NCEA Level 3 Calculus cheat sheet
NCEA Level 3 Calculus · Keep the key ideas within reach.
Differentiation and Integration
Power rule: d/dx(x^n) = n·x^(n−1) for all real n; integrate with ∫x^n dx = x^(n+1)/(n+1) + C, valid only when n ≠ −1, since that case gives ln|x| + C.
Chain rule: if y = f(g(x)) then dy/dx = f′(g(x))·g′(x). For parametric curves, dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0.
Product and quotient rules: (uv)′ = u′v + uv′; (u/v)′ = (u′v − uv′)/v². Quotient rule is avoided by rewriting as a product with a negative index.
Derivatives to memorise: d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x, d/dx(e^x) = e^x, d/dx(ln x) = 1/x.
Standard integrals: ∫sin x dx = −cos x + C, ∫cos x dx = sin x + C, ∫sec²x dx = tan x + C, ∫e^x dx = e^x + C, ∫(1/x) dx = ln|x| + C.
Definite integrals: ∫ from a to b of f(x) dx = F(b) − F(a) where F′ = f. Reversing limits changes the sign; splitting at c gives ∫a→c + ∫c→b.
Area between curves: integrate (upper − lower) between intersection points found by solving f(x) = g(x). Area is never negative; a sign change in the integrand means splitting the interval.
Chain rule in reverse (integration by substitution): ∫f′(g(x))g′(x) dx = f(g(x)) + C. For definite integrals also convert the limits to g(a) and g(b).
Stationary points occur where f′(x) = 0. Use the sign of f″ at the point: f″ > 0 gives a minimum, f″ < 0 gives a maximum, f″ = 0 is inconclusive and needs a sign table.
Conic Sections and Coordinate Geometry
Circle with centre (a, b) and radius r: (x − a)² + (y − b)² = r². General form x² + y² + Dx + Ey + F = 0 is converted by completing the square.
Parabola y² = 4ax has vertex at the origin, focus (a, 0) and directrix x = −a; every point is equidistant from focus and directrix. Swapping x and y gives x² = 4ay.
Ellipse x²/a² + y²/b² = 1 (a > b) has centre at origin, semi-major axis a, semi-minor axis b, foci at (±c, 0) where c² = a² − b².
Hyperbola x²/a² − y²/b² = 1 has foci (±c, 0) with c² = a² + b² and asymptotes y = ±(b/a)x. Note the plus sign in c², unlike the ellipse.
Translated conics: replacing x by (x − h) and y by (y − k) moves the centre or vertex to (h, k) without changing a, b or the shape.
General second-degree curve Ax² + Bxy + Cy² + Dx + Ey + F = 0: the discriminant B² − 4AC classifies it as parabola (0), ellipse (< 0) or hyperbola (> 0).
Gradient of a line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁). Parallel lines have equal m; perpendicular lines satisfy m₁m₂ = −1.
Distance from point (x₁, y₁) to line ax + by + c = 0 is |ax₁ + by₁ + c| / √(a² + b²). The perpendicular distance from a focus to a tangent of a parabola equals the focal distance relation.
Tangent to y² = 4ax at (at², 2at) is ty = x + at²; tangent to x² + y² = r² at (x₁, y₁) is x·x₁ + y·y₁ = r².
Trigonometric Functions and Identities
Pythagorean identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ. The second follows by dividing the first by cos²θ.
Compound angles: sin(A ± B) = sin A cos B ± cos A sin B; cos(A ± B) = cos A cos B ∓ sin A sin B; tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B).
Double angles: sin 2θ = 2 sin θ cos θ; cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ; tan 2θ = 2tan θ/(1 − tan²θ).
General solutions: sin θ = k gives θ = nπ + (−1)^n·arcsin k; cos θ = k gives θ = 2nπ ± arccos k; tan θ = k gives θ = nπ + arctan k, with n an integer.
Exact values: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3; sin 45° = cos 45° = 1/√2; sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.
Reciprocal and quotient definitions: tan θ = sin θ/cos θ, cot θ = cos θ/sin θ, sec θ = 1/cos θ, cosec θ = 1/sin θ. Undefined where the denominator is zero.
Amplitude and period: y = A sin(Bx + C) + D has amplitude |A|, period 2π/|B|, phase shift −C/B and vertical shift D. For tan, the period is π/|B|.
Solving equations like a cos θ + b sin θ = c: rewrite as R sin(θ + α) with R = √(a² + b²) and tan α = a/b, then solve the single sine equation.
Inverse trig ranges: arcsin returns [−π/2, π/2], arccos returns [0, π], arctan returns (−π/2, π/2); solutions outside these ranges must be generated from the general solution.
Complex Numbers and Algebra
Complex number z = a + bi has real part a and imaginary part b, with i² = −1. Equality requires both parts equal; z = 0 only when a = b = 0.
Conjugate z̄ = a − bi. Products: z·z̄ = a² + b², a real non-negative number equal to |z|². Division multiplies numerator and denominator by the conjugate of the denominator.
Modulus and argument: |z| = √(a² + b²), arg z = arctan(b/a) adjusted for the quadrant. Polar form z = r(cos θ + i sin θ) = r cis θ = re^(iθ).
Multiplication and division in polar form: multiply moduli and add arguments; divide moduli and subtract arguments. Powers follow De Moivre: (r cis θ)^n = r^n cis(nθ).
Roots: the n distinct nth roots of r cis θ are r^(1/n) cis((θ + 2kπ)/n) for k = 0, 1, …, n − 1; they lie equally spaced on a circle of radius r^(1/n).
Quadratic formula works over the complex numbers: for az² + bz + c = 0, z = (−b ± √(b² − 4ac))/(2a). Negative discriminant gives a conjugate pair of non-real roots.
If coefficients are real, non-real roots occur in conjugate pairs; the sum of roots is −b/a and the product is c/a (Viète's formulas).
Factor and remainder theorems: (x − k) is a factor of P(x) exactly when P(k) = 0; the remainder on dividing by (x − k) is P(k).
Geometric interpretation: |z − z₀| = r is a circle centred at z₀ with radius r; |z − z₀| = |z − z₁| is the perpendicular bisector of the segment joining z₀ and z₁.