Calculus and its applications
Calculus questions apply differentiation and integration to rates of change, maxima and minima, areas and kinematics with variable acceleration. The emphasis is on setting up the model correctly before differentiating or integrating.
Differentiation
Use the product, quotient and chain rules as needed. Stationary points satisfy dy/dx = 0; classify them with the second derivative or a sign change. For related rates, differentiate a relation with respect to time, for example dV/dt = (dV/dr)(dr/dt).
Integration
A definite integral gives net area; displacement is ∫v dt and velocity is ∫a dt. With variable acceleration, a = dv/dt = v dv/ds. Always add a constant for an indefinite integral and use initial conditions to fix it.
Kinematics with calculus
Differentiate displacement to get velocity and velocity to get acceleration. Integrate acceleration to get velocity, then integrate again for displacement. Turning points occur where v = 0; distance travelled may require splitting the time interval when velocity changes sign.
Worked example
A particle has v = 3t² − 8t + 5 m/s and s = 0 when t = 0. Find s at t = 3 and when a = 0. Integrate: s = t³ − 4t² + 5t + C; s(0) = 0 gives C = 0. At t = 3, s = 27 − 36 + 15 = 6 m. Differentiate: a = 6t − 8; a = 0 when t = 4/3 s.
Traps
Omitting the constant of integration; treating displacement as distance when v changes sign; forgetting the chain rule in related rates; checking dy/dx = 0 without classifying the point; mixing units of time.