Applications of Differentiation | A-Maths Study Guide

Secondary 4 Additional Mathematics · Study guide

Turn gradients into equations of lines, classify stationary points, optimise models and connect rates of change.

Concept → worked examples → practice25 min read · practice linked below

Translate the question into a derivative condition

Differentiation becomes useful when its value answers a geometric or physical question. A tangent uses a gradient at a point; a turning point uses a zero gradient; a rate question connects changes in different quantities.

What are you finding?
Question	Mathematical condition
Tangent gradient at $x=a$	Evaluate $f^{\prime}(a)$ .
Stationary point	Solve $f^{\prime}(x)=0$ , then find $y$ and classify.
Increasing or decreasing	Inspect the sign of $f^{\prime}$ on the domain.
Largest or smallest model value	Find feasible stationary values; compare any included endpoints.
Rate with respect to time	Use $dy/dt=(dy/dx)(dx/dt)$ .
Approximate small change	Use $\Delta y\approx f^{\prime}(x)\Delta x$ .

Tangents and normals

At $(a,f(a))$ , the tangent has gradient $m=f^{\prime}(a)$ and equation $y-f(a)=m(x-a)$ . A normal is perpendicular to the tangent. If $m\ne0$ is finite, its gradient is $-1/m$ , since the gradients of two perpendicular non-vertical lines multiply to $-1$ .

Example 1: find both lines and check the point

Find the tangent and normal to $y=x^2$ at $x=1$ . First find the point: $y=1$ , so it is $(1,1)$ . Then $y^{\prime}=2x$ , giving tangent gradient 2 and normal gradient $-1/2$ .

\text{Tangent: }y-1=2(x-1),\quad y=2x-1.

\text{Normal: }y-1=-\tfrac12(x-1),\quad y=-\tfrac12x+\tfrac32.

Both lines pass through $(1,1)$ , and their gradient product is $-1$ .

Example 2: a horizontal tangent has a vertical normal

Find the tangent and normal to $y=x^2+3$ at its stationary point. Since $y^{\prime}=2x=0$ , the point is $(0,3)$ .

The tangent is $y=3$ . Its normal is the vertical line $x=0$ . Do not calculate $-1/0$ : a vertical line has no finite gradient and cannot be written as $y=mx+c$ .

Increasing, decreasing and stationary points

Where $f^{\prime}(x)>0$ , the graph rises locally; where $f^{\prime}(x)<0$ , it falls locally. A zero derivative gives a horizontal tangent, but does not by itself prove a maximum or minimum. On either side of a stationary point, a sign change $+\to-$ gives a local maximum, and $-\to+$ gives a local minimum.

If the second derivative is continuous near a stationary point, $f^{\prime\prime}(a)>0$ proves a local minimum and $f^{\prime\prime}(a)<0$ proves a local maximum. If it is zero, the test is inconclusive. An inflection point involves a change in concavity, not just a zero second derivative.

Example 3: classify two stationary points and state the intervals

For $f(x)=x^3-3x$ , find and classify the stationary points.

f^{\prime}(x)=3(x-1)(x+1)=0\quad\Rightarrow\quad x=-1,1.

The points are $(-1,2)$ and $(1,-2)$ . Since $f^{\prime\prime}(x)=6x$ , the first is a local maximum and the second a local minimum.

The derivative is positive for $x<-1$ or $x>1$ , and negative for $-1<x<1$ . Thus the curve is strictly increasing on $( -\infty,-1]$ and $[1,\infty)$ , and strictly decreasing on $[-1,1]$ . The included stationary endpoints do not stop these intervals being monotonic. If the question asks where the gradient is strictly positive, exclude $-1$ and 1.

Example 4: zero second derivative: two different outcomes

Classify the stationary point at the origin for $f(x)=x^3$ and $g(x)=x^4$ .

For $f$ , $f^{\prime}=3x^2$ is positive on both sides of 0, while $f^{\prime\prime}=6x$ changes from negative to positive. The origin is a stationary point of inflection, and the function is strictly increasing across it.

For $g$ , $g^{\prime}=4x^3$ changes from negative to positive at 0, so the origin is a minimum. Both second derivatives equal zero at 0. Therefore $f^{\prime\prime}(0)=0$ alone cannot distinguish the outcomes.

Optimisation: build one feasible-variable model

Write the quantity to be optimised as a function of one variable, using the constraint to eliminate others. State the feasible domain before solving the derivative equation. Then prove the candidate is a maximum or minimum. On a closed interval, also compare endpoint values; for an open endpoint, a limiting value may never be attained.

Example 5: maximum area with a fixed fence length

A rectangular pen against a straight wall uses 20 m of fencing for its other three sides. Let each side perpendicular to the wall be $x$ m and the third side be $y$ m.

2x+y=20,\qquad A=xy=x(20-2x),\qquad 0<x<10.

A^{\prime}=20-4x=0\Rightarrow x=5,\qquad A^{\prime\prime}=-4<0.

Hence $y=10$ m and the maximum area is $50\text{ m}^2$ . The candidate is feasible. Also $A=50-2(x-5)^2\le50$ , proving it is the greatest area throughout the domain.

Example 6: minimum material for a closed cylinder

A closed cylinder has volume $54\pi\text{ cm}^3$ . Find the radius and height that minimise its surface area.

\pi r^2h=54\pi\Rightarrow h=\frac{54}{r^2},\quad r>0.

S=2\pi r^2+2\pi rh=2\pi r^2+\frac{108\pi}{r}.

S^{\prime}=4\pi r-\frac{108\pi}{r^2}=\frac{4\pi(r^3-27)}{r^2}.

The only positive stationary radius is $r=3$ . The derivative is negative for $0<r<3$ and positive for $r>3$ , so this is the global minimum. Then $h=6$ cm and $S=54\pi\text{ cm}^2$ . Both circular ends were included because the cylinder is closed.

Related rates: retain the time variable and the sign

When $y$ depends on $x$ , and $x$ depends on time, the chain rule gives $dy/dt=(dy/dx)(dx/dt)$ . Write the geometric relationship before substituting a particular instant. A decreasing quantity has a negative rate; the phrase “rate of decrease” normally asks for its positive magnitude.

Example 7: a changing sphere

Air enters a spherical balloon at $32\pi\text{ cm}^3/\text{s}$ . Find how fast its radius and surface area increase when $r=4$ cm.

V=\tfrac43\pi r^3,\qquad \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

At that instant, $32\pi=4\pi(4)^2(dr/dt)$ , so $dr/dt=1/2\text{ cm/s}$ . For $A=4\pi r^2$ ,

\frac{dA}{dt}=8\pi r\frac{dr}{dt}=8\pi(4)\left(\frac12\right)=16\pi\text{ cm}^2/\text{s}.

The radius rate is not constant just because the volume rate is constant: the factor $4\pi r^2$ changes as the balloon grows.

Use the constant shape ratio before differentiating. The water radius and depth both change with time.

Example 8: a falling level with net outflow

A conical tank has radius-to-height ratio $r/h=1/2$ . Water enters at $3\text{ cm}^3/\text{s}$ and leaves at $5\text{ cm}^3/\text{s}$ . Find the rate of change of water depth when $h=4$ cm.

Similar triangles give $r=h/2$ at every depth. Hence

V=\tfrac13\pi r^2h=\frac{\pi h^3}{12},\qquad \frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.

The net rate is $dV/dt=3-5=-2$ . Substituting the instant gives $-2=4\pi(dh/dt)$ , so $dh/dt=-1/(2\pi)\text{ cm/s}$ . The depth decreases at $1/(2\pi)\text{ cm/s}$ .

Small increments and extreme gradients

A smooth curve is close to its tangent near a point. Thus $f(x+\Delta x)\approx f(x)+f^{\prime}(x)\Delta x$ for a sufficiently small change within the domain. This is an approximation, not an equality. To optimise the gradient, treat $f^{\prime}(x)$ as the quantity to optimise.

Example 9: estimate a small change with units and an error check

Estimate $\sqrt{9.12}$ using $f(x)=\sqrt{x}$ near $x=9$ .

f(9)=3,\quad f^{\prime}(9)=\frac16,\quad \Delta x=0.12.

\sqrt{9.12}\approx3+\frac16(0.12)=3.02.

The square-root curve is concave down for $x>0$ , so the tangent estimate is slightly too large. Indeed $3.02^2=9.1204>9.12$ . The estimated increase in the square root is 0.02, not 0.12.

Example 10: minimum gradient is not minimum height

Find the minimum gradient of $y=x^3-6x^2+9x+2$ , and the point where it occurs.

y^{\prime}=3x^2-12x+9=3(x-2)^2-3.

The smallest gradient is $-3$ , occurring at $x=2$ . Substitution into the original curve gives $y=4$ , so the point is $(2,4)$ . Equivalently, $y^{\prime\prime}=6x-12$ changes from negative to positive at 2, so the gradient first decreases then increases. The curve itself is not stationary there because its gradient is $-3\ne0$ .

Explore when turning points merge

For $y=x^3-3kx$ , the derivative is $3(x^2-k)$ . Positive $k$ gives two stationary inputs; zero gives a stationary inflection; negative $k$ gives no stationary points.

Parameter k:1

Marked points are stationary. When k = 0, the two points meet at the origin.

k = 1. Local maximum (−1, 2); local minimum (1, −2).

Common mistakes and useful checks

Reporting only an input: a stationary point needs both coordinates and its nature.
Assuming a zero second derivative proves inflection: inspect sign changes or concavity.
Confusing positive gradient with strict increase: $x^3$ is strictly increasing even though its derivative is zero at 0.
Ignoring feasibility: a negative length or excluded endpoint cannot be an optimum.
Substituting too early: preserve the variable relationship before differentiating a rate equation.
Dropping signs or units: distinguish a negative rate of change from a positive rate of decrease.

Continue with topic practice

Use the Secondary 4 differentiation applications practice bank for exam questions and their worked solutions. Attempt a question on paper before comparing methods. This guide supports Secondary 4 learning and revision of the full Secondary Additional Mathematics course.