Maths Methods 1&2 · Alea · Test Monday

Plain English & method

CalculusThe branch of maths that studies rates of change — how fast one quantity changes as another one changes., in one sentence, is the study of rates of change — and this is the very first, simplest version of that idea, with no differentiation required yet.

The average rate of change between two points P and Q on a curve is just the gradientHow steep a line is — its slope. Calculated as "rise over run": how far up divided by how far across. of the straight line joining P and Q (a chord) — the same "rise over run" slope formula from linear graphs, applied to two points that happen to sit on a curve instead of a straight line. Every time you see "gradient" from here on, read it as "steepness" or "slope" — they all mean exactly the same thing.

This tells you the overall trend between the two points — how much the function changed on average per unit of x — but says nothing about what happened in between (the function could have gone up, down, sped up, or slowed down along the way; the average rate of change only sees the two endpoints).

Key ideas (from class)

Class notes with a graph example (points P and Q on a curve) and a function-rule example, both computing the chord gradient.

Worked examples

Worked examples from class: reading the average rate of change off a labelled graph, and calculating it directly from a function's rule between two given x-values.

Plain English & method

Unlike the average rate of change (16C), the instantaneous rate of changeThe rate of change at one single exact instant/point, rather than averaged over an interval. asks about the rate of change at one exact single point — this is the true starting idea behind everything in Chapter 17 (differentiation), just before any of the algebraic shortcuts get introduced.

A tangentA straight line that touches a curve at one point and has the same gradient as the curve there. is a straight line that touches the curve at one point and has the same gradient as the curve at that exact point. Two ways to find the instantaneous rate of change at a point:

• Draw the tangent by hand at the point of interest, then calculate its gradient over a convenient interval (rise over run using two points on the tangent line itself). Quick, but only ever an estimate — how accurate it is depends entirely on how carefully the tangent is drawn.
• Use a short chord instead — pick a second point on the curve very close to the point of interest, and calculate the average rate of change (16C's method) between them. A chordA straight line segment connecting two points that both sit on the curve. between two very close points has almost the same gradient as the tangent at that point, so this gives a good numerical estimate without needing to draw anything.

This second method — shrinking the chord until it's almost the tangent — is exactly the seed idea that grows into the formal definition of the derivative in Chapter 17 (see 17B's "Building it from scratch" section for the full version of this idea, using dx instead of a fixed small gap).

Key ideas (from class)

Class notes and graph examples covering tangent-drawing and the chord-approximation method, including a worked question comparing average vs instantaneous rate at different points on the same graph.

Worked examples

Worked examples: comparing the average rate of change over a wide interval against the instantaneous rate of change at specific points within it, and estimating a curve's gradient at a point using the gradient of a nearby short chord.

Plain English & method

This lesson is the first appearance of kinematicsThe study of motion — here, always a single particle moving along a straight line. in the course — the same topic that returns properly, with full calculus, in 18F. Here, the ideas are introduced using only averages and graph-reading, before derivatives are available as a tool.

• Displacement describes an object's position relative to a fixed origin O. Positive displacement means the object is to the right of (or ahead of) the origin; negative displacement means it's to the left of (or behind) the origin.
• Velocity is the rate of change of displacement with respect to time — literally 16C/16D's "rate of change" idea, just with x(t) (position) as the function instead of a generic f(x).
• Acceleration is the rate of change of velocity with respect to time — the same idea again, one layer up.

Average velocity over a time interval is calculated exactly like 16C's average rate of change: (change in position) ÷ (change in time), i.e. \(\dfrac{x(t_2) - x(t_1)}{t_2 - t_1}\).

Reading motion graphs: on a position-time graph, the gradient at any point is the velocity at that instant. On a velocity-time graph, the gradient at any point is the acceleration at that instant. This is exactly why 18F later differentiates position to get velocity, and differentiates velocity to get acceleration — this lesson is building the graphical intuition for that before the algebra is introduced.

Key ideas (from class)

Class notes covering displacement/velocity/acceleration definitions, worked position-time graph description questions, and average-velocity calculation examples.

Worked examples

Worked examples: calculating average velocity over different time intervals from a position function, describing a bicycle trip from its position-time graph and sketching the matching velocity-time graph, and reading instantaneous velocity plus its sign (positive/negative) directly off a velocity function or graph.

Plain English & method

Differentiation turns a function into a new function — its derivativeThe function giving the gradient (steepness) of the original curve at every point x. — that tells you the gradient of the original curve at every point. You almost never need first-principles limits to find it: there is a small toolbox of rules that let you differentiate a polynomial in seconds.

The power rule — the one rule that does 90% of the work: to differentiate \(ax^n\), bring the power down as a multiplier, then subtract 1 from the power.

Two shortcuts that follow immediately:

• The derivative of a sum is the sum of the derivatives — differentiate each term separately and add: d/dx[f(x) + g(x)] = f′(x) + g′(x).
• The derivative of a constant (a plain number with no x, like +7) is always 0 — a constant doesn't change, so it has zero gradient contribution.

Reading the formal notation

Every symbol in the formal definition of a derivative has one specific job — once you can name each part, the notation stops looking like a wall of unfamiliar signs:

Your class notes write this using dx and dy instead of h (see the "Building it from scratch" box below); VCE exams and most textbooks use h. They're the same idea with different letters — if you see either form, recognise it as "take a tiny step, measure the resulting change, and see what the ratio settles down to as that step shrinks to nothing."

Building it from scratch (the Calculus Made Easy way)

Key ideas (from class)

The key-ideas screenshot below is straight from your class notes — it states the power rule and the sum/constant rules exactly as you'll use them on the test.

Worked examples

Two worked examples from your Haileybury notes: differentiating a simple polynomial term-by-term, and a case with a constant term dropping to zero.

Plain English & method

The power rule from 17B still works exactly the same way when the power is negative — bring the power down, subtract 1. The only new thing is domainThe set of x-values a function is actually allowed to take.: a negative power means x is in a denominator (e.g. \(x^{-2} = 1/x^2\)), and you can never divide by zero, so x = 0 has to be excluded from the domain of both the function and its derivative.

Rewriting the domain rule concretely from your notes:

• If n is positive, the domain of \(x^n\) is all real numbers (no restriction).
• If n is negative, the domain of \(x^n\) is \(x \neq 0\) (x can be anything except zero).

Key ideas (from class)

Class notes cover differentiating negative-power terms and stating the correct domain alongside the answer.

Worked examples

Worked examples: finding a derivative with a negative power, evaluating a derivative at a given x-value, and finding a gradient at a specific point on the curve.

Plain English & method

This section asks you to go the other way: instead of computing f′(x) with algebra, you read the gradient straight off a picture of f(x) and sketch f′(x) from that.

The rule of thumb: walk along the curve of f(x) left to right and ask "is it going up, flat, or going down right now?"

• Where f(x) is increasing (sloping up), f′(x) is positive — plot it above the x-axis.
• Where f(x) is decreasing (sloping down), f′(x) is negative — plot it below the x-axis.
• Where f(x) has a stationary pointA point where the gradient is exactly zero — the tangent is horizontal. (a turning point, momentarily flat), f′(x) = 0 — plot it exactly on the x-axis.

The steeper the curve at a point, the further f′(x) sits from zero at that x-value; the flatter the curve, the closer f′(x) sits to zero.

Angle of a tangent: since gradient = tan(θ), where θ is the angle the tangent makes with the positive x-axis, you can also work backwards from an angle to a gradient value using m = tan(θ), then solve f′(x) = m.

Worked examples

Screenshot from class showing the sign-of-gradient table (increasing/decreasing/stationary) used to build up the shape of the derivative graph.

Worked examples: sketching a derivative graph from a given polynomial graph, and finding points where the tangent makes a given angle with the x-axis or is parallel to a given line.

Plain English & method

A limitThe value a function approaches as x gets closer and closer to some number, whether or not the function is actually defined there. asks: as x sneaks up on some value a, what value does f(x) sneak up on? Notation: \(\lim_{x \to a} f(x)\).

Method 1 — direct substitution. If the function is defined at x = a, just plug a in. Most limits are this easy.

Method 2 — factorise first. If substituting gives the undefined form 0/0, that is a signal — not an error — that x = a is a common root of the top and bottom, so a matching factor can be cancelled before you substitute again.

Other substitution outcomes:

• Substituting gives (nonzero number)/0 → the limit does not exist (the function shoots off to infinity there).
• Substituting gives 0/(nonzero number) → the limit is 0.

Continuity and one-sided limits. A function is continuousYou can draw it without lifting your pen — no jumps, holes, or breaks. at x = a only if the limit from the left and the limit from the right both exist and are equal to each other (and to f(a)). If the left-hand limit and right-hand limit disagree, the function has a discontinuity at that point and \(\lim_{x \to a} f(x)\) does not exist overall — even though each one-sided limit might exist separately.

Building it from scratch (the Calculus Made Easy way)

Worked examples

Class screenshots covering the substitution method and the 0/0-factorise method, plus the graph example showing left/right limits disagreeing at a discontinuity.

Worked examples: two limits solved by factorising a 0/0 form, and reading left-hand vs right-hand limits off a graph with a break at x = 1.

Plain English & method

A function is differentiableIts derivative exists at that point — the curve has one single well-defined gradient there. at a point if the derivative actually exists there, formally: the limit that defines f′(x) exists at that x-value. For that to happen, the graph of f(x) has to be smooth at that point — no breaks, no sharp corners/cusps, and no vertical tangent.

Practically, look for three trouble spots:

• A break or hole in the graph (a discontinuity, straight from 17F) — not differentiable there, because the function isn't even continuous.
• A sharp corner/cusp (like the point of |x| at x = 0) — the gradient coming in from the left doesn't match the gradient leaving to the right, so there's no single gradient value to call the derivative.
• A vertical tangent — the gradient would be infinite, which isn't a real number, so the derivative doesn't exist there either.

When sketching f′(x), mark any of these x-values with an open dot on the derivative graph — the derivative graph has a genuine gap there, even if f(x) itself is unbroken (as at a sharp corner).

Worked examples

Class screenshots showing the identifying-features table for differentiability, plus example graphs of piecewise functions with a corner point.

Worked examples: sketching f′(x) with an open dot at a non-differentiable point, and finding the domain of f′(x) for a piecewise-defined function.

Plain English & method

The derivative gives you the gradient of the curve at a point. Once you have a gradient and a point, finding the equation of a line is just \(y - y_1 = m(x - x_1)\).

The tangent is the straight line that just touches the curve at that point and has the same gradient as the curve there: \(m_{tangent} = f'(x_1)\).

The normal is the line through the same point that is perpendicularMeeting at a right angle (90°). to the tangent. Two perpendicular gradients always multiply to −1, so:

Method, every time:

1. Differentiate f(x) to get f′(x).
2. Substitute the given x-value into f′(x) to get the gradient at that point.
3. For the normal, flip and negate that gradient (m → −1/m).
4. Substitute the point and the gradient into \(y - y_1 = m(x - x_1)\) and simplify.

Key ideas (from class)

Class key-ideas screenshot plus two worked examples on this exact method.

Worked examples

Worked examples straight from Haileybury notes: the tangent equation at (1, −1), the normal equation at the same point, and finding where tangents to a curve are parallel to a given line.

Plain English & method

Two different, easily-confused ideas share the words "rate of change" in VCE Methods — know exactly which one a question is asking for.

Average rate of change (AROC) over an interval [a, b] is just the gradient of the straight line joining the two endpoints of the curve — how much the function changed overall, divided by how far x moved:

Instantaneous rate of change (IROC) at a single point x = a is the derivative evaluated there — the gradient of the tangent at that exact instant:

Sign tells you direction: f′(a) > 0 means the quantity is increasing at that instant; f′(a) < 0 means it's decreasing.

Key ideas (from class)

Class notes with the AROC formula and a worked balloon-volume problem (leaking balloon, volume decreasing over time).

Worked examples

Worked examples: computing AROC over two different intervals and the IROC at a specific time for the same function, plus a real-world rate problem (a leaking balloon's volume) including checking when the model stops making physical sense.

Plain English & method

A stationary pointA point where the tangent to the curve is exactly horizontal — the gradient is momentarily zero. is any point on the curve where the gradient is zero — the tangent there is parallel to the x-axis (perfectly flat).

Method:

1. Differentiate f(x) to get f′(x).
2. Set f′(x) = 0 and solve for x — this may need factorising, the quadratic formula, or CAS.
3. Substitute each x-value back into the original f(x) (not f′(x)!) to get the matching y-coordinate.

A cubic or quartic can have several stationary points — solving f′(x) = 0 finds all of them at once, since it's just solving an equation.

Building it from scratch (the Calculus Made Easy way)

Key ideas (from class)

Class notes and a labelled graph showing three stationary points (A, B, C) on one curve.

Worked examples

Worked examples: finding stationary point coordinates for two different polynomials, and a reverse problem — given that a curve has a stationary point at a known x-value, find an unknown coefficient in the function's rule.

Plain English & method

Finding a stationary point (18C) only tells you where the gradient is zero — it doesn't say what kind of point it is. There are three types:

• Local maximumA stationary point that is higher than the points immediately around it — a "peak". — gradient goes from positive to negative as x increases through the point (curve rises, flattens, falls).
• Local minimumA stationary point that is lower than the points immediately around it — a "trough". — gradient goes from negative to positive (curve falls, flattens, rises).
• Stationary point of inflectionA flat "shelf" in the curve where it keeps going in the same overall direction on both sides. — gradient has the same sign on both sides (positive→positive or negative→negative) — the curve flattens for an instant but keeps heading the same way.

Method — the sign diagram:

1. Find the stationary point(s) as in 18C by solving f′(x) = 0.
2. Pick a test x-value just to the left of the stationary point and one just to the right (not equal to the stationary x-value itself).
3. Substitute each test value into f′(x) and record only its sign (+ or −), not the number.
4. Read the pattern left→right against the three bullet points above to name the point.

Building it from scratch (the Calculus Made Easy way)

Key ideas (from class)

Class notes and worked examples building sign diagrams, plus a full sketch (intercepts + stationary points) example.

Worked examples

Worked examples: classifying every stationary point of a given function using a sign diagram, sketching a full graph with y-intercepts, x-intercepts and stationary points all marked, and finding the interval(s) where a function is increasing.

Plain English & method

This is where calculus starts solving real problems: "what dimensions give the biggest volume", "what's the cheapest way to build this". The calculus (18C + 18D) is the easy part — the hard part is turning the words into a function of one variable.

Standard method:

1. Draw a diagram and label the changing quantities with variables.
2. Write down the quantity you want to optimise (volume, area, cost…) as a function — it will usually start with two variables, connected by a constraint the question gives you (e.g. a fixed amount of material/area).
3. Use the constraint to eliminate one variable, leaving a single-variable function, e.g. V(x).
4. Work out the valid domain for x from the physical setup — lengths can't be negative, and there's often an upper limit too.
5. Differentiate, set the derivative to 0, solve, and check it's a maximum or minimum with a sign diagram (18D).
6. Answer the actual question asked — often "what is the maximum value" (substitute back into the original function), not just the x-value that produces it.

Endpoint maxima/minima — the twist to watch for. Calculus only finds local turning points. If the domain is restricted (e.g. 0 ≤ x ≤ 0.3), the true absolute maximumThe single highest value the function actually reaches anywhere across its whole domain — not just at a local turning point. or absolute minimumThe single lowest value the function actually reaches anywhere across its whole domain. might occur at the edge of the domain instead of at a stationary point. Always check the function's value at both endpoints of the domain and compare them against any stationary points found inside it.

Key ideas (from class)

Class notes with a canvas-shelter volume problem and a folded-box volume problem, including the endpoint-restriction twist on the box problem.

Worked examples

Two full worked applications from your notes: a canvas shelter with 24 m² of material (maximise enclosed volume), and a 2 m square sheet of metal folded into an open box (maximise volume, then re-solve with a height restriction of 0.3 m forcing an endpoint check).

Plain English & method

KinematicsThe study of motion — here, always a single particle moving along a straight line. in this course means one particle moving along a straight line, and differentiation is the tool that connects position, velocity, and acceleration.

Two quantities that sound similar but aren't the same:

• DisplacementPosition relative to a fixed origin O — it's a vector, so it has a sign/direction. is a vector (has direction) — it can be negative, meaning "on the other side of the origin".
• Distance is a scalar (no direction) — it's always positive, and it's the total length actually travelled, which is not the same as the net displacement if the particle changes direction along the way.

Reading the sign of velocity and acceleration:

• v(t) > 0 → moving in the positive direction; v(t) < 0 → moving in the negative direction; v(t) = 0 → instantaneously at rest.
• a(t) > 0 → speeding up in the positive sense; a(t) < 0 → slowing down (or speeding up in the negative sense); a(t) = 0 → travelling at constant velocity.

Key ideas (from class)

Class notes covering the displacement/velocity/acceleration definitions, the displacement-time and velocity-time graph gradients, and worked examples.

Worked examples

Multi-part worked examples from your notes on particles moving in a straight line: finding when velocity/acceleration is zero and what the other quantity equals at that instant, finding total distance travelled, and working out for how long the velocity stays negative.

Plain English & method

Note: the Canvas page for this lesson on your school's system is broken — it's an accidental duplicate of the 18F kinematics page (same images, same content), so there was no usable class material to pull from here. The explanation below is written from scratch to cover the standard VCE Methods topic this lesson title refers to, so this gap doesn't leave you unprepared if it does come up.

A "family of functions" just means a group of related graphs built from one base function by transforming it — shifting it, stretching it, or flipping it. The question this lesson answers: if you know how f(x) was transformed into a new function, what happens to its derivative?

The key fact: transformations applied vertically (up/down shifts, vertical stretches) and transformations applied horizontally (left/right shifts, horizontal stretches) affect the derivative differently.

• Vertical translation — g(x) = f(x) + c: adding a constant shifts the whole graph up or down but doesn't change its steepness anywhere, so g′(x) = f′(x) exactly, unchanged. (This is the same fact as "the derivative of a constant is 0" from 17B, just viewed graphically.)
• Vertical dilation — g(x) = k·f(x): stretching the graph vertically by factor k also stretches every gradient by the same factor: g′(x) = k·f′(x).
• Horizontal translation — g(x) = f(x − h): shifting the graph sideways by h shifts its derivative graph sideways by the same amount: g′(x) = f′(x − h).
• Horizontal dilation — g(x) = f(ax): scaling x by a factor also scales the derivative by that same factor (this one follows from the chain rule in 20A: differentiating f(ax) with u = ax gives g′(x) = a·f′(ax)).
• Reflection in the x-axis — g(x) = −f(x): flips the sign of the gradient everywhere: g′(x) = −f′(x).

Practically, exam questions in this style usually show you a graph of f(x) (or describe a transformation in words) and ask you to identify or sketch the matching transformation of f′(x) — which is really just 17D's "read the gradient off the graph" skill, applied to a shifted or scaled picture instead of the original.

Worked examples

No usable key-ideas image exists for this lesson (see note above).

No class worked examples exist for this lesson (see note above) — see the original worked example below instead.

Plain English & method

Some functions, like \(y = (3x^2 + 1)^{-1}\) or \(y = (x^3 + 1)^3\), are one function nested inside another — a "function of a function", called a composite functionA function built by feeding the output of one function into another, e.g. \(y = (x^3+1)^3\) is "cube it" applied to "\(x^3+1\)".. Expanding a bracket like \((x^3+1)^3\) by hand to differentiate term-by-term is slow and error-prone. The chain rule differentiates composite functions directly, without expanding anything.

The method: substitute an inner function as u.

1. Spot the "inside" expression (what's inside the brackets, or under a root, or in a denominator) and call it u.
2. Rewrite y purely in terms of u — this is now a simple power of u.
3. Differentiate y with respect to u (dy/du), and separately differentiate u with respect to x (du/dx).
4. Multiply the two derivatives together — that product is dy/dx.

Worked example, straight from the class video (20A): differentiate \(y = (x^3 + 1)^3\).

• Let \(u = x^3 + 1\), so \(y = u^3\).
• \(\dfrac{dy}{du} = 3u^2\) and \(\dfrac{du}{dx} = 3x^2\).
• Chain rule: \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx} = 3u^2 \times 3x^2 = 9x^2(x^3+1)^2\).

Second worked example, also from the class video: find the gradient of \(y = \dfrac{16}{3x^2+1}\) at the point \((1, 4)\).

• Rewrite as a negative power first: \(y = 16(3x^2+1)^{-1}\).
• Let \(u = 3x^2+1\), so \(y = 16u^{-1}\).
• \(\dfrac{dy}{du} = -16u^{-2}\) and \(\dfrac{du}{dx} = 6x\).
• \(\dfrac{dy}{dx} = -16u^{-2} \times 6x = -96x(3x^2+1)^{-2}\).
• Substitute \(x=1\) (so \(u = 3(1)^2+1 = 4\)): \(\dfrac{dy}{dx} = -96(1)(4)^{-2} = -\dfrac{96}{16} = -6\). The gradient at \((1,4)\) is \(-6\).

Building it from scratch (the Calculus Made Easy way)

Worked examples

No textbook key-ideas screenshot was posted for this lesson on Canvas (it's video-only) — the formula and both worked examples above were reconstructed from the class lecture video.

The two worked examples above are directly from the 20A lecture video, reconstructed frame-by-frame.

Plain English & method

A rational powerA power written as a fraction, e.g. x^(1/2), which is the same thing as a root: x^(1/2) = √x. is just a fraction sitting in the exponent — and the good news is the exact same power rule from 17B/17C still applies with no changes: bring the power down, subtract 1 from it. The only extra care needed is fraction arithmetic when subtracting 1 from a fractional power.

Worked example, straight from the class video (20B): differentiate \(f(x) = 4x^{2/3}\).

• Bring down the power: \(4 \times \dfrac{2}{3} = \dfrac{8}{3}\).
• Subtract 1 from the power: \(\dfrac{2}{3} - 1 = \dfrac{2}{3} - \dfrac{3}{3} = -\dfrac{1}{3}\).
• \(f'(x) = \dfrac{8}{3}x^{-1/3}\), which can also be written as \(\dfrac{8}{3x^{1/3}}\) or \(\dfrac{8}{3\sqrt[3]{x}}\).

Second worked example, also from the class video: differentiate \(f(x) = x^{1/5}\).

• Bring down the power: \(1 \times \dfrac{1}{5} = \dfrac{1}{5}\).
• Subtract 1 from the power: \(\dfrac{1}{5} - 1 = \dfrac{1}{5} - \dfrac{5}{5} = -\dfrac{4}{5}\).
• \(f'(x) = \dfrac{1}{5}x^{-4/5}\).

When the inside isn't just x — e.g. differentiating something like \((x^2+1)^{1/2}\) (which is \(\sqrt{x^2+1}\)) — the power is fractional and there's a composite function, so 20A's chain rule and 20B's fractional power rule combine: treat it exactly like a chain-rule problem (\(u = x^2+1\)), just with a fractional power on u.

Worked examples

No textbook key-ideas screenshot was posted for this lesson on Canvas either (also video-only) — the rule and both worked examples above were reconstructed from the class lecture video.

Both worked examples above are directly from the 20B lecture video, reconstructed frame-by-frame.