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O-Level Additional Mathematics · Secondary 3 and 4

Applications of Differentiation

Once you can differentiate, you can use the derivative to answer real questions about curves and changing quantities. This topic covers the equations of tangents and normals, increasing and decreasing functions, stationary points and their nature, maximum and minimum problems, and connected rates of change. The questions are often set in context, such as the volume of a container or the area of a field, so you need to build the expression as well as differentiate it. The second derivative helps you decide whether a stationary point is a maximum or a minimum.

What the syllabus expects

  • Find the equations of the tangent and the normal to a curve at a given point
  • Determine where a function is increasing or decreasing using the sign of dy/dx
  • Find stationary points and determine their nature using the first derivative test or the second derivative test
  • Solve practical problems on maximum and minimum values
  • Solve problems on connected rates of change using the chain rule

Tangents and normals

At the point where x = a, the gradient of the tangent is the value of dy/dx at x = a. The normal is the line perpendicular to the tangent, so if the tangent has gradient m, the normal has gradient -1/m. Use y - y₁ = m(x - x₁) to write each equation. For example, for y = x³ - 2x at the point (2, 4), dy/dx = 3x² - 2, which is 10 at x = 2. The tangent is y - 4 = 10(x - 2), which is y = 10x - 16. The normal has gradient -1/10, so y - 4 = -1/10(x - 2), which simplifies to x + 10y = 42. If the tangent is horizontal, the normal is vertical, with an equation of the form x = a.

Increasing, decreasing and stationary points

A function is increasing where dy/dx > 0 and decreasing where dy/dx < 0. A stationary point is where dy/dx = 0. To find its nature, use the second derivative: if d²y/dx² < 0 the point is a maximum, and if d²y/dx² > 0 it is a minimum. If d²y/dx² = 0, this test tells you nothing, so use the first derivative test: check the sign of dy/dx just before and just after the point. A change from positive to negative means a maximum, negative to positive means a minimum, and no change of sign means a point of inflexion. For y = x³ - 3x² - 9x + 5, dy/dx = 3x² - 6x - 9 = 3(x - 3)(x + 1), so the stationary points are at x = 3 and x = -1. Since d²y/dx² = 6x - 6, which is 12 at x = 3 and -12 at x = -1, the point (3, -22) is a minimum and (-1, 10) is a maximum. The function is decreasing for -1 < x < 3.

Maximum and minimum problems

Practical problems follow a pattern. Write the quantity to be maximised or minimised in terms of one variable, using the given constraint to remove any second variable. Differentiate, set the derivative equal to zero and solve. Then confirm the nature with the second derivative and answer the question that was asked, with units. For example, a farmer has 80 m of fencing to make a rectangular pen against a straight wall, fencing only three sides. If the two sides perpendicular to the wall are x m each, the side parallel to the wall is (80 - 2x) m, so the area is A = x(80 - 2x) = 80x - 2x². Then dA/dx = 80 - 4x = 0 gives x = 20, and d²A/dx² = -4 < 0, so this is a maximum. The largest area is 20 × 40 = 800 m².

Connected rates of change

When two quantities both change with time, the chain rule links their rates: dy/dt = dy/dx × dx/dt. Identify the rate you are given, the rate you want, and an equation linking the two variables. For example, the radius of a circular ripple increases at 3 cm/s. The area is A = πr², so dA/dr = 2πr. When r = 10 cm, dA/dt = dA/dr × dr/dt = 2π(10) × 3 = 60π cm²/s, which is about 188 cm²/s. Units help you check: (cm²/cm) × (cm/s) gives cm²/s. Substitute the particular value, such as r = 10, only after differentiating, never before.

Worked example

A closed cylinder has radius r cm, height h cm and volume 250π cm³. Show that its total surface area A cm² is given by A = 2πr² + 500π/r. Find the value of r for which A is stationary, show that this gives a minimum, and find the minimum surface area in terms of π.

  1. Volume: πr²h = 250π, so h = 250/r².
  2. Surface area: A = 2πr² + 2πrh = 2πr² + 2πr(250/r²) = 2πr² + 500π/r, as required.
  3. Differentiate: dA/dr = 4πr - 500π/r².
  4. For a stationary value, 4πr - 500π/r² = 0, so 4r³ = 500, r³ = 125 and r = 5.
  5. Second derivative: d²A/dr² = 4π + 1000π/r³. At r = 5 this is 4π + 8π = 12π > 0, so A is a minimum.
  6. Minimum A = 2π(25) + 500π/5 = 50π + 100π = 150π.

Answer: r = 5 cm, and the minimum surface area is 150π cm², about 471 cm² (the height is then 10 cm)

Common mistakes

  • Using the tangent gradient for the normal. The normal gradient is the negative reciprocal, so a tangent gradient of 10 gives a normal gradient of -1/10.
  • Stopping after solving dy/dx = 0. Questions usually also ask for the y-coordinate and the nature of the point, so substitute x back into y (not into dy/dx) and find d²y/dx².
  • Substituting a particular value such as r = 10 before differentiating in a rates of change problem. This turns a variable into a constant and gives a derivative of zero.
  • Leaving two variables in the expression to be maximised. Use the constraint, such as a fixed volume or length of fencing, to write the quantity in terms of one variable first.
  • Deciding that a point is a point of inflexion just because d²y/dx² = 0. The second derivative test is inconclusive then, so check the sign of dy/dx on either side.
  • Giving the value of x when the question asks for the maximum area or minimum cost. Substitute back to find the quantity asked for, with units.

Exam tips

  • In 'show that' maximum problems the expression is given, so even if you cannot show it, use the given expression to finish the rest of the question.
  • Write the chain rule in full for rates of change, such as dr/dt = dV/dt ÷ dV/dr, and put units on the answer. It makes the method clear even if a number slips.
  • To show that a function is increasing for all x, show that dy/dx is always positive, often by completing the square. For example, 3x² - 6x + 5 = 3(x - 1)² + 2, which is always at least 2.

Practice questions

1. Find the equation of the normal to the curve y = x² - 4x + 1 at the point where x = 3.

Show answer

x + 2y + 1 = 0. At x = 3, y = 9 - 12 + 1 = -2 and dy/dx = 2x - 4 = 2. The normal gradient is -1/2, so y + 2 = -1/2(x - 3), giving 2y + 4 = -x + 3, which is x + 2y + 1 = 0.

2. Find the coordinates of the stationary points of y = 2x³ - 3x² - 12x + 4 and determine their nature.

Show answer

Maximum at (-1, 11) and minimum at (2, -16). dy/dx = 6x² - 6x - 12 = 6(x - 2)(x + 1), which is zero at x = 2 and x = -1. Then y(2) = 16 - 12 - 24 + 4 = -16 and y(-1) = -2 - 3 + 12 + 4 = 11. d²y/dx² = 12x - 6 is 18 > 0 at x = 2 and -18 < 0 at x = -1.

3. A spherical balloon is inflated so that its volume increases at a constant rate of 100 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm. The volume of a sphere is V = 4/3 πr³.

Show answer

1/π cm/s, about 0.318 cm/s. dV/dr = 4πr², which is 100π when r = 5. Then dr/dt = dV/dt ÷ dV/dr = 100/(100π) = 1/π.

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