NCEA Level 3 Calculus common mistakes
NCEA Level 3 Calculus · Understand the traps and how to avoid them.
Differentiating sin(3x) or e^(2x) as 3cos(3x) but then forgetting the inner derivative on a second term such as sin(x²), writing cos(x²) instead of 2x·cos(x²).
Candidates memorise the outer derivative pattern (sin → cos, e → e) and apply it mechanically, treating the bracket as a single variable. The chain rule is learned as a formula for simple linear arguments and is not generalised to non-linear inner functions, so the inner derivative is dropped whenever it is not a constant.
What to do instead
Force the habit of naming the inner function before differentiating: write u = x², then dy/dx = dy/du · du/dx, and physically write the du/dx factor (2x) on every line until it becomes automatic. Drill a mixed set where the inner function is linear, quadratic, and trigonometric, and mark each answer with the inner derivative circled so omissions are visible.
Integrating x·cos(x) by writing (x²/2)·sin(x), i.e. integrating each factor separately instead of using integration by parts.
The product rule for differentiation is well rehearsed, but its inverse (integration by parts) is not recognised as the tool for a product of unlike functions. Candidates default to the power rule and the standard integral table, applying them term by term because the integrand superficially looks like a simple product.
What to do instead
Before integrating any product, classify the factors: polynomial × trigonometric, polynomial × exponential, or logarithm × polynomial all signal integration by parts. Write the formula ∫u dv = uv − ∫v du explicitly and choose u by LIATE (log, inverse trig, algebraic, trig, exponential). Practise at least ten products so the classification becomes a reflex rather than a guess.
Omitting the constant of integration on an indefinite integral, or writing + c on a definite integral after evaluating limits.
Candidates treat + c as cosmetic notation rather than a mathematical necessity, and under exam pressure the final line is written quickly. The distinction between an indefinite integral (a family of antiderivatives) and a definite integral (a number) is not firmly held, so the symbol is added or dropped by habit rather than meaning.
What to do instead
State the rule in words on every practice paper: indefinite integrals always carry + c, definite integrals never do because the constant cancels. When a question says 'find the antiderivative', underline the word and check the final line for + c before moving on; when limits are present, evaluate F(b) − F(a) and confirm the answer is a number.
Finding dy/dx for a curve and then substituting the x-value to get the tangent gradient, but forgetting to substitute the same x into the original equation to find the y-coordinate of the point of contact.
The gradient is the visible output of differentiation, so candidates stop once they have a number. The tangent line equation requires both a gradient and a point, and the point must lie on the curve, not on the tangent. The two-step structure of the problem is not held in working memory.
What to do instead
Use a fixed template for every tangent or normal question: (1) differentiate, (2) substitute x to get m, (3) substitute x into the original y to get the point, (4) write y − y₁ = m(x − x₁). Number the steps in the margin and tick each one off. For normals, add step (2b): m_normal = −1/m_tangent, and check the sign.
Completing the square for a circle or ellipse and losing the constant, e.g. writing x² + 4x + y² − 6y = 12 as (x+2)² + (y−3)² = 12 instead of = 25.
Candidates add the squared half-coefficients inside the brackets but forget that the same amount must be added to the right-hand side to preserve equality. The balancing step is the least visible part of the procedure, so it is the first to be dropped, especially when working quickly.
What to do instead
Always write the intermediate line with the additions shown explicitly: x² + 4x + 4 + y² − 6y + 9 = 12 + 4 + 9. Then simplify to (x+2)² + (y−3)² = 25 and read off centre (−2, 3) and radius 5. Check by expanding the final form back to the original; if the constant differs, the balancing step was missed.
Reading the radius of x²/9 + y²/16 = 1 as 9 and 16 rather than 3 and 4, or labelling the major axis as horizontal when b² > a².
The standard form is memorised as x²/a² + y²/b² = 1, and candidates equate the denominators directly with the semi-axes, skipping the square root. The convention that a is the larger semi-axis is confused with the position of the term, so the orientation of the ellipse is misread.
What to do instead
Write the standard form with the squares visible (x²/3² + y²/4² = 1) before extracting a and b, and take the square root explicitly. State which denominator is larger and therefore which axis is major: here 16 > 9, so the major axis is vertical with semi-axis 4. Sketch the ellipse with intercepts marked to confirm orientation.
Solving sin(2x) = 1/2 and giving only x = 15° (or π/12) instead of all solutions in the required interval, such as 15°, 75°, 195°, 255°.
Candidates apply the inverse sine once and stop, treating the equation as linear. The periodic nature of sine and the doubling of the argument both multiply the number of solutions, and the general solution form is not applied systematically.
What to do instead
Use the general solution: 2x = 30° + 360°k or 2x = 150° + 360°k, then divide by 2 to get x = 15° + 180°k or x = 75° + 180°k. List k = 0, ±1, ±2 and keep only values inside the stated interval. Always check the number of solutions against the period of the function (here period 180°, so two solutions per 180°).
Simplifying sin(A + B) as sin A + sin B, or cos(2x) as 2cos(x) when applying a double-angle identity.
The notation sin(A+B) looks like a distributive operation, and candidates transfer algebraic habits into trigonometry. The compound-angle and double-angle identities are memorised as isolated formulas and not recognised as the correct expansion, so a plausible-looking but invalid simplification is used.
What to do instead
Write the correct expansions on a reference card and test any suspected identity numerically with a simple angle (e.g. A = B = 30°): sin 60° = 0.866 but sin 30° + sin 30° = 1, so the distributive version is false. For double angles, practise converting cos 2x into 1 − 2sin²x or 2cos²x − 1 depending on which form simplifies the equation.
Multiplying complex numbers in polar form by multiplying the moduli but also multiplying the arguments, instead of adding them.
The rules for multiplication and division in polar form are easily transposed, and candidates reason by analogy with the modulus rule (multiply moduli, so multiply arguments). The geometric meaning of argument addition as rotation is not connected to the algebraic rule.
What to do instead
Anchor the rule geometrically: multiplying by a complex number of argument θ rotates by θ, so arguments add; dividing rotates back, so arguments subtract. Write z₁z₂ = r₁r₂ cis(θ₁ + θ₂) and z₁/z₂ = (r₁/r₂) cis(θ₁ − θ₂) side by side, and verify with i·i = −1: arguments 90° + 90° = 180°, giving modulus 1 and argument 180°, which is −1.
Solving a quadratic with complex roots and writing only one root, or discarding the negative discriminant as 'no solution'.
Candidates carry the real-number convention that a negative discriminant means the equation cannot be solved, and stop rather than introducing i. When both roots are found, the conjugate pair is not recognised, so the second root is omitted or the ± is dropped.
What to do instead
State explicitly: if Δ < 0, the roots are a conjugate pair involving i. Apply the quadratic formula with the ± intact, simplify √(negative) as i√(positive), and present both roots. Check that the two roots are conjugates and that their sum equals −b/a and product equals c/a, which catches sign and omission errors.