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Worked examples: VCE Specialist Mathematics Formula Sheets
Read the VCAA Specialist Mathematics formulas with worked examples in complex numbers, calculus, statistics, vectors and kinematics.
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Formula index with worked examples
Mensuration · circle segment
Formula or reference: A = r²(θ − sin θ)/2
Units and conditions: θ in radians and A in square units.
Worked example: For r = 2 cm and θ = π/2, A = (π − 2) cm², the quarter-circle area less the triangle area.
Mensuration · cylinder, cone, pyramid and sphere
Formula or reference: Cylinder: V = πr²h. Cone: V = πr²h/3. Pyramid: V = Ah/3. Sphere: V = 4πr³/3.
Units and conditions: Use perpendicular height h, base area A and cubic units.
Worked example: At r = 2 cm and h = 3 cm, the cylinder has volume 12π cm³ and the cone 4π cm³. A pyramid with A = 12 cm² and h = 3 cm has volume 12 cm³. A sphere of radius 3 cm has volume 36π cm³.
Trigonometry · triangle formulas
Formula or reference: A = bc sin A₀/2, where A₀ is the included angle. a/sin A₀ = b/sin B = c/sin C. c² = a² + b² − 2ab cos C.
Units and conditions: Match lengths to their opposite angles. Use square units for area.
Worked example: A right triangle with legs 3 cm and 4 cm has area 6 cm² and hypotenuse 5 cm by the cosine rule. Its angle opposite the 3 cm leg satisfies sin A₀ = 3/5.
Complex numbers · modulus
Formula or reference: |x + iy| = √(x² + y²)
Units and conditions: The modulus is non-negative.
Worked example: For z = −3 + 4i, |z| = √(9 + 16) = 5.
Complex numbers · polar form and principal argument
Formula or reference: z = x + iy = r(cos θ + i sin θ) = r cis θ. For z ≠ 0, −π < Arg(z) ≤ π.
Units and conditions: r = |z| and θ is in radians. Arg(z) denotes the principal value.
Worked example: For z = −1 + i, r = √2 and Arg(z) = 3π/4, giving √2 cis(3π/4).
Complex numbers · products and quotients
Formula or reference: z₁z₂ = r₁r₂ cis(θ₁+θ₂). z₁/z₂ = (r₁/r₂) cis(θ₁−θ₂) for z₂ ≠ 0.
Units and conditions: Use the same angle units for both factors.
Worked example: For z₁ = 6 cis(π/3) and z₂ = 2 cis(π/6), the product is 12i and the quotient is 3 cis(π/6).
Complex numbers · powers
Formula or reference: If z = r cis θ, then zⁿ = rⁿ cis(nθ) for integer n, with z ≠ 0 for negative n.
Units and conditions: Use the same angle convention throughout.
Worked example: (2 cis(π/6))³ = 8 cis(π/2) = 8i.
Statistics · linear transformations
Formula or reference: E(aX+b) = aE(X)+b. Var(aX+b) = a²Var(X).
Units and conditions: Shifting changes the mean but leaves variance unchanged. Variance uses squared units.
Worked example: If E(X) = 4 and Var(X) = 9, then 2X+3 has mean 11 and variance 36.
Statistics · weighted and independent sums
Formula or reference: E(ΣaᵢXᵢ) = ΣaᵢE(Xᵢ). For independent variables, Var(ΣaᵢXᵢ) = Σaᵢ²Var(Xᵢ).
Units and conditions: Independence is needed for this variance rule. The expectation rule is linear regardless.
Worked example: For independent X and Y with means 3 and 5 and variances 2 and 4, 2X−Y has mean 1 and variance 12.
Statistics · identically distributed observations
Formula or reference: For independent identically distributed Xi with mean μ and variance σ², E(ΣXi) = nμ and Var(ΣXi) = nσ². E(X̄) = μ and Var(X̄) = σ²/n.
Units and conditions: Use n independent observations with common mean and finite variance.
Worked example: For four observations with μ = 10 and σ² = 9, the sum has mean 40 and variance 36. Their mean has expectation 10 and variance 2.25.
Statistics · approximate confidence interval for a mean
Formula or reference: x̄ ± z s/√n
Units and conditions: Use the critical z for the requested confidence level and check the conditions for the approximation.
Worked example: With x̄ = 12, s = 4 and n = 100, the approximate 95% interval is 12 ± 1.96 × 0.4, or (11.216, 12.784).
Calculus · powers and linear powers
Formula or reference: d(xⁿ)/dx = nx^(n−1). ∫ xⁿ dx = x^(n+1)/(n+1) + C. ∫(ax+b)ⁿ dx = (ax+b)^(n+1)/[a(n+1)] + C.
Units and conditions: For the power antiderivatives n ≠ −1, and a ≠ 0 in the linear form.
Worked example: d(x⁴)/dx = 4x³. Also ∫x³ dx = x⁴/4 + C and ∫(2x+1)³ dx = (2x+1)⁴/8 + C.
Calculus · exponential and logarithm
Formula or reference: d(e^(ax))/dx = ae^(ax). ∫e^(ax) dx = e^(ax)/a + C. d(ln x)/dx = 1/x. ∫1/x dx = ln|x| + C. ∫1/(ax+b) dx = ln|ax+b|/a + C.
Units and conditions: Use a ≠ 0 where dividing by a, x > 0 for ln x, and intervals away from denominator zeros for the integrals.
Worked example: The derivative of e^(2x) is 2e^(2x) and its antiderivative is e^(2x)/2 + C. At x = 2, (ln x)′ = 1/2. An antiderivative of 1/(3x+1) is ln|3x+1|/3 + C.
Calculus · sine and cosine
Formula or reference: d[sin(ax)]/dx = a cos(ax), d[cos(ax)]/dx = −a sin(ax). ∫sin(ax) dx = −cos(ax)/a + C and ∫cos(ax) dx = sin(ax)/a + C.
Units and conditions: Use radians and a ≠ 0 for the antiderivative forms.
Worked example: For a = 2, the derivatives at x = 0 are 2 and 0. The antiderivatives are −cos(2x)/2 + C and sin(2x)/2 + C.
Calculus · tangent and cotangent
Formula or reference: d[tan(ax)]/dx = a sec²(ax), d[cot(ax)]/dx = −a cosec²(ax). ∫sec²(ax) dx = tan(ax)/a + C. ∫cosec²(ax) dx = −cot(ax)/a + C.
Units and conditions: Use radians and work between the corresponding poles. Use a ≠ 0 for the integral forms.
Worked example: At a = 2 and x = π/8, the tangent and cotangent derivatives are 4 and −4. The integral of cosec²(2x) is −cot(2x)/2 + C.
Calculus · secant and cosecant
Formula or reference: d[sec(ax)]/dx = a sec(ax)tan(ax). d[cosec(ax)]/dx = −a cosec(ax)cot(ax). ∫sec(ax)tan(ax) dx = sec(ax)/a + C. ∫cosec(ax)cot(ax) dx = −cosec(ax)/a + C.
Units and conditions: Use radians and stay where the functions are defined. For the integrals a ≠ 0.
Worked example: At a = 2 and x = π/8 the derivatives are 2√2 and −2√2. Differentiating sec(2x)/2 and −cosec(2x)/2 recovers the two integrands.
Calculus · inverse-trigonometric derivatives
Formula or reference: d[arcsin(ax)]/dx = a/√(1−a²x²). d[arccos(ax)]/dx = −a/√(1−a²x²). d[arctan(ax)]/dx = a/(1+a²x²).
Units and conditions: Use radians and |ax| < 1 for the first two derivatives.
Worked example: At x = 0 and a = 3 the derivatives are 3, −3 and 3.
Calculus · inverse-trigonometric integrals
Formula or reference: ∫1/√(a²−x²) dx = arcsin(x/a) + C. ∫−1/√(a²−x²) dx = arccos(x/a) + C. ∫a/(a²+x²) dx = arctan(x/a) + C.
Units and conditions: Use a > 0 and |x| < a for the first two integrands.
Worked example: For a = 2, the antiderivative of 2/(4+x²) is arctan(x/2) + C. The matching square-root forms give arcsin(x/2) + C and arccos(x/2) + C.
Calculus · product, quotient and chain
Formula or reference: (uv)′ = u′v + uv′. (u/v)′ = (vu′ − uv′)/v². dy/dx = (dy/du)(du/dx).
Units and conditions: For the quotient rule v ≠ 0.
Worked example: Differentiate x sin x to get sin x + x cos x. For x/(1+x), the quotient rule gives 1/(1+x)². For cos(x²), the chain rule gives −2x sin(x²).
Calculus · integration by parts
Formula or reference: ∫u v′ dx = uv − ∫v u′ dx
Units and conditions: Choose a differentiable u and an integrable v′.
Worked example: For ∫xeˣ dx, take u = x and v = eˣ. This gives xeˣ − eˣ + C.
Calculus · Euler’s method
Formula or reference: For y′ = f(x,y), x(next) = x + h and y(next) = y + h f(x,y), starting at the given initial point.
Units and conditions: Use the stated step h and evaluate f at the current point.
Worked example: With y′ = x + y, y(0) = 2 and h = 0.1, the first step gives x = 0.1 and y ≈ 2.2.
Calculus · parametric arc length
Formula or reference: L = ∫√[(dx/dt)²+(dy/dt)²] dt over the parameter interval.
Units and conditions: Use matching units for x and y. Length is non-negative.
Worked example: For x = 3t and y = 4t, 0 ≤ t ≤ 2, the speed in the parameter is 5 and arc length is 10.
Calculus · Cartesian surfaces of revolution
Formula or reference: About the x-axis: integrate 2π|y|√(1+(dy/dx)²) dx. About the y-axis: integrate 2π|x|√(1+(dx/dy)²) dy.
Units and conditions: Use the distance to the axis as the radius. Output is surface area excluding end caps.
Worked example: Rotating y = 2 over 0 ≤ x ≤ 3 gives 12π square units. Rotating x = 2 over 0 ≤ y ≤ 3 about the y-axis gives the same result.
Calculus · parametric surfaces of revolution
Formula or reference: Integrate 2πr(t)√[(dx/dt)²+(dy/dt)²] dt, where r(t) = |y(t)| about the x-axis or |x(t)| about the y-axis.
Units and conditions: Integrate over the chosen parameter interval. Output is in square units.
Worked example: For x = 2t, y = 3, 0 ≤ t ≤ 1, rotation about the x-axis gives surface area 12π. For x = 3, y = 2t on the same interval, rotation about the y-axis also gives 12π.
Kinematics · acceleration
Formula or reference: a = d²x/dt² = dv/dt = v dv/dx = d(v²/2)/dx
Units and conditions: Use consistent length and time units. The position form applies when velocity can be treated as a differentiable function of position.
Worked example: For v = 2x on the motion segment, v dv/dx = 4x, so at x = 3 the acceleration is 12 in the chosen units.
Kinematics · constant acceleration
Formula or reference: v = u + at. s = ut + at²/2. v² = u² + 2as. s = (u + v)t/2.
Units and conditions: Choose a positive direction and use m, s and signed components.
Worked example: For u = 2 m/s, a = 2 m/s² and t = 3 s, the final velocity is 8 m/s and the displacement is 15 m.
Vectors · position, magnitude and derivative
Formula or reference: r(t) = (x(t),y(t),z(t)). |r(t)| = √(x²+y²+z²). r′(t) = (x′(t),y′(t),z′(t)).
Units and conditions: Differentiate each component with respect to the same parameter.
Worked example: For r(t) = (3t,4t,t²), |r(1)| = √26 and r′(1) = (3,4,2).
Vectors · dot and cross products
Formula or reference: u · v = u₁v₁ + u₂v₂ + u₃v₃ = |u||v|cos θ. u × v = (u₂v₃−u₃v₂, u₃v₁−u₁v₃, u₁v₂−u₂v₁).
Units and conditions: The dot product is scalar. The order fixes the direction of the cross product.
Worked example: For u = (1,0,0) and v = (1,1,0), u · v = 1 and θ = 45°. Their cross product is (0,0,1).
Vectors · line equations
Formula or reference: Vector form r(t) = a + tb. Parametric coordinates are x = a₁ + tb₁, y = a₂ + tb₂ and z = a₃ + tb₃.
Units and conditions: Use a point a and a non-zero direction b.
Worked example: With a = (1,2,3) and b = (1,0,−1), t = 2 gives (3,2,1).
Vectors · plane equations
Formula or reference: r(s,t) = r₀ + su + tv. Parametric coordinates follow by taking each component. Cartesian form is ax + by + cz = d, with normal (a,b,c).
Units and conditions: Use independent directions u and v or a non-zero normal.
Worked example: Taking r₀ = (0,0,2), u = (1,0,0) and v = (0,1,0) gives (x,y,z) = (s,t,2), so the Cartesian equation is z = 2.
Circular functions · Pythagorean identities
Formula or reference: cos² x + sin² x = 1. 1 + tan² x = sec² x. 1 + cot² x = cosec² x.
Units and conditions: Each identity holds where the functions on both sides are defined.
Worked example: At x = π/4, the squared sine and cosine sum to 1, while 1 + tan² x and 1 + cot² x both equal 2.
Circular functions · sum and difference
Formula or reference: sin(x±y) = sin x cos y ± cos x sin y. cos(x±y) = cos x cos y ∓ sin x sin y. tan(x±y) = (tan x ± tan y)/(1 ∓ tan x tan y).
Units and conditions: Use the upper signs together or the lower signs together. The tangent forms require non-zero denominators.
Worked example: For x = 45° and y = 30°, sin(x+y) = (√6+√2)/4, cos(x−y) = (√6+√2)/4 and tan(x−y) = 2−√3.
Circular functions · double angles
Formula or reference: sin(2x) = 2 sin x cos x. cos(2x) = cos² x − sin² x = 2cos² x − 1 = 1 − 2sin² x. tan(2x) = 2tan x/(1−tan² x).
Units and conditions: Use consistent angle units. Apply the tangent formula where both sides are defined.
Worked example: For x = 30°, sin(2x) = √3/2, cos(2x) = 1/2 and tan(2x) = √3.
Circular functions · squared sine and cosine
Formula or reference: sin²(ax) = [1 − cos(2ax)]/2. cos²(ax) = [1 + cos(2ax)]/2.
Units and conditions: Keep angle units consistent.
Worked example: If ax = π/3, cos(2ax) = −1/2. The squared sine is 3/4 and the squared cosine is 1/4.
What to work out beyond the sheet
When a question asks for a principal argument, first find the correct quadrant and then choose the equivalent angle in the required interval.
Subject resources · Official past papers · ATAR calculator guide