Further Mathematics · Lesson 15

Applications of calculus: rates, maxima/minima, areas, volumes, kinematics

WAECNECO

What the syllabus asks

WAEC: applications of differentiation (gradients, equations of tangents and normals, rates of change, connected rates, small changes, stationary points and their nature, maximum and minimum problems, kinematics: velocity and acceleration); applications of integration (area under a curve, area between curves, volumes of solids of revolution, distance from velocity)

NECO: applications of differentiation (gradients, equations of tangents and normals, rates of change, connected rates, small changes, stationary points and their nature, maximum and minimum problems, kinematics: velocity and acceleration); applications of integration (area under a curve, area between curves, volumes of solids of revolution, distance from velocity)

What you will learn

Find the gradient of a curve and the equations of tangents and normals.
Locate stationary points and determine their nature.
Solve maximum and minimum problems from real situations.
Solve rates-of-change and connected-rates problems, and estimate small changes.
Find velocity and acceleration from displacement, and displacement from velocity.
Calculate areas under curves and between curves.
Calculate volumes of solids of revolution about the x-axis.

Key points

Gradient at a point = value of dy/dx there; tangent y − y₁ = m(x − x₁); normal gradient −1/m.
Stationary points: dy/dx = 0; find y from the original equation.
d²y/dx² < 0 maximum, > 0 minimum; if 0, check the sign of dy/dx either side.
Optimisation: one variable, differentiate, set to 0, confirm, answer the question with units.
Connected rates: dA/dt = (dA/dr)(dr/dt); differentiate before substituting.
Small changes: δy ≈ (dy/dx)δx; y = kxⁿ gives about n times the percentage change.
Kinematics: v = ds/dt, a = dv/dt; at rest v = 0; integrate back using initial conditions.
Distance needs splitting where v = 0; displacement does not.
Area = ∫ₐᵇ y dx (take modulus below the axis); between curves ∫(upper − lower) dx.
Volume about the x-axis: π∫ₐᵇ y² dx.

Start here: Gradients, tangents and normals

Calculus began as a way to answer one question: how steep is a curve at a particular point? The derivative dy/dx gives that steepness, the gradient, at any point.

The tangent at a point is the straight line that just touches the curve there and has the same gradient. The normal is the line at right angles to the tangent through the same point. Since perpendicular gradients multiply to −1, the normal's gradient is −1/m.

Worked example 15.1

Find the equations of the tangent and normal to y = x³ − 3x² + 4 at x = 1.

Point: y = 1 − 3 + 4 = 2, so (1, 2).
dy/dx = 3x² − 6x; at x = 1, m = 3 − 6 = −3. This is the tangent's gradient.
Tangent: y − 2 = −3(x − 1) → y = −3x + 5.
Normal gradient = −1/(−3) = 1/3. Normal: y − 2 = (1/3)(x − 1) → 3y − 6 = x − 1 → 3y = x + 5.

Tangent y = −3x + 5; normal 3y = x + 5

Worked example 15.2

Find the point on y = x² − 4x + 7 where the tangent is parallel to the line y = 2x − 1.

Parallel lines have equal gradients, so we need dy/dx = 2.
2x − 4 = 2 → x = 3; y = 9 − 12 + 7 = 4.

(3, 4)

Why does this matter outside the classroom? An engineer designing a road over a hill needs the gradient at each point to set safe speed limits, and a doctor reading a growth chart looks at how steeply weight is rising, not just the weight itself. In every case the derivative turns a curve into a number you can act on.

⚠️ Common mistake: using the derivative expression (3x² − 6x) as the gradient without substituting the x-value.

Try it yourself. Find the gradient of y = 2x² + 3x at x = −1, and the equation of the tangent there.

Show solution

dy/dx = 4x + 3 = −1; point (−1, −1); tangent y + 1 = −(x + 1) → y = −x − 2.

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Practice questions

1. Find the equation of the tangent to y = x² − 3x + 2 at x = 3.

Show worked answer

Point (3, 2); m = 2(3) − 3 = 3; y = 3x − 7.

2. Find and classify the stationary points of y = 2x³ − 3x² − 12x + 1.

Show worked answer

y′ = 6(x − 2)(x + 1); (−1, 8) max, (2, −19) min.

3. A rectangle has area 64 cm². Find the least possible perimeter.

Show worked answer

P = 2x + 128/x; P′ = 0 → x = 8; 32 cm.

4. The radius of a circle increases at 0.2 cm/s. Find the rate of increase of area when r = 5 cm.

Show worked answer

2π(5)(0.2) = 2π cm²/s.

5. s = t³ − 9t² + 24t. Find the acceleration when the particle is first at rest.

Show worked answer

v = 3(t − 2)(t − 4); first rest t = 2; a = 6t − 18 = −6 m/s².

Key words

Derivative: The gradient function dy/dx, giving the rate of change of y with respect to x.
Tangent: A straight line touching a curve at a point, with the same gradient as the curve there.
Normal: The line perpendicular to the tangent at the point of contact.
Stationary point: A point where dy/dx = 0: a maximum, minimum or point of inflexion.
Turning point: A stationary point where the curve changes from rising to falling or vice versa.
Second derivative: d²y/dx², the derivative of dy/dx; used to classify stationary points.
Rate of change: How fast one quantity changes with another, usually with time (d/dt).
Connected rates: Rates linked by the chain rule, e.g. dV/dt = (dV/dr)(dr/dt).
Displacement: Position measured from a fixed point, with direction; symbol s.
Velocity: Rate of change of displacement, v = ds/dt.
Acceleration: Rate of change of velocity, a = dv/dt.
Definite integral: ∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f.

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