LC Applied Maths common mistakes
LC Applied Maths · Understand the traps and how to avoid them.
When resolving forces for a particle on an inclined plane, the candidate takes the component of weight along the plane as W·cos θ and the normal component as W·sin θ (or resolves friction along the wrong axis), then carries this through the whole force equation.
The candidate memorises 'cos for the plane' from one worked example without re-deriving from the angle geometry: the angle between the weight vector and the normal to the plane equals the plane's inclination, so the perpendicular component is W·cos θ and the parallel component is W·sin θ. Under exam pressure the pattern is recalled rather than reconstructed.
What to do instead
Before writing equations, draw the weight vector and the normal to the plane, mark the inclination θ at the base, and derive both components from the right triangle every time. Sanity-check limiting cases: at θ = 0 the parallel component must vanish (so it must contain sin θ) and at θ = 90° the normal component must vanish (so it must contain cos θ).
The candidate uses u = 0 for a projectile launched at an angle, or applies a single equation such as v² = u² + 2as along the line of motion while treating the motion as one-dimensional, so range, time of flight and maximum height come out wrong.
Projectile motion is two independent constant-acceleration problems sharing one time variable, but candidates collapse it into a single SUVAT chain. They also forget that at the highest point the vertical velocity is zero while the horizontal velocity is not, and that speed at impact is not zero.
What to do instead
Always set up two columns - horizontal (a = 0, so x = u cos α · t) and vertical (a = −g) - and link them only through t. Write the vertical velocity as u sin α − gt explicitly and impose the correct condition for each quantity: v_y = 0 for max height, y = 0 for time of flight, and combine components for speed. Check that the answer's dimensions and limiting cases (α = 0, α = 90°) behave sensibly.
For a body in equilibrium under three non-parallel forces, the candidate takes moments about a point that is not the intersection of two unknown forces, producing equations containing several unknowns, or omits a force (often the reaction at a hinge or the weight of a uniform rod) from the moment sum.
Candidates take moments about the 'obvious' end of the rod out of habit rather than choosing the point that eliminates unknowns. They also forget that a uniform rod's weight acts at its midpoint and that a hinge reaction has two components unless its direction is known.
What to do instead
State the principle first: three non-parallel coplanar forces in equilibrium are concurrent. Take moments about the intersection of the two unknown forces so only one unknown appears. For a hinged rod, draw the reaction as an unknown force at an unknown angle, or, better, take moments about the hinge to eliminate it entirely.
In a ladder or rod against a wall problem, the candidate writes friction as μN at the wall as well as the floor, or assumes limiting friction at both contacts simultaneously when the question specifies slipping at only one end, and so over-constrains the system and gets the wrong angle or coefficient.
Candidates apply 'friction = μN' as a reflex at every contact instead of recognising it as a limiting-case condition that holds only where slipping is about to occur. They also confuse the normal reactions at the two contacts (the floor reaction balances weight; the wall reaction balances the floor friction).
What to do instead
At each contact write F ≤ μN and only set F = μN where the problem states slipping is imminent. Then count unknowns against equations: two force resolutions plus one moment equation give three equations, so at most three unknowns can be found. If you have more, you have wrongly assumed limiting friction somewhere.
The candidate treats centripetal force as an extra outward or inward force to be added to the force list, writing equations like T = mg + mv²/r regardless of where the particle is on the circle, or omits the radial component of weight in a vertical circle.
The phrase 'centripetal force' suggests a new force rather than a resultant. Candidates also fail to resolve weight along the radius, which changes sign between the top and bottom of a vertical circle, so a single memorised formula is applied everywhere.
What to do instead
Never put 'centripetal force' on a free-body diagram. Instead resolve all real forces along the inward radius and set their sum equal to mv²/r, and set the sum of tangential components equal to m·(dv/dt). Do this separately at the top and bottom of a vertical circle, where the weight component along the radius is +mg and −mg respectively.
In simple harmonic motion the candidate uses ω = 2πf and ω = √(k/m) or ω = √(g/l) interchangeably without checking units, or substitutes a displacement x into v = ωx instead of v = ω√(A² − x²), and reports the maximum speed at the endpoint rather than at the centre.
The symbol ω is met in several contexts (circular motion, SHM, waves) and candidates transplant the formula that comes to mind. The square-root structure of the velocity–displacement relation is also easily dropped under time pressure.
What to do instead
Anchor SHM on the defining equation a = −ω²x and derive v² = ω²(A² − x²) once from it. Then check the physical extremes: v must be zero at x = ±A and maximal, equal to ωA, at x = 0; acceleration must be zero at the centre and maximal at the endpoints. Confirm ω's units are rad s⁻¹ before substituting.
Applying Newton's law of gravitation, the candidate measures r from the surface of a planet rather than from its centre, or equates gravitational force to mg using the surface value of g at an altitude where g has changed.
Diagrams show the planet as a circle and the distance to the surface is visually salient, so r = radius + height is replaced by r = height. Candidates also treat g as a universal constant rather than a local value that falls off as 1/r².
What to do instead
Write r = R + h explicitly and substitute the numerical value before simplifying. When a problem involves altitude, use GM/r² with the full r, or g' = g(R/(R+h))², and reserve g = 9.8 m s⁻² for points near the surface. Check that the answer tends to the surface value as h → 0.
When differentiating or integrating, the candidate drops the chain rule (writing d/dx of sin kx as cos kx), forgets the constant of integration, or applies the power rule to a function of a function such as (3x + 1)⁵ without the inner derivative.
Under exam timing, candidates pattern-match to the nearest familiar derivative rather than identifying the composite structure. The arbitrary constant is treated as bookkeeping and omitted, which then makes a later 'find the particular solution' step impossible.
What to do instead
Before differentiating, name the inner function u(x) out loud in writing; then apply d/dx f(u) = f′(u)·u′. After every indefinite integral, write '+ C' as a compulsory step and use any given condition to fix it. Verify by differentiating your answer back to the integrand - a ten-second check that catches most slips.
In a related-rates or max/min problem, the candidate differentiates with respect to the wrong variable (treating dV/dt as dV/dr), or sets f′(x) = 0 and reports that x as the answer without checking whether it is a maximum, minimum, or neither, and without checking the endpoints of a closed interval.
The chain rule link between the rates is the conceptual step, and candidates skip writing it. For optimisation, the habit of 'stationary point = answer' ignores that the question asks for a maximum value or that the optimum may lie at an endpoint.
What to do instead
For related rates, write the chain-rule equation explicitly, e.g. dV/dt = (dV/dr)(dr/dt), before substituting any numbers. For optimisation, tabulate f′'s sign either side of each stationary point (or use f″) and also evaluate f at the interval's endpoints; then state the conclusion in the units the question asks for.
In probability, the candidate adds probabilities of non-mutually-exclusive events without subtracting the intersection, or multiplies P(A)P(B) for events that are not independent, and in statistics uses n instead of n − 1 in the sample variance or reads a z-table for the wrong tail.
The addition and multiplication rules are learned as bare formulas, and the conditions 'mutually exclusive' and 'independent' are not checked against the problem wording. In statistics, the distinction between population and sample formulas is glossed over, and tail direction is lost when the question is phrased as 'at least' or 'greater than'.
What to do instead
Before computing, write down whether the events can co-occur (if yes, subtract P(A∩B)) and whether one event's occurrence changes the other's probability (if yes, use P(A)P(B|A)). For variance, state whether the data are a sample (divide by n − 1) or a population (divide by n). For normal problems, sketch the curve, shade the required region, and convert to a standardised value before touching tables.