Further Applications of Calculus

Further applications of calculus is the biggest focus area in Extension 1: it was worth 17 marks on NESA's sample paper for the 2024 syllabus, and the matching content was worth about 11–18 marks in recent HSC papers. Expect a volume of revolution, a related rates question and a differential equation in Section II, plus slope field and growth questions in multiple choice.

• Multiplicity of zeros: a zero of multiplicity $m>1$ of $P(x)$ is a zero of multiplicity $m-1$ of $P^{\prime }(x)$; the shape of the graph at each zero; graphing factored polynomials.
• Further rates of change: the chain rule for composite rates, related rates problems, and $\frac{dQ}{dt}=k(Q-P)$ with solution $Q=P+A{e}^{kt}$ (Newton's law of cooling and limiting values).
• Areas and volumes: areas between curves; volumes of solids of revolution about the $x$- or $y$-axis, including the region between two curves.
• Differential equations: order, solutions and initial value problems, slope fields, and solving $\frac{dy}{dx}=f(x)$, $\frac{dy}{dx}=g(y)$, $\frac{dy}{dx}=f(x)g(y)$ and the logistic equation.

If $P(x)=(x-a{)}^{m}Q(x)$ with $Q(a)\ne 0$, the product rule gives

The bracket is $mQ(a)\ne 0$ at $x=a$, so $a$ is a zero of $P^{\prime }(x)$ of multiplicity $m-1$. So a repeated zero of $P$ is also a zero of $P^{\prime }$: to find one, solve $P^{\prime }(x)=0$ and test those values in $P(x)$.

Shape at the zero: multiplicity 1 crosses the axis; even multiplicity touches the axis at a turning point; odd multiplicity greater than 1 gives a horizontal point of inflection.

Worked example 1: finding a double zero

Show that $P(x)=x^3-3x+2$ has a double zero, and factorise it.

$P^{\prime }(x)=3x^2-3=0$ gives $x=\pm 1$. Test: $P(1)=1-3+2=0$ but $P(-1)=-1+3+2=4$. So $x=1$ is a zero of both $P$ and $P^{\prime }$, a double zero. Then $P(x)=(x-1{)}^2(x+c)$, and comparing constant terms gives $c=2$:

The graph touches the axis at $x=1$ and crosses at $x=-2$.

Related rates

When two quantities both change with time and are linked by an equation, the chain rule connects their rates:

Method: name the variables, write the equation linking them, differentiate (or use the chain rule), then substitute the values at the moment asked about. Substitute only after differentiating.

Worked example 2: a sliding ladder

A 5 m ladder leans against a vertical wall. The foot slides away from the wall at 0.4 m/s. How fast is the top sliding down when the foot is 3 m from the wall?

Let the foot be $x$ m from the wall and the top $y$ m up. Then $x^2+y^2=25$, so $y=\sqrt{25-x^2}$ and

When $x=3$, $y=4$:

The top slides down at 0.3 m/s (the negative sign means $y$ is decreasing).

Limited growth and decay

When a quantity $Q$ changes at a rate proportional to its difference from a fixed value $P$:

Verify by differentiating: $\frac{dQ}{dt}=kA{e}^{kt}=k(Q-P)$. If $k<0$, $Q\to P$ as $t\to \infty$, so the line $Q=P$ is a horizontal asymptote of the graph of $Q$ against $t$. In Newton's law of cooling, $P$ is the room temperature. If $A=0$ then $Q=P$ for all time.

Quick example. A dish at 185°C cools in a 25°C kitchen. With $T=25+160e^{kt}$ and $T=105$ after 10 minutes, $e^{10k}=\frac{80}{160}=\frac{1}{2}$, so $k=-\frac{\ln 2}{10}$. It reaches 45°C when $e^{kt}=\frac{20}{160}=\frac{1}{8}$, which is three halvings: $t=30$ minutes.

The area between curves is ${\int }_a^b[f(x)-g(x)]dx$ where $f\ge g$, as in Advanced, and in Extension 1 it can be set in a practical context. Rotating a region about an axis makes a solid whose cross-sections are discs (or washers).

For rotation about the $y$-axis, rewrite the curve as $x$ in terms of $y$ and use $y$-limits. For the region between two curves, subtract the inner radius squared from the outer radius squared: $\pi \int (R^2-r^2)$, never $\pi \int (R-r{)}^2$.

Worked example 3: rotating about the y-axis

The region between $y=x^2$ and $y=x$ for $0\le x\le 1$ is rotated about the $y$-axis. Find the volume.

The curves meet at $(0,0)$ and $(1,1)$. In terms of $y$: the parabola is $x=\sqrt{y}$ (outer, further from the $y$-axis) and the line is $x=y$ (inner).

The volume is $\frac{\pi }{6}$ cubic units.

A differential equation links a function and its derivatives. Its order is the highest derivative that appears. A solution is a function, and there are usually infinitely many (a family with a constant $C$). An initial condition picks out one: an initial value problem.

A slope field draws a short segment of gradient $\frac{dy}{dx}$ at many points. Solution curves follow the segments. Use it to match a field to an equation: if segments are the same along vertical lines, $\frac{dy}{dx}$ depends only on $x$; along horizontal lines, only on $y$; where segments are flat, $\frac{dy}{dx}=0$.

Methods

• $\frac{dy}{dx}=f(x)$: integrate directly.
• $\frac{dy}{dx}=g(y)$: flip to $\frac{dx}{dy}=\frac{1}{g(y)}$, integrate, then make $y$ the subject if possible.
• $\frac{dy}{dx}=f(x)g(y)$: separate to $\int \frac{dy}{g(y)}=\int f(x)dx$.
• $\frac{dQ}{dt}=kQ$ and $\frac{dQ}{dt}=k(Q-P)$ give the exponential solutions above.

For example, $\frac{dy}{dx}=(1+y^2)\cos x$ with $y(0)=0$ separates to $\int \frac{dy}{1+y^2}=\int \cos xdx$, so ${\tan }^{-1}y=\sin x+C$. The condition gives $C=0$, so $y=\tan (\sin x)$.

The logistic equation

A population with carrying capacity $C$ grows by

Separate and use partial fractions, $\frac{C}{P(C-P)}=\frac{1}{P}+\frac{1}{C-P}$:

Solving for $P$ gives the logistic function, where $A=\frac{C-P_0}{P_0}$:

Worked example 4: using a logistic model

A fish population follows $\frac{dP}{dt}=0.4P\left(1-\frac{P}{500}\right)$ with $P(0)=100$ ($t$ in years). Find $P$ as a function of $t$ and when the population reaches 400.

$A=\frac{500-100}{100}=4$, so $P=\frac{500}{1+4e^{-0.4t}}$. Set $P=400$:

The population reaches 400 after about 6.9 years, and approaches 500 in the long run.

• Substituting the "moment" values before differentiating in a related rates problem.
• Rotating about the $y$-axis but integrating with respect to $x$, or using $x$-limits.
• Writing $\pi \int (R-r{)}^2$ instead of $\pi \int (R^2-r^2)$ for the region between two curves.
• Leaving out $+C$ when solving a differential equation, or adding it only at the end after taking exponentials.
• Dropping the absolute value in $\ln |Q-P|$ and getting the wrong sign of $A$ when $Q<P$.
• Confusing order (highest derivative) with degree (highest power).
• Saying a double zero of $P$ is a zero of $P^{\prime \prime }$; it is a zero of $P^{\prime }$ only.

• Related rates: write "By the chain rule, $\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}$" as your first line. It earns a mark and organises the working.
• Show that a function satisfies a differential equation: differentiate it and rewrite the result in terms of the function. Don't solve from scratch unless asked.
• In volume questions, sketch the region and the axis of rotation, find the intersection points, then write the integral. Exact answers in terms of $\pi$ unless told otherwise.
• For a logistic or cooling model, interpret the constants in context: $P$ or $C$ is the long-term value and $k$ controls how fast it is approached.
• For slope fields, test a few points (where is $\frac{dy}{dx}=0$? positive? negative?) to eliminate options quickly.

• Polynomials (ME11-PO): multiplicity and the shape of polynomial graphs, now proved with calculus.
• Further calculus skills (ME12-CS): substitution, inverse trig integrals and ${\sin }^2$ integrals appear inside volume and differential equation questions.
• Advanced, Applications of Calculus (MA12-AC): exponential growth and decay and rates of change.
• Advanced, Integral Calculus (MA12-IC): areas between curves.
• Inverse trigonometric functions (ME12-IT): separable equations involving $1+y^2$ give ${\tan }^{-1}$.