Game plan 7 · Unit 4

Rates of change and differential equations

Classify the form first. The form decides whether you integrate, flip or separate.

Reach for it when

You see $\frac{dy}{dx}$, “the rate of change of”, related quantities changing together, an inverse trig function to differentiate, a slope field, “verify that … is a solution”, or a growth, decay, cooling or logistic context.

Part 1

Related rates

Game plan 7.1

Related rates

1Write the chain: the rate you want equals the rate you have times a link, such as $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}$.
2The link comes from geometry. Volume and surface area formulas for cones, pyramids and spheres are on the sheet under Measurement. Write the formula in one variable before you differentiate, using similar triangles for cones.
3Differentiate the link, put the chain together, then substitute the values at the instant in the question. The numbers go in last.

Habit

In related rates, substitute the values at the instant only after all the differentiation is done.

In the Topic Guides

Unit 4 Topic 3.1 · Related rates

Part 2

Implicit differentiation

Game plan 7.2

Implicit differentiation

1Differentiate both sides term by term with respect to $x$. Every $y$-term picks up a $\frac{dy}{dx}$ through the chain rule, for example $\frac{d}{dx}\left(y^2\right)=2y\frac{dy}{dx}$. Products of $x$ and $y$ need the product rule.
2Collect the $\frac{dy}{dx}$ terms on one side, factorise, and solve.
3For the gradient at a point, substitute both coordinates.

In the Topic Guides

Unit 4 Topic 3.1 · Implicit differentiation

Part 3

Separable differential equations

The form	What to do
$\dfrac{dy}{dx}=f(x)$	Integrate directly
$\dfrac{dy}{dx}=g(y)$	Flip it: $\dfrac{dx}{dy}=\dfrac{1}{g(y)}$, then integrate with respect to $y$
$\dfrac{dy}{dx}=f(x)\,g(y)$	Separate: $\displaystyle\int\frac{1}{g(y)}\,dy=\int f(x)\,dx$

Game plan 7.3

Solving a separable differential equation

1Classify the form, using the table above.
2Integrate both sides, with one constant.
3Apply the initial or boundary condition to find $c$, then solve for $y$ explicitly where you can.
4State any domain restrictions the context forces.

Watch for

Use one constant of integration, and apply the condition before you solve for $y$. Using two constants, or applying the condition too late, both lose marks.

In the Topic Guides

Unit 4 Topic 3.2 · Solving differential equations

Part 4

Models you’ll be given

Game plan 7.4

Logistic, cooling and decay models

1The question gives you the differential equation. The syllabus guarantees it, so never memorise these equations.
2Solve by separating. The logistic equation separates into partial fractions (Game plan 6.1, check 3).
3Interpret. For the logistic model, give the carrying capacity (the horizontal asymptote) and the S-shaped approach. For Newton’s law of cooling, the temperature approaches the surrounding temperature. For radioactive decay, find the half-life by setting the amount to $\frac12$ of the initial value and solving with logs.
4“Show that … can never exceed”: finish with an inequality, not an equals sign. Point to the exponential term and say why it is always positive, for example $e^{-kt}\gt0\Rightarrow y\lt L$.

In the Topic Guides

Unit 4 Topic 3.2 · Models you’ll be given

Part 5

Slope fields

Game plan 7.5

Working with slope fields

1Sketching a field from an equation: substitute each point’s coordinates into $\frac{dy}{dx}$ and draw a short segment at that gradient. The segments show direction only, and their length means nothing.
2Sketching a solution through a point: start at the point and draw one curve that follows the gradients across the field. Stay parallel to the nearby segments, and never cross them.
3Interpreting or matching a field: work through the four quadrants in turn, then the $x$-axis, then the $y$-axis. In each, ask what the equation makes the gradient do. Look for where it is zero, undefined, positive and negative.

$\frac{dy}{dx}=\frac{4-y}{2}$. There is no $x$ in the equation, so the gradient never changes along a horizontal line. It is zero along $y=4$, positive below and negative above. The solution through $(-4,-2)$ follows the segments towards that line.

$\frac{dy}{dx}=\frac yx$. The gradient is zero along the $x$-axis and undefined along the $y$-axis. It is positive in quadrants 1 and 3 and negative in 2 and 4, and the quadrant check settles it.

Matching a slope field to its equation comes up often in multiple choice. Test the four quadrants and the two axes against each option. The first disagreement rules it out.

In the Topic Guides

Formulas

Formula panel

Each formula is labelled with how to treat it in the exam.

The chain, product and quotient rules, all the standard derivatives, and the measurement formulas

On the sheet

The derivatives of $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$. The $\cos^{-1}$ one carries a negative sign, so read it off rather than assuming it matches.

On the sheet

The flip: $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}$

Memorise

$\dfrac{dy}{dx}=ky\Rightarrow y=Ae^{kx}$. Separating takes four lines, and knowing where you’re heading makes it easy.

Derive

The logistic model $\frac{dy}{dx}=y(a-by)$, Newton’s law of cooling and radioactive decay. Solve them, don’t memorise them. Spotting the logistic form early tells you partial fractions are coming.

Supplied

QCAA language

What the question says, and what it means

Question type	What you’ll see	What it means
Related rates	Two rates and a geometric object	Game plan 7.1. Write the chain first
Implicit	An equation mixing $x$ and $y$, “gradient of the curve”	Game plan 7.2
Solving a differential equation	$\frac{dy}{dx}=\dots$, “general solution”, “particular solution”	Game plan 7.3. Classify, then separate
Model	“Logistic”, “Newton’s law of cooling”, a given equation in context	Game plan 7.4
Slope field	A grid of short segments	Game plan 7.5
Verify	“Verify” or “show that $y=\dots$ satisfies”	Differentiate the given function and substitute into both sides. Never solve it.

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