Introduction to Calculus

Welcome to Calculus!

Calculus is the mathematics of change and accumulation. In this unit you will learn to find the gradient of a curve at any point (differentiation), the equations of tangent and normal lines, turning points, and how to estimate areas under curves using the trapezium rule.

Power Rule: if y = xⁿ then dy/dx = nxⁿ⁻¹

Three big ideas:
1. Differentiation — find the gradient function dy/dx
2. Tangents & Normals — find equations of lines touching curves
3. Trapezium Rule — estimate area under a curve numerically

Gradient of a Curve

Power rule, dy/dx, gradient at a point

Tangents & Normals

Equations, turning points, 2nd derivative

Trapezium Rule

Area estimation, over/underestimate

Worked Examples

6 fully worked calculus problems

Curve Explorer

Interactive tangent line visualiser

Practice (25q)

Self-marking practice questions

Learn 1 — Gradient of a Curve

A straight line has a constant gradient Δy/Δx everywhere. A curve has a changing gradient — to find the gradient at a specific point we need the tangent at that point.

The Power Rule (Differentiation)

If y = xⁿ then dy/dx = n·xⁿ⁻¹ | Bring the power down, reduce by 1

Function y	Derivative dy/dx	Reason
x²	2x	Power 2→ 2×x¹ = 2x
x³	3x²	Power 3→ 3×x²
x⁴	4x³	Power 4→ 4×x³
5	0	Constant → gradient = 0
4x	4	4x¹ → 4×1×x⁰ = 4
3x²	6x	Multiply coefficient: 3×2=6

Differentiating a polynomial — do it term by term:
y = 4x² − 3x + 2
dy/dx = 4×2x − 3×1 + 0 = 8x − 3

y = 2x³ + 5x² − 7x + 1
dy/dx = 6x² + 10x − 7

Finding the Gradient at a Point

Substitute the x-value into dy/dx.

Example: y = x² − 4x + 1. Find gradient at x = 3.
dy/dx = 2x − 4
At x = 3: gradient = 2(3) − 4 = 6 − 4 = 2

Memory hook: "Bring down the power, knock it off by 1."
y = x⁵ → dy/dx = 5x⁴ → dy/dx of that = 20x³ (second derivative)

Learn 2 — Tangents, Normals, and Turning Points

Equation of a Tangent

Find dy/dx (differentiate)
Substitute x = a to get gradient m
Find y-coordinate: substitute x = a into original equation
Use y − y₁ = m(x − x₁)

Example: Find tangent to y = x² + 2 at x = 1.
dy/dx = 2x → at x=1: m = 2
y-coordinate: y = 1² + 2 = 3 → point is (1, 3)
Tangent: y − 3 = 2(x − 1) → y = 2x + 1

Equation of a Normal

The normal is perpendicular to the tangent at the same point.

Gradient of normal = −1 / (gradient of tangent)

Example: Normal to y = x³ at x = 2.
dy/dx = 3x² → at x=2: tangent gradient = 12
Normal gradient = −1/12
y-coordinate: y = 2³ = 8 → point (2, 8)
Normal: y − 8 = −(1/12)(x − 2)

Turning Points (Stationary Points)

At a turning point: dy/dx = 0

Differentiate to get dy/dx
Set dy/dx = 0 and solve for x
Substitute x back into y for coordinates
Find d²y/dx² for nature: >0 → minimum; <0 → maximum

Example: y = x² − 6x + 5
dy/dx = 2x − 6 = 0 → x = 3
y = 9 − 18 + 5 = −4 → turning point at (3, −4)
d²y/dx² = 2 > 0 → minimum ✓

Nature test: d²y/dx² > 0 (positive) → like a smile → minimum
d²y/dx² < 0 (negative) → like a frown → maximum

Learn 3 — Area Under Curve (Trapezium Rule)

The trapezium rule estimates the area under a curve by dividing it into strips, each approximated as a trapezium.

A ≈ h/2 × [y₀ + 2y₁ + 2y₂ + … + 2yₙ₋₁ + yₙ]

where h = width of each strip = (b−a)/n, and y₀, y₁, …, yₙ are the y-values at x₀, x₁, …, xₙ.

Pattern: first + last + 2×(all middle)
y₀ and yₙ appear once; all others (y₁ to yₙ₋₁) are multiplied by 2.

Worked Structure

Example: Estimate area under y=x² from x=0 to x=4 with 4 strips.
h = (4−0)/4 = 1
x: 0, 1, 2, 3, 4
y: 0, 1, 4, 9, 16
A ≈ (1/2)[0 + 2(1) + 2(4) + 2(9) + 16]
A ≈ (1/2)[0 + 2 + 8 + 18 + 16] = (1/2)(44) = 22

Over or Underestimate?

Concave up (∪ shape): Trapezium rule gives an OVERESTIMATE
Concave down (∩ shape): Trapezium rule gives an UNDERESTIMATE
More strips = more accurate result

Application: Kinematics

Velocity-time graph: Area under v-t graph = displacement
Use trapezium rule on a v-t graph to estimate distance/displacement when the curve is not simple.

Trap for students: Never double the FIRST (y₀) or LAST (yₙ) y-values. They appear exactly once in the formula.

Example 1 — Differentiate y = 3x² − 5x + 2, find gradient at x = 4

dy/dx = 3×2x − 5×1 + 0 = 6x − 5

At x = 4: gradient = 6(4) − 5 = 24 − 5 = 19

Example 2 — Equation of tangent to y = x² + 2x at x = 1

dy/dx = 2x + 2. At x=1: m = 2(1)+2 = 4

y-coordinate: y = 1² + 2(1) = 3. Point: (1, 3)

Tangent: y − 3 = 4(x − 1) → y = 4x − 1

Example 3 — Equation of normal to y = x³ at x = 2

dy/dx = 3x². At x=2: tangent gradient m = 3(4) = 12

Normal gradient = −1/12

y-coordinate: y = 2³ = 8. Point: (2, 8)

Normal: y − 8 = −(1/12)(x − 2) → y = −x/12 + 1/6 + 8

Simplified: y = −x/12 + 49/6

Example 4 — Turning point of y = 2x² − 8x + 3

dy/dx = 4x − 8. Set equal to 0: 4x − 8 = 0 → x = 2

y = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5. Turning point: (2, −5)

d²y/dx² = 4 > 0 → MINIMUM

Example 5 — Trapezium rule: area under y = x² from x = 0 to x = 3, 3 strips

h = (3−0)/3 = 1. x-values: 0, 1, 2, 3

y-values: y₀=0, y₁=1, y₂=4, y₃=9

A ≈ (1/2)[y₀ + 2y₁ + 2y₂ + y₃] = (1/2)[0 + 2 + 8 + 9] = (1/2)(19) = 9.5

Note: exact area = 9, so this is an overestimate (curve is concave up)

Example 6 — Kinematics: v = 3t² − 6t, distance in first 3 seconds, trapezium rule (t=0,1,2,3)

h = 1. t-values: 0, 1, 2, 3

v(0)=0, v(1)=3−6=−3, v(2)=12−12=0, v(3)=27−18=9

Area ≈ (1/2)[0 + 2(−3) + 2(0) + 9] = (1/2)[0 − 6 + 0 + 9] = (1/2)(3) = 1.5

Note: negative velocities indicate reverse motion; total distance needs careful handling of signs

Common Mistakes in Calculus

Mistake 1: Differentiating a constant to get itself
If y = 5, then dy/dx = 0 (NOT 5). The gradient of a horizontal line is always 0.

Mistake 2: Power rule — forgetting to reduce power by 1
y = x³ → dy/dx = 3x² (NOT 3x³ or x²/3). Bring the power DOWN then REDUCE it by 1.

Mistake 3: Not finding y₁ before writing tangent equation
To write y − y₁ = m(x − x₁), you need BOTH x₁ and y₁. Always substitute x₁ into y to find y₁ first.

Mistake 4: Normal gradient = tangent gradient
Normal gradient = −1/m (negative reciprocal). If tangent gradient is 4, normal gradient is −1/4.

Mistake 5: Doubling first/last in trapezium rule
The formula is: h/2 × [y₀ + 2y₁ + ... + 2yₙ₋₁ + yₙ]. Only the MIDDLE values (y₁ to yₙ₋₁) are doubled. y₀ and yₙ are NOT doubled.

Key Formulas — Calculus

Power Ruley=xⁿ → dy/dx=nxⁿ⁻¹

Constant Ruley=c → dy/dx=0

Linear Ruley=ax → dy/dx=a

Sum Ruled/dx[f+g] = f'+g'

Tangent Gradientm = dy/dx at x=a

Normal Gradientm_n = −1/m_t

Line Equationy−y₁=m(x−x₁)

Turning Pointdy/dx = 0

Min Testd²y/dx² > 0

Max Testd²y/dx² < 0

Trapezium Ruleh/2[y₀+2y₁+…+2yₙ₋₁+yₙ]

Strip widthh = (b−a)/n

Situation	Method
Find gradient at point x=a	Differentiate, substitute x=a
Find tangent equation	Get m, get y₁, use y−y₁=m(x−x₁)
Find normal equation	Normal gradient = −1/m, same point
Find turning point	Set dy/dx=0, solve for x, find y
Nature of turning point	Compute d²y/dx²: +ve=min, −ve=max
Estimate area under curve	Trapezium rule with n strips

Curve Explorer — Interactive Tangent

Enter polynomial coefficients for y = ax² + bx + c, then move the slider to see the tangent at any point.

a (x²): b (x): c:

x = 0

Gradient and tangent equation appear here.

Exercise 1 — Differentiation

Q1. Differentiate y = x⁵. Enter dy/dx.

Q2. Differentiate y = 6x³. Enter dy/dx.

Q3. Differentiate y = 4x² − 3x + 7. Enter dy/dx.

Q4. Differentiate y = 2x³ + 5x² − 1. Enter dy/dx.

Q5. Differentiate y = 10. Enter dy/dx.

Q6. Differentiate y = x⁴ − 2x. Enter dy/dx.

Exercise 2 — Find Gradient at a Point

Q1. y = x². Find gradient at x = 3.

Q2. y = 3x² − 2x. Find gradient at x = 2.

Q3. y = x³ + 1. Find gradient at x = −1.

Q4. y = 4x² − 5. Find gradient at x = 0.

Q5. y = 2x³ − 3x² + x. Find gradient at x = 1.

Q6. y = x⁴ − 2x². Find gradient at x = 2.

Exercise 3 — Tangent Equations

Find the equation of the tangent to each curve at the given point. Enter in the form y=mx+c — enter the value of c.

Q1. y = x² at x = 2. Enter the tangent equation c-value (y = 4x + c, find c).

Q2. y = x² + 3 at x = 1. Gradient = 2. y-coordinate = 4. Enter c in y = 2x + c.

Q3. y = 2x² − x at x = 3. Find gradient m.

Q4. y = x³ at x = −1. Find gradient m = 3x². What is m?

Q5. y = x² − 4x + 5 at x = 2. The gradient is 0. What type of point is this?

Enter: turning

Q6. y = 3x² + 2x at x = 1. Enter gradient m.

Exercise 4 — Turning Points

Q1. y = x² − 6x + 5. Find x-coordinate of turning point.

Q2. y = x² − 6x + 5. Find y-coordinate of turning point.

Q3. y = 2x² − 8x + 3. Is the turning point a maximum or minimum?

Q4. y = −x² + 4x − 1. Find x-coordinate of turning point.

Q5. y = x³ − 3x. Find the x-values of both turning points. Enter the positive one.

Q6. y = −2x² + 8x. At what x does the maximum occur?

Exercise 5 — Trapezium Rule

Q1. Estimate area under y = x² from x=0 to x=2 using 2 strips (h=1). x: 0,1,2 — y: 0,1,4. Apply trapezium rule.

Q2. Trapezium rule with h=1, y₀=2, y₁=5, y₂=10. Estimate the area.

Q3. Is the trapezium rule an overestimate or underestimate for a concave up (∪) curve?

Q4. Estimate area under y = 2x from x=0 to x=4 using 4 strips. y-values: 0, 2, 4, 6, 8. Apply trapezium rule (h=1).

Q5. On a v-t graph, v-values at t=0,1,2,3 are 0, 4, 6, 4. Estimate displacement using trapezium rule with h=1.

Q6. Trapezium rule: y₀=1, y₁=3, y₂=5, y₃=3, y₄=1, h=0.5. Estimate area.