Stationary points
At a stationary point the gradient is zero: dy/dx = 0. To classify it, use the second derivative: if d²y/dx² > 0 it is a minimum; if d²y/dx² < 0 it is a maximum; if it is zero, check the gradient on either side (it may be a point of inflexion).
Find and classify the stationary points of y = x³ − 3x.
- Differentiate and set to zero.
- Second derivative.
- At x = 1 it is 6 > 0; at x = −1 it is −6 < 0.
Answer: Minimum at (1, −2) and maximum at (−1, 2).
Optimisation: an open box
A box with no lid is made from a square sheet of side 12 cm by cutting a square of side x from each corner and folding up the sides. The volume is V = x(12 − 2x)². Then dV/dx = 12(x − 2)(x − 6) = 0 gives x = 2 (the maximum, since x = 6 gives no box). The largest volume is 2 × 8² = 128 cm³.
Choose the size of the corner squares cut out of a 12 cm sheet.
Rates of change
If a quantity y depends on time t, then dy/dt is its rate of change. With related rates, use the chain rule: dV/dt = dV/dr × dr/dt. For a sphere V = (4/3)πr³, so dV/dr = 4πr².
Small changes
Series expansions
Maclaurin series expresses a function as a power series. The standard ones are: eˣ = 1 + x + x²/2! + x³/3! + …; sin x = x − x³/3! + x⁵/5! − …; cos x = 1 − x²/2! + x⁴/4! − …; ln(1 + x) = x − x²/2 + x³/3 − … for −1 < x ≤ 1.
Key terms
- Stationary point
- A point where dy/dx = 0.
- Second derivative
- The derivative of the derivative.
- Optimisation
- Finding a maximum or minimum.
- Maclaurin series
- A power series expansion about x = 0.
Summary
- Stationary points satisfy dy/dx = 0; the second derivative classifies them.
- Optimisation problems are solved by differentiating the quantity.
- δy ≈ (dy/dx)δx for small changes.
- Maclaurin series give polynomial approximations.