac{3}{x ^{2}} = 3 x^{- 2}$ 3/x² = 3x⁻².
The derivative at $x = a$ x = a, written $f^{'} (a)$ f'(a), is the gradient of the curve there.

Worked example Calculus · differentiation

Finding a gradient

Question

f (x) = 2 x^{3} - 5 x + \frac{4}{x}

. Find

f^{'} (x)

and the gradient of the curve

y = f (x)

at

x = 2

.

1
Rewrite with powers: $f (x) = 2 x^{3} - 5 x + 4 x^{- 1}$ f(x) = 2x³ − 5x + 4x⁻¹
2
Differentiate each term: $f^{'} (x) = 6 x^{2} - 5 - 4 x^{- 2}$ f'(x) = 6x² − 5 − 4x⁻²
3
Substitute $x = 2$ x = 2: $f^{'} (2) = 6 (4) - 5 - \frac{4}{4} = 24 - 5 - 1$ f'(2) = 6(4) − 5 − 4/4 = 24 − 5 − 1

Answer

f^{'} (x) = 6 x^{2} - 5 - \frac{4}{x ^{2}}

, and the gradient at

x = 2

is 18.

⚠️ Watch out

\frac{4}{x}

is not

4 x

. Rewrite it as

4 x^{- 1}

before you differentiate.

Quick check

?Find

\frac{d y}{d x}

when

y = \frac{3}{x ^{2}}

.

Calculus · differentiation

Increasing, decreasing, tangents and normals

If $f^{'} (x) > 0$ f'(x) > 0 the function is increasing; if $f^{'} (x) < 0$ f'(x) < 0 it is decreasing; if $f^{'} (x) = 0$ f'(x) = 0 the curve is momentarily flat.
The tangent at $x = a$ x = a has gradient $m = f^{'} (a)$ m = f'(a) and passes through $(a, f (a))$ (a, f(a)): $y - f (a) = m (x - a)$ y − f(a) = m(x − a).
The normal is perpendicular to the tangent, so its gradient is $- \frac{1}{m}$ −1/m.
On a GDC paper you can find $f^{'} (a)$ f'(a) numerically with your GDC, then write the line by hand.

Worked example Calculus · differentiation

Tangent and normal

Question

Find the equations of the tangent and the normal to

y = x^{3} - 4 x + 1

at the point where

x = 2

.

1
Find the point: $y = 8 - 8 + 1 = 1$ y = 8 − 8 + 1 = 1, so the point is $(2, 1)$ (2, 1)
2
Differentiate: $\frac{d y}{d x} = 3 x^{2} - 4$ dy/dx = 3x² − 4, so at $x = 2$ x = 2 the gradient is $12 - 4 = 8$ 12 − 4 = 8
3
Tangent: $y - 1 = 8 (x - 2)$ y − 1 = 8(x − 2), so $y = 8 x - 15$ y = 8x − 15
4
Normal: gradient $- \frac{1}{8}$ −1/8, so $y - 1 = - \frac{1}{8} (x - 2)$ y − 1 = −1/8(x − 2), which gives $8 y - 8 = - x + 2$ 8y − 8 = −x + 2

Answer

tangent

y = 8 x - 15

; normal

x + 8 y - 10 = 0

⚠️ Watch out

the normal's gradient is

- \frac{1}{m}

, not

- m

and not

\frac{1}{m}

.

Calculate

?The curve

y = x^{2} - 2 x + 3

has a normal at the point where

x = 3

. Find the

x

-coordinate of the point where this normal crosses the

x

-axis.

Sort it

?

f (x) = x^{3} - 12 x

. Decide what

f

is doing at each value of

x

.

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?A car's distance is

s = 2 t^{2}

metres after

t

seconds. How fast is it going at exactly 3 s?

Mission 2

A tap's flow rate keeps changing. How much filled the bath?

Make a guess: A tap's flow rate keeps changing. How much filled the bath?

Calculus · integration

Anti-differentiation

Integration reverses differentiation. If $\frac{d y}{d x} = f (x)$ dy/dx = f(x), then $y = \int f (x) d x$ y = ∫ f(x) dx.
For integer $n \neq = - 1$ n ≠ −1:

\int a x^{n} d x = \frac{a x ^{n + 1}}{n + 1} + C

Raise the power by 1, then divide by the new power. Integrate a sum term by term.
Add the constant of integration $C$ C: every constant differentiates to 0, so the reverse step cannot know it.
A boundary condition (a known point on the curve) fixes the value of $C$ C.

Worked example Calculus · integration

Using a boundary condition

Question

\frac{d y}{d x} = 6 x^{2} - 4 x + 1

, and

y = 5

when

x = 2

. Find

y

in terms of

x

.

1
Integrate each term: $y = \frac{6 x ^{3}}{3} - \frac{4 x ^{2}}{2} + x + C = 2 x^{3} - 2 x^{2} + x + C$ y = 6x³/3 − 4x²/2 + x + C = 2x³ − 2x² + x + C
2
Substitute $x = 2$ x = 2, $y = 5$ y = 5: $5 = 16 - 8 + 2 + C$ 5 = 16 − 8 + 2 + C
3
Solve: $5 = 10 + C$ 5 = 10 + C, so $C = - 5$ C = −5

Answer

y = 2 x^{3} - 2 x^{2} + x - 5

⚠️ Watch out

the 1 integrates to

x

, not to 0. Only differentiating sends a constant to 0.

Quick check

?Find

\int (6 x^{2} + 3 x^{- 2}) d x

.

Calculate

?

f^{'} (x) = 4 x - 3

and

f (1) = 6

. Find

f (3)

.

Calculus · integration

Definite integrals and area

The area between $y = f (x)$ y = f(x) and the $x$ x-axis from $x = a$ x = a to $x = b$ x = b, where $f (x) > 0$ f(x) > 0, is the definite integral

Area = \int_{a b} f (x) d x

Write the integral first, with its limits, then use your GDC to evaluate it. Examiners award a mark for the correct expression.
If $f$ f is a rate of change, the area under its graph is the total change: the area under a velocity–time graph is distance, and under a flow-rate graph it is volume.

Worked example Calculus · integration

Area under a curve

Question

Find the area of the region enclosed by

y = 6 + x - x^{2}

and the

x

-axis.

1
Find where the curve meets the $x$ x-axis: $6 + x - x^{2} = 0$ 6 + x − x² = 0, so $(3 - x) (2 + x) = 0$ (3 − x)(2 + x) = 0, giving $x = - 2$ x = −2 and $x = 3$ x = 3
2
Check the curve is above the axis between them: at $x = 0$ x = 0, $y = 6 > 0$ y = 6 > 0
3
Write the expression: $Area = \int_{- 2 3} (6 + x - x^{2}) d x$ Area = ∫₋₂³ (6 + x − x²) dx
4
Evaluate with your GDC: $20.833 \dots$ 20.833…

Answer

area

= 20.8

square units (3 s.f.)

⚠️ Watch out

the limits are the

x

-intercepts, not the

y

-intercept.

Calculate

?Water flows into a tank at a rate of

r (t) = 12 + 0.6 t - 0.03 t^{2}

litres per minute, for

0 \leq t \leq 10

minutes. Use your GDC to find how much water flows in during these 10 minutes.

litres

Calculus · integration

The trapezoidal rule

When you only have a table of values, split the area into $n$ n strips of equal width $h = \frac{b - a}{n}$ h = (b − a)/n and treat each strip as a trapezium.

\int_{a b} y d x \approx \frac{h}{2} [y_{0} + y_{n} + 2 (y_{1} + y_{2} + \dots + y_{n - 1})]

First and last values once, every middle value twice.
If the curve is concave-down (bends like a hill), the trapezia sit under it and the rule underestimates; if it is concave-up, the rule overestimates.

Calculate

?A tap's flow rate is read every minute.

Time (min)	0	1	2	3	4
Rate (litres per minute)	10	13	15	16	16

Use the trapezoidal rule with these five values to estimate the volume of water, in litres, that flows in the 4 minutes.

Mission check

?You know a tap's flow rate (litres per minute) at every moment. How do you find how much water filled the bath?

Mission 3

What shape of can uses the least metal for 330 ml?

Make a guess: What shape of can uses the least metal for 330 ml?

Calculus · stationary points

Stationary points

A stationary point is where $f^{'} (x) = 0$ f'(x) = 0: the tangent is horizontal.
Local maximum: $f^{'}$ f' changes from positive to negative. Local minimum: $f^{'}$ f' changes from negative to positive.
To find them, solve $f^{'} (x) = 0$ f'(x) = 0, then test the sign of $f^{'}$ f' just either side.
On a GDC paper, your GDC can find a local maximum or minimum point directly.
A local maximum is not always the greatest value on a domain: check the endpoints too.

Worked example Calculus · stationary points

Finding and classifying stationary points

Question

Find the stationary points of

f (x) = x^{3} - 3 x^{2} - 9 x + 5

and determine their nature.

1
Differentiate: $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ f'(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1)
2
Solve $f^{'} (x) = 0$ f'(x) = 0: $x = - 1$ x = −1 or $x = 3$ x = 3, giving $f (- 1) = 10$ f(−1) = 10 and $f (3) = - 22$ f(3) = −22
3
Test signs: $f^{'} (- 2) = 15 > 0$ f'(−2) = 15 > 0, $f^{'} (0) = - 9 < 0$ f'(0) = −9 < 0, $f^{'} (4) = 15 > 0$ f'(4) = 15 > 0
4
Read off: positive to negative at $x = - 1$ x = −1; negative to positive at $x = 3$ x = 3

Answer

local maximum

(- 1, 10)

; local minimum

(3, - 22)

⚠️ Watch out

always find the

y

-coordinate as well. "The point" means both coordinates.

Calculate

?

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

. Find the

y

-coordinate of the local maximum point.

Calculus · stationary points

Optimisation

1. Name the variables and write the quantity to optimise (area, cost, volume…).
2. Use the constraint (a fixed volume, a fixed length) to write it in one variable.
3. Differentiate and solve $\frac{d A}{d x} = 0$ dA/dx = 0 (or use your GDC on a GDC paper).
4. Show it is a maximum or minimum, then answer the question asked, with units, in context.

Worked example Calculus · stationary points

The largest open box

Question

An open box is made from a 30 cm square sheet by cutting a square of side

x

cm from each corner and folding up the sides. Find the maximum volume.

1
Write the volume: $V = x (30 - 2 x)^{2}$ V = x(30 − 2x)², for $0 < x < 15$ 0 < x < 15
2
Differentiate: $\frac{d V}{d x} = (30 - 2 x) (30 - 6 x)$ dV/dx = (30 − 2x)(30 − 6x)
3
Solve $\frac{d V}{d x} = 0$ dV/dx = 0: $x = 5$ x = 5 or $x = 15$ x = 15; $x = 15$ x = 15 gives $V = 0$ V = 0, so take $x = 5$ x = 5
4
Check: $V^{'} (4) = 132 > 0$ V'(4) = 132 > 0 and $V^{'} (6) = - 108 < 0$ V'(6) = −108 < 0, so $x = 5$ x = 5 gives a maximum
5
Evaluate: $V = 5 \times 2 0^{2}$ V = 5 × 20²

Answer

maximum volume

= 2000

cm³ (when

x = 5

cm)

⚠️ Watch out

reject values of

x

outside the domain or giving a zero box.

Calculate

?A closed cylindrical can of radius

r

cm and height

h

cm holds 330 cm³. Show that its surface area is

A = 2 π r^{2} + \frac{660}{r}

, then find the value of

r

that minimises

A

. Give your answer to 3 significant figures.

cm

Calculate

?

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

for

- 3 \leq x \leq 3

. Find the greatest value of

f (x)

in this domain.

Mission check

?A closed cylindrical can holds 330 ml. Which shape uses the least metal?

Recap

Big ideas

The derivative $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x) is the gradient of the curve (the limit of chord gradients) and the rate of change of $y$ y with $x$ x.
For integer $n$ n, $a x^{n}$ axⁿ differentiates to $an x^{n - 1}$ anxⁿ⁻¹; rewrite $\frac{a}{x ^{n}}$ a/xⁿ as $a x^{- n}$ ax⁻ⁿ first. $f^{'} (x) > 0$ f'(x) > 0 means increasing, $f^{'} (x) < 0$ f'(x) < 0 decreasing.
Tangent at $x = a$ x = a: gradient $m = f^{'} (a)$ m = f'(a) through $(a, f (a))$ (a, f(a)). Normal: gradient $- \frac{1}{m}$ −1/m through the same point.
$\int a x^{n} d x = \frac{a x ^{n + 1}}{n + 1} + C$ ∫ axⁿ dx = (axⁿ⁺¹)/(n+1) + C for $n \neq = - 1$ n ≠ −1; a boundary condition fixes $C$ C.
The area under $y = f (x) > 0$ y = f(x) > 0 from $a$ a to $b$ b is $\int_{a b} f (x) d x$ ∫_a^b f(x) dx: write the integral, then evaluate it with your GDC. The area under a rate graph is the total change.
Trapezoidal rule: $\frac{h}{2} [y_{0} + y_{n} + 2 (y_{1} + \dots + y_{n - 1})]$ h/2[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)] with equal strip widths.
Stationary points: solve $f^{'} (x) = 0$ f'(x) = 0. Classify by the sign of $f^{'}$ f' either side: $+$ + to $-$ − is a maximum, $-$ − to $+$ + a minimum.
Optimisation: use the constraint to get one variable, differentiate, set to zero, justify, answer in context. Check endpoints for the greatest or least value.

🏆

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You've worked through Calculus (AI SL) for IB Diploma Mathematics: Applications & Interpretation SL. 🎉

Mission 1A car's distance graph curves. How fast is it at exactly 3 s?
Mission 2A tap's flow rate keeps changing. How much filled the bath?
Mission 3What shape of can uses the least metal for 330 ml?

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Specification points and sources for this lesson are listed at the bottom of the page.

Calculus (AI SL)

A car's distance graph curves. How fast is it at exactly 3 s?

The gradient of a curve as a limit

Differentiating axnaxⁿ

Finding a gradient

Increasing, decreasing, tangents and normals

Tangent and normal

A tap's flow rate keeps changing. How much filled the bath?

Anti-differentiation

Using a boundary condition

Definite integrals and area

Area under a curve

The trapezoidal rule

What shape of can uses the least metal for 330 ml?

Stationary points

Finding and classifying stationary points

Optimisation

The largest open box

Big ideas

Mini-lesson complete!

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Differentiating $a x^{n}$ axⁿ