Applications of Calculus

Applications of calculus is the most heavily examined focus area in Mathematics Advanced: 18 marks on NESA's sample paper for the 2024 syllabus, and the old-syllabus equivalent averaged about 16 marks a year. The last few questions of the paper are often long optimisation or curve-sketching problems, so practise full setting-out.

• Differentiability: a function is differentiable at $x=a$ if the limit defining $f^{\prime }(a)$ exists (a non-vertical tangent). Differentiable implies continuous, but not the other way round.
• Turning points and inflections: first and second derivatives, the first and second derivative tests, concavity, points of inflection.
• Curve sketching: stationary points, inflections, intercepts, symmetry, asymptotes, end behaviour; sketching $y=f^{\prime }(x)$ and $y=f^{\prime \prime }(x)$ from the graph of $y=f(x)$.
• Optimisation: local versus global extrema, checking endpoints and discontinuities, modelling real problems.
• Rates of change: interpreting $\frac{dQ}{dt}$ and $\frac{d^{2}Q}{d{t}^2}$, integrating a rate to get a quantity, and exponential growth and decay with $\frac{dQ}{dt}=kQ$.
• Motion in a straight line: displacement, velocity and acceleration linked by differentiation and integration.

If this limit exists, $f$ is differentiable at $a$. A function that is differentiable at $a$ must be continuous there. The converse is false: a function can be continuous but not differentiable at a sharp corner (as in $y=|x|$ at $x=0$, where the left and right gradients differ) or where the tangent is vertical (as in $y=\sqrt[3]{x}$ at $x=0$). Jumps, holes and asymptotes are discontinuities, so the function is not differentiable there either.

A stationary point is where $f^{\prime }(x)=0$. Classify it with either test.

First derivative test. Check the sign of $f^{\prime }(x)$ just either side:

Concavity. $f^{\prime \prime }(x)$ is the gradient of $f^{\prime }(x)$. If $f^{\prime \prime }(x)>0$ the curve is concave up (gradient increasing); if $f^{\prime \prime }(x)<0$ it is concave down.

Second derivative test. At a stationary point, $f^{\prime \prime }(a)>0$ means a local minimum and $f^{\prime \prime }(a)<0$ means a local maximum. If $f^{\prime \prime }(a)=0$ the test fails, so go back to the first derivative test.

A point of inflection is where the concavity changes. Find where $f^{\prime \prime }(x)=0$, then check that the second derivative changes sign either side. $f^{\prime \prime }(a)=0$ alone is not enough: $y=x^4$ has $y^{\prime \prime }=0$ at $x=0$ but no inflection there.

Worked example 1

Sketch $f(x)=x^3+3x^2-9x-2$, showing stationary points and the point of inflection.

Stationary points.

so $x=1$ or $x=-3$. Then $f(1)=-7$ and $f(-3)=25$.

Nature. $f^{\prime \prime }(x)=6x+6$. At $x=1$: $f^{\prime \prime }(1)=12>0$, so $(1,-7)$ is a local minimum. At $x=-3$: $f^{\prime \prime }(-3)=-12<0$, so $(-3,25)$ is a local maximum.

Inflection. $f^{\prime \prime }(x)=0$ at $x=-1$, and $f^{\prime \prime }$ changes from negative to positive there (it is linear), so there is a point of inflection at $(-1,9)$.

Other features. The $y$-intercept is $(0,-2)$. As $x\to \infty$, $f(x)\to \infty$, and as $x\to -\infty$, $f(x)\to -\infty$ (positive cubic).

Plot these five points and join them with a smooth cubic: rising to the maximum, falling through the inflection to the minimum, then rising. The point of inflection is exactly halfway between the turning points, which is true for every cubic.

Sketching the derivative graph

Given only the graph of $y=f(x)$:

• Stationary points of $f$ become $x$-intercepts of $f^{\prime }$.
• Where $f$ is increasing, $f^{\prime }$ is above the axis; where decreasing, below.
• Points of inflection of $f$ become turning points of $f^{\prime }$.
• Concave up sections of $f$ are where $f^{\prime }$ is increasing.

A global maximum is the largest value of $f(x)$ over the whole domain; a local maximum is only the largest nearby. On a restricted domain, the global maximum or minimum can occur at a stationary point, an endpoint or a point where $f^{\prime }$ is undefined. Always compare all three.

Method for word problems:

1. Draw a diagram and define variables with units.
2. Write the quantity to optimise in terms of one variable, using the constraint to eliminate the other. Most questions do this part for you as a "show that".
3. State the domain allowed by the context.
4. Differentiate, solve $\frac{dQ}{dx}=0$, and justify the max or min (second derivative or table of signs).
5. Check endpoints if the domain is closed, then answer the question that was asked, with units.

Worked example 2

A rectangle has its base on the $x$-axis and its top two corners on the parabola $y=12-x^2$. Find the largest possible area.

Let the top right corner be $(x,12-x^2)$ with $x>0$. By symmetry the width is $2x$, so

The domain is $0<x<\sqrt{12}$ (the corner must be above the axis).

$\frac{d^{2}A}{d{x}^2}=-12x=-24<0$ at $x=2$, so this is a maximum. It is the only stationary point in the domain and $A\to 0$ at both ends, so it is the global maximum:

If $Q$ is a function of time $t$, then $\frac{dQ}{dt}$ is its rate of change and $\frac{d^{2}Q}{d{t}^2}$ tells you whether that rate is increasing or decreasing. Interpret signs in words: "the volume is decreasing, but at a decreasing rate" means $\frac{dV}{dt}<0$ and $\frac{d^{2}V}{d{t}^2}>0$.

If you are given the rate, integrate to get the quantity, then use an initial condition to find the constant. The total change from $t=a$ to $t=b$ is ${\int }_a^b\frac{dQ}{dt}dt$.

Exponential growth and decay

When the rate of change of a positive quantity is proportional to the quantity itself:

where $A$ is the initial value ($Q$ at $t=0$). If $k>0$ this is growth; if $k<0$, decay. You should be able to verify the solution by substitution: $\frac{d}{dt}(A{e}^{kt})=kA{e}^{kt}=kQ$. The graph of $Q=A{e}^{kt}$ has a horizontal asymptote at $Q=0$.

Worked example 3

A bacteria culture has 500 cells at $t=0$ and 1200 cells after 3 hours, growing according to $\frac{dN}{dt}=kN$.

(a) Find $k$. $N=500e^{kt}$ and $N(3)=1200$:

(b) When does the culture reach 5000 cells?

(c) How fast is it growing when $t=4$? First $N(4)=500e^{4k}\approx 1607$. Then

cells per hour. Keep $k$ unrounded in your calculator for each part.

For a particle moving along a line with displacement $x$ at time $t$:

Going the other way, integrate acceleration to get velocity and velocity to get displacement, using initial conditions for the constants.

Key language:

• At rest: $v=0$.
• At the origin: $x=0$.
• Moving in the positive direction: $v>0$.
• Speeding up: $v$ and $a$ have the same sign. Slowing down: opposite signs.
• Speed is $|v|$. Distance travelled counts every metre, so split the time interval wherever $v=0$.

Worked example 4

A particle starts at $x=2$ m with velocity $v=6t-3t^2$ m/s. Find its displacement as a function of $t$ and the distance it travels in the first 3 seconds.

Displacement. Integrate: $x=3t^2-t^3+C$. At $t=0$, $x=2$, so $C=2$:

When does it change direction? $v=3t(2-t)=0$ at $t=0$ and $t=2$, and $v$ changes sign at $t=2$.

Positions. $x(0)=2$, $x(2)=6$, $x(3)=2$.

Distance. It moves from 2 out to 6, then back to 2:

Its displacement after 3 seconds is 0 (back where it started), but the distance travelled is 8 m. The acceleration is $a=6-6t$, which is negative for $t>1$, so from $t=2$ onwards velocity and acceleration are both negative and the particle is speeding up.

• Calling a point an inflection just because $f^{\prime \prime }(a)=0$, without checking the sign change.
• Stopping at a local maximum when the question asks for the greatest value on an interval: check the endpoints.
• Not stating or using the domain in optimisation (for example, rejecting a negative length).
• Not justifying that a stationary point is a maximum or minimum. "Max" with no test loses a mark.
• Computing distance travelled as $x(b)-x(a)$ when the particle turns around in between.
• Rounding $k$ early in exponential problems, so later answers drift.
• Confusing "the rate is decreasing" ($Q^{\prime \prime }<0$) with "the quantity is decreasing" ($Q^{\prime }<0$).

• Find and determine the nature: solve $f^{\prime }(x)=0$, give both coordinates, and show a test (a table of signs or $f^{\prime \prime }$ with its sign stated).
• Sketch: label stationary points, inflections and intercepts with coordinates, show asymptotes as dashed lines, and use a sensible scale. A clean, well-labelled sketch is worth as much as the algebra.
• Show that in optimisation: start from the diagram or constraint and reach the given expression exactly. If you can't, use the given expression for later parts.
• Justify / explain: in rate and motion questions, refer to the sign of the relevant derivative in words, in context.
• Late-paper questions often combine topics (for example, a trig or log function in an optimisation). Fluency with MA12-DC derivatives saves time here.

• MA12-DC Differential calculus: every derivative used here.
• MA12-IC Integral calculus: integrating rates of change and velocity.
• MA11-DI Introduction to differentiation: first principles, increasing and decreasing functions, tangents.
• MA12-SS Sequences and series: $Q=A{e}^{kt}$ is the continuous version of geometric growth.
• MA12-RV Random variables: finding where a probability density function is largest uses these same tests.