Ch 2 · 14.5–14.9 Applications of Differentiation · O Level Additional Mathematics 4037
Topic 2 of 4 in Calculus — O Level Additional Mathematics 4037 — 78 lessons.
Introduction
With derivatives in hand you can answer real questions: what size of box holds the most? How fast is the water level rising? If a measurement is out by a little, how far out is the answer? This lesson applies differentiation to tangents, maxima and minima, rates of change and small changes. (Syllabus 14.5–14.9.)
Key ideas — Tangents and normals
At x = a, the tangent has gradient m = f'(a) and passes through (a, f(a)). The normal is perpendicular: gradient -1/m. Use y - y_1 = m(x - x_1).
🧑🏫 Teacher's explanation
Imagine you are driving a car along a winding mountain road. If you suddenly lost all traction and your car flew off the road in a perfectly straight line, that straight path is the tangent. It represents the instantaneous direction of your movement at that exact moment. The normal is simply the line that is perfectly perpendicular (at 90^°) to that path, like a flagpole standing straight up from the road's surface.
📖 Definitions
🔍 Worked example — Easy
Find the equation of the tangent to the curve y = x^2 + 3x at the point where x = 1.
🔍 Worked example — Standard
Find the equation of the normal to the curve y = √(x) at the point where x = 4.
🔍 Worked example — Exam-style
A curve has the equation y = 2x^3 - 9x^2 + 12x + 5. Find the coordinates of the stationary points and determine their nature using the second derivative test.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Stationary points and their nature
Stationary points have dy/dx = 0. To decide their nature, and justify it:
🧑🏫 Teacher's explanation
Imagine you are hiking on a hilly landscape. A 'stationary point' is any spot where the ground is perfectly flat for a split second—you aren't going up or down. This happens when the gradient, or dy/dx, is exactly 0.
📖 Definitions
🔍 Worked example — Easy
Find the nature of the stationary point for the curve y = x^2 - 4x + 5.
🔍 Worked example — Standard
A curve is defined by y = 3x^3 - 9x + 2. Find the coordinates of the stationary points and determine their nature.
🔍 Worked example — Exam-style
A rectangular area is to be enclosed by a fence of length 40 m. Find the maximum area possible.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Connected rates of change
Link rates with the chain rule:
🧑🏫 Teacher's explanation
Imagine you are blowing up a spherical balloon. As you blow air in, the volume V increases. But as the volume increases, the radius r also increases. Because the radius and volume are 'connected', they don't change independently; the speed at which the volume grows depends directly on how fast the radius is expanding at that exact moment.
📖 Definitions
🔍 Worked example — Easy
A spherical balloon is being inflated such that its radius r increases at a constant rate of 0.2 cm s^-1. Find the rate at which the volume V is increasing when r = 5 cm. (Volume of a sphere V = 4/3π r^3)
🔍 Worked example — Standard
The area A of a circle is related to its radius r by A = π r^2. If the area is increasing at a rate of 10 cm^2 s^-1, find the rate at which the radius is increasing when r = 4 cm.
🔍 Worked example — Exam-style
A cube has a volume V and side length x. The side length is increasing at a rate of 3 cm s^-1. Find the rate of change of the volume when the total surface area of the cube is 144 cm^2.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Small increments and approximations
For a small change delta x in x, the change in y is approximately
🧑🏫 Teacher's explanation
Imagine you are walking up a very steep hill. If you take one tiny, microscopic step forward, the change in your height is almost exactly the same as if you had walked along a straight ramp that matches the slope of the hill at that exact spot. In calculus, we use the gradient (the derivative) to turn a complex, curvy function into a simple, straight line for a very short distance.
📖 Definitions
🔍 Worked example — Easy
Given y = x^2 + 3x, find the approximate change in y when x increases from 5 to 5.02.
🔍 Worked example — Standard
Find the approximate value of √(16.04) using differentiation.
🔍 Worked example — Exam-style
A function is defined by f(x) = 1/x. Find the approximate change in f(x) when x increases from 2 to 2+p, where p is a very small positive value.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Practical maxima and minima
🧑🏫 Teacher's explanation
Imagine you are designing a rectangular garden against a long brick wall. You have a fixed amount of fencing, say 40 meters, and you want to capture as much grass as possible. If you make the garden very wide, it becomes very shallow; if you make it very deep, it becomes very narrow. There is a 'sweet spot' where the area is at its absolute largest. This 'sweet spot' is what we call a maximum.
📖 Definitions
🔍 Worked example — Easy
Find the maximum area of a rectangle with a perimeter of 20 cm.
🔍 Worked example — Standard
A closed cylindrical can is to have a volume of 100 cm^3. Find the radius r that minimizes the surface area S.
🔍 Worked example — Exam-style
A wire of length 12 cm is bent into the shape of a rectangle. Find the dimensions of the rectangle that maximize the area.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: stationary points
y = x^3 - 6x^2 + 9x + 1. dy/dx = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) = 0 at x = 1 and x = 3. d^2y/dx^2 = 6x - 12: at x = 1 it is -6 < 0, so (1, 5) is a maximum; at x = 3 it is 6 > 0, so (3, 1) is a minimum.
🔍 Worked example — Easy
Find the stationary points of the curve y = x^2 - 4x + 5.
🔍 Worked example — Standard
Determine the nature of the stationary points of y = x^3 - 3x + 2 using the second derivative test.
🔍 Worked example — Exam-style
A particle moves along a path defined by y = 2x^3 - 9x^2 + 12x + 5. Find the coordinates of the local maximum and local minimum.
⚠️ Common errors
Worked examples — Example 2: normal to an exponential curve
y = e^2x at x = 0: point (0, 1), gradient 2, so the normal has gradient -tfrac12: y = 1 - tfrac12x.
🔍 Worked example — Easy
Find the equation of the tangent to the curve y = e^3x at the point where x = 0.
🔍 Worked example — Standard
Find the equation of the normal to the curve y = e^2x at the point where x = 0.
🔍 Worked example — Exam-style
A curve is defined by y = ln(x) + 2. Find the equation of the normal to the curve at the point where x = 1.
⚠️ Common errors
Worked examples — Example 3: small increments
y = 2x^3 - 3x. dy/dx = 6x^2 - 3 = 21 at x = 2. If x increases from 2 to 2 + p, delta y ≈ 21p.
🔍 Worked example — Easy
Given y = x^2 + 5x, find the approximate change in y when x increases from 3 to 3.01.
🔍 Worked example — Standard
If y = 2x^3 - 3x, find the approximate change in y when x increases from 2 to 2 + p, where p is a small positive value.
🔍 Worked example — Exam-style
The volume of a sphere is V = 4/3π r^3. Find the approximate increase in volume when the radius r increases from 10 cm to 10.02 cm.
⚠️ Common errors
Worked examples — Example 4: connected rates
The volume of a cube increases at 12 cm³ s⁻¹. When the side is 2 cm, dV/dx = 3x^2 = 12, so dx/dt = 12/12 = 1 cm s⁻¹.
🔍 Worked example — Easy
A cube's volume V is given by V = x^3, where x is the side length. If the volume is increasing at a rate of 12 cm^3 s^-1, find the rate at which the side length x is increasing when x = 2 cm.
🔍 Worked example — Standard
The area of a circle is A = π r^2. If the area is increasing at a constant rate of 10 cm^2 s^-1, find the rate of change of the radius r when r = 5 cm.
🔍 Worked example — Exam-style
A spherical balloon is being inflated such that its volume V = 4/3π r^3 increases at a rate of 5 cm^3 s^-1. Find the rate at which the surface area S = 4π r^2 is increasing when the radius r = 3 cm.
⚠️ Common errors
Worked examples — Example 5: the largest box
Squares of side x cm are cut from the corners of a 12 cm square sheet, which is folded into an open box. V = x(12 - 2x)^2 = 4x^3 - 48x^2 + 144x. dV/dx = 12x^2 - 96x + 144 = 12(x - 2)(x - 6) = 0. x = 6 gives no box, so x = 2. d^2V/dx^2 = 24x - 96 = -48 < 0: maximum. V = 128 cm³.
🔍 Worked example — Easy
A rectangle has a perimeter of 20 cm. Find the dimensions that maximize the area.
🔍 Worked example — Standard
Squares of side x cm are cut from the corners of a 12 cm square sheet, which is folded into an open box. Find the value of x that maximizes the volume.
🔍 Worked example — Exam-style
A cylinder is inscribed in a sphere of radius R. Find the height h of the cylinder that maximizes its volume.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) point (2, 2), gradient 1: y = x. (2) dy/dx = (1 - x)e^-x = 0 at x = 1; first-derivative test: positive for x < 1, negative for x > 1, so a maximum at (1, tfrac1e ). (3) dA/dt = 2π r × 0.5 = 4π ≈ 12.6 cm² s⁻¹. (4) dy/dx = 1/2√(x) = 0.05; delta y ≈ 0.05. (5) sides x and 20 - x: A = 20x - x^2, A' = 20 - 2x = 0 at x = 10, A'' = -2 < 0 (maximum); the sides are 10 and 10, a square.