Differential Equations with Separable Variables, and Setting Them Up from Rates of Change

~65 min · WMA14 · 6.4

WMA14 · 6.4 · 65 min

Every mark this topic loses tends to go before the calculus even starts. Separating variables, integrating, and applying a boundary condition is genuinely routine once the equation is correctly set up — but a single real WMA14 examiner report, on exactly the kind of rates-of-change tank question this lesson is built around, names four places where it actually goes wrong: the wrong volume formula for the shape in the question, half of a two-term rate quietly dropped, a constant integrated as if it were a fraction of the variable, and a final answer reported in the wrong unit. None of those four is a calculus mistake, and this lesson spends as much time on the setup as on the solving.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What a differential equation is, and what "separable" means

A differential equation is simply an equation that relates a function to its own derivative. "First order" means only the first derivative appears — $\frac{dy}{dx}$ , never $\frac{d^2y}{dx^2}$ — which is exactly what spec 6.4 names it: "Analytical solution of simple first order differential equations with separable variables." P4 restricts to a specific, checkable kind of first-order equation: one where the right-hand side, once written as $\frac{dy}{dx} = (\text{something})$ , splits into a function of x alone multiplied by a function of y alone — $\frac{dy}{dx} = f(x)g(y)$ . In $\frac{dy}{dx} = 6x^2 / y^3$ , that split is $f(x) = 6x^2$ and $g(y) = 1/y^3$ : everything on the right is either a pure x-expression or a pure y-expression, multiplied together, never added or entangled.

Not every first-order equation is separable — $\frac{dy}{dx} = x + y$ is first order but cannot be split into a product of an x-only factor and a y-only factor, no matter how it's rearranged. Solving an equation like that needs a genuinely different technique. This course makes no claim about whether or where the wider Pearson syllabus revisits that technique — WMA14's own spec content list (verified against the spec PDF itself) has exactly one entry for solving a differential equation analytically, and it is scoped to the separable case only. (Spec 5.2, covered later in this lesson, is a second WMA14 entry that also touches differential equations — but it is about forming one from a rates-of-change context, not solving it, so it does not weaken this point.)

Spec 6.4's guidance is explicit and short: "General and particular solutions will be required." A general solution is a whole family of curves — one member for every possible value of the arbitrary constant C that appears when you integrate. A particular solution is the ONE member of that family picked out by a boundary condition: a specific pair of values, one for x and one for y, that the curve is told to pass through. Both are genuinely examined, separately, and stopping at the general solution when a boundary condition has been given in the question answers only half of what was asked.

Because P4 is cumulative — the spec's own prerequisites note states plainly that "a knowledge of the specifications for P1, P2 and P3, their prerequisites and associated formulae, is assumed and may be tested" — the actual INTEGRATION inside a separable equation (an $\int y^{-3}\,dy$ here, an $\int \frac{1}{4x+5}\,dx$ there) is assumed fluent from P1–P3 already. The genuinely new content in 6.4 is the act of separating and the discipline of applying a boundary condition correctly, not the integration technique itself.

Mechanism

Separating variables — why moving dy and dx around is legal, and why only one constant survives

\frac{dy}{dx} = f(x)g(y), rearranged and integrated as $\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$ , looks like it treats dy and dx as ordinary numbers you can shuffle across an equals sign — which they are not. Here is why the shuffle is legal anyway. Let $G(y)$ be any of $\frac{1}{g(y)}$ , so that $G'(y) = \frac{1}{g(y)}$ by definition. Now differentiate $G(y)$ with respect to x using the chain rule: $\frac{d}{dx}\left[G(y)\right] = G'(y)\cdot\frac{dy}{dx} = \frac{1}{g(y)}\cdot\frac{dy}{dx}$ . But the original equation says $\frac{dy}{dx} = f(x)g(y)$ , so substituting that in: $\frac{d}{dx}\left[G(y)\right] = \frac{1}{g(y)}\cdot f(x)g(y) = f(x)$ — the $g(y)$ 's cancel completely, leaving a plain function of x on both sides of a genuine, ordinary equation: $\frac{d}{dx}\left[G(y)\right] = f(x)$ . Integrating both sides with respect to x, in the completely standard sense (no shuffling required), gives $G(y) = \int f(x)\,dx + C$ . And by G's own definition, $G(y)$ IS $\int \frac{1}{g(y)}\,dy$ . So $\int \frac{1}{g(y)}\,dy = \int f(x)\,dx + C$ — exactly the "separate and integrate each side" recipe, but now derived rather than pattern-matched, and it also explains the one-constant rule directly: the constant only ever entered on ONE side of this derivation, the x-side, because the y-side is just $G(y)$ with nothing left to integrate independently. Writing a separate $+C_1$ on the y-side and a separate $+C_2$ on the x-side and then combining them into one C at the end gives the identical final equation — the two-constant version is not wrong, it is simply doing one extra step of bookkeeping to reach the same place.

Worked, in full

Solve $\dfrac{dy}{dx} = \dfrac{6x^2}{y^3}$ given that $y = 2$ when $x = 1$ — the general solution, then the particular one (VERIDIAN-original equation, built and checked to demonstrate spec 6.4's own "general and particular" requirement in one continuous problem, not copied from any past paper)

01
Recognise the equation is separable, and rearrange so every y-term (including dy) sits on one side and every x-term (including dx) sits on the other. $\dfrac{dy}{dx} = \dfrac{6x^2}{y^3}$ is $\dfrac{dy}{dx} = 6x^2 \times \dfrac{1}{y^3}$ — a function of x alone times a function of y alone — so it separates as $y^3\,dy = 6x^2\,dx$ .
Earns: M1 — separates the variables correctly: every y-term (and dy) on one side, every x-term (and dx) on the other, with none of either left stranded on the wrong side.
02
Integrate each side with respect to its own variable: $\displaystyle\int y^3\,dy = \int 6x^2\,dx \;\Rightarrow\; \dfrac{y^4}{4} = 2x^3 + C$ . This is the general solution — spec 6.4's guidance is explicit that both general and particular solutions will be required, and this is the general one: a whole family of curves, one member for every value of C.
Earns: M1 — attempts to integrate both sides, each power raised by one. A1 — both integrals correct, with a single +C. (Each side technically produces its own constant of integration; the two always combine into one, so carrying only one from the start loses nothing — see the mechanism block above.)
03
The boundary condition given in the question, $y=2$ when $x=1$ , selects exactly one member of that family. Substitute: $\dfrac{2^4}{4} = 2(1)^3 + C \;\Rightarrow\; 4 = 2 + C \;\Rightarrow\; C = 2$ .
Earns: M1 — substitutes the given x and y into the general solution to find C, dependent on having reached a genuine general solution in stage 2.
04
State the particular solution with C resolved: $\dfrac{y^4}{4} = 2x^3 + 2$ , i.e. $y^4 = 8x^3 + 8 = 8(x^3+1)$ .
Earns: A1 — correct value of C substituted back in and the equation rearranged into a clean particular-solution form. Either form (unresolved fraction, or cleared of fractions) is acceptable unless a mark scheme for a specific printed target says otherwise.
05
Where a question asks for y explicitly rather than left implicit, take the fourth root of both sides, keeping only the branch consistent with the boundary condition: $y = \sqrt[4]{8(x^3+1)}$ — the positive root, since $y=2>0$ at the boundary condition, so the negative fourth root is rejected without needing to check it.
Earns: A1 — rearranged to give y explicitly, with the correct branch chosen and justified against the boundary condition rather than simply assumed.

Complete it yourself

Complete the chain — the particular solution of $\dfrac{dy}{dx} = \dfrac{2y}{x+1}$ given $y = 3$ when $x = 0$

01
Separate the variables: every y-term (with dy) on one side, every x-term (with dx) on the other. $\dfrac{dy}{dx} = \dfrac{2y}{x+1} \;\Rightarrow\; \dfrac{dy}{y} = \dfrac{2}{x+1}\,dx$ .
02
Integrate each side: $\displaystyle\int \dfrac{1}{y}\,dy = \int \dfrac{2}{x+1}\,dx \;\Rightarrow\; \ln|y| = 2\ln|x+1| + C$ . This is the general solution — one arbitrary constant, one family of curves.

In your own words

In one sentence: why does separating $\dfrac{dy}{dx} = f(x)g(y)$ into $\dfrac{dy}{g(y)} = f(x)\,dx$ only ever need ONE arbitrary constant, even though each side is integrated separately?

Forming the equation before you solve it — connected rates of change (spec 5.2)

Spec 5.2's own guidance is short and specific: "Questions involving connected rates of change may be set." Almost no WMA14 question on this topic simply hands you $\frac{dy}{dx} = \ldots$ to solve. Instead, a real-world context describes how ONE quantity changes with time — water flows into a tank at a constant rate, a radius increases at 2 cm per second — and asks how a DIFFERENT, related quantity changes with time. The bridge between the two rates is always the same three-step recipe: (1) find the equation connecting the two quantities themselves, from the geometry of the situation; (2) differentiate that relationship; (3) apply the to connect the rate you were given to the rate you actually want, exactly the mechanism the second prequestion above tested.

The first place this goes wrong is step (1), before any calculus at all — the geometry itself. A real WMA14 examiner report, on a cylindrical water-tank rates-of-change question, records candidates "using the volume of the cylinder as V = (4/3)πr²h" (Oct 2021, Q9) — the volume of a SPHERE, $\frac{4}{3}\pi r^3$ , misapplied (and mistyped, with an h where the formula has no h at all) to a cylinder, whose actual volume is $\pi r^2 h$ . This is not a formula you can look up mid-exam: WMA14-verified-facts.md §3 confirms the exam formula booklet's P1 mensuration entry gives only a sphere's SURFACE area and a cone's curved SURFACE area — no solid's volume appears in the booklet at all, for any shape. Every volume formula in this topic — cylinder, cone, sphere alike — has to be recognised correctly from memory, from the shape actually described in the question.

The second place it goes wrong is step (2) itself, and a real WMA14 report on a DIFFERENT rates-of-change question — a cone, tested via its surface area rather than its volume, spec 5.1/5.2 territory rather than a full differential equation to solve — records the exact failure: "treating l in the given S formula as a constant" (Jan 2024, Q4), where l (the cone's slant height) was actually related to the radius via Pythagoras and had to be substituted as a function of r BEFORE differentiating S with respect to r, not held fixed. The same report notes plainly that "the concept of rates of change was shown to be not well understood by most" on that question. The general lesson generalises past just this one shape: sometimes a quantity in a geometric formula genuinely IS a constant with respect to the variable you're differentiating (a cylinder's cross-sectional radius, if the tank's sides are vertical, really doesn't change as the water rises) — and sometimes a quantity that looks fixed is actually itself a function of the variable you're differentiating, and has to be substituted in before you differentiate, not treated as a label.

The third place it goes wrong is the rate itself, whenever it is genuinely built from two competing effects — an inflow AND an outflow, growth AND decay — and one of the two terms is silently dropped. The same Oct 2021 Q9 report records candidates "using either dV/dt = 0.6π or dV/dt = −0.15πh" — each one is literally half of what should have been a single two-term rate, dV/dt = 0.6π − 0.15πh (inflow minus outflow), with one entire term thrown away. The worked example below is built specifically so that dropping either term is a live, checkable trap, not a hypothetical one.

Marked, line by line

A cylindrical tank has a constant cross-sectional radius of 5 m. Water flows in at a constant rate of $2\pi$ m³ per minute. At the same time, water leaks out through a valve in the base at a rate proportional to the depth of water, equal to $0.5\pi h$ m³ per minute, where h metres is the depth at time t minutes after the tank starts empty. (a) Show that $\dfrac{dh}{dt} = \dfrac{4-h}{50}$ . (4) (b) Given that the tank starts empty, solve the differential equation to find t in terms of h. (5) (c) Find, to the nearest minute, the time taken for the depth to reach 2 m, stating an appropriate unit with your answer. (2) — VERIDIAN-original question, built specifically to walk through the real documented traps on this topic (a cylinder's volume formula, a genuine two-term rate, a constant integrated with respect to time, a final answer needing its unit checked) rather than copying any past-paper question's own numbers, which the research bank does not supply for this topic. The per-line mark allocations are modelled on the verified mark-scheme conventions in WMA14-verified-facts.md §4 (M/A/B definitions, dependency, ag/"show that" gating), not transcribed from a real mark scheme, which for an original question does not exist.

11 marks available

(a) — 4 marks

“Show that” — the answer is already printed above

Matching the printed result isn’t the same as deriving it — real examiner reports describe scripts that adjust flawed working just to still land on it. Write your own full working below before checking it against the mark scheme.

(b) — 5 marks

101
Separate the variables: $\dfrac{dh}{4-h} = \dfrac{dt}{50}$ .
Separates correctly: every h-term (with dh) on one side, every t-term (with dt) on the other.
M1
102
Integrate the left side with respect to h: $\displaystyle\int \dfrac{1}{4-h}\,dh = -\ln|4-h|$ . (The minus sign appears because $\frac{d}{dh}(4-h) = -1$ ; it is the same integral as the more familiar $\int\frac{1}{h-4}\,dh = \ln|h-4|$ , differing only in that sign.)
Correctly integrates a reciprocal-of-linear expression in h, including the sign that comes from the negative coefficient of h inside the bracket.
M1
103
Integrate the right side with respect to t: $\displaystyle\int \dfrac{1}{50}\,dt = \dfrac{t}{50}$ , and combine into one equation with a single constant: $-\ln|4-h| = \dfrac{t}{50} + C$ .
Correctly integrates a plain constant (1/50) with respect to t, giving t/50 — not a logarithm. A real WMA14 examiner report on this exact kind of tank question records the wrong-antiderivative version of this line directly: 'integrating 1/320 to ln(320t)' (Oct 2021, Q9) — the analogous error here would be writing ln(50t) instead of t/50.
A1
104
Apply the boundary condition: the tank starts empty, so $t=0$ when $h=0$ . $-\ln|4-0| = 0 + C \;\Rightarrow\; C = -\ln 4$ .
Substitutes the given boundary condition to find C, dependent on having reached a genuine general solution in the line above.
dM1
105
$-\ln|4-h| = \dfrac{t}{50} - \ln 4 \;\Rightarrow\; \ln 4 - \ln(4-h) = \dfrac{t}{50} \;\Rightarrow\; \ln\!\left(\dfrac{4}{4-h}\right) = \dfrac{t}{50} \;\Rightarrow\; t = 50\ln\!\left(\dfrac{4}{4-h}\right)$ . (The modulus can be dropped: h is a depth strictly between 0 and 4, so $4-h$ is always positive.)
Correct particular solution for t in terms of h, with the logarithms combined using ln a − ln b = ln(a/b) before the equation is rearranged for t.
A1

201
Substitute $h=2$ into the part (b) answer: $t = 50\ln\!\left(\dfrac{4}{4-2}\right) = 50\ln 2 \approx 34.657$ .
Substitutes h = 2 into their own part (b) result (ft their equation), correctly.
M1
202
$t \approx 35$ minutes (to the nearest minute).
Correct value, awrt 35, WITH the correct unit stated. The unit is not decoration: a real examiner report on this exact question type records a numerically-correct final answer, 208, losing marks by 'giving the units for the answer 208 as seconds rather than minutes' (Oct 2021, Q9) — every quantity in this problem was defined in minutes from the very first line, and the final answer has to be reported in the same unit, not converted or mislabelled at the last step.
A1

Diagram — The family of solutions to dh/dt = (4−h)/50, and the one member the boundary condition picks out

x-axis: t (minutes) · y-axis: h (metres)

Particular solution: h = 4(1 − e^(−t/50)), through (0, 0): The tank starts empty (h=0 at t=0) and fills, rising fastest at the start when the outflow is smallest, and levelling off as h approaches 4 — the depth at which the outflow exactly equals the inflow, so dh/dt reaches zero. This is the ONE curve the marked-solution above actually solves for; it is picked out of the whole family by the boundary condition h=0 at t=0.
Another family member: h = 4 + 2e^(−t/50), through (0, 6): A DIFFERENT tank, started at depth 6 m instead of empty, obeying the exact same differential equation. Because 6 m is above the equilibrium depth of 4 m, the outflow briefly exceeds the inflow and the depth falls — but it falls toward the same equilibrium, never below it. This curve solves the identical equation dh/dt=(4−h)/50; only its boundary condition differs.

Equilibrium depth, h = 4: Where dh/dt = 0 — the inflow (2π) exactly balances the outflow (0.5π×4 = 2π). Every member of the family, whatever depth it starts at, approaches this same horizontal line and never crosses it — a curve starting below rises toward it, a curve starting above falls toward it, and neither ever quite reaches it in finite time. Marked at t=100, sitting between both plotted curves at that time (h≈3.459 and h≈4.271), to show the asymptote they are converging toward rather than a point either curve actually touches.
Boundary condition (0, 0): The one piece of information that turns the whole family into a single curve. Without it, dh/dt=(4−h)/50 is satisfied by infinitely many curves — one for every value of the constant C — and 'the general solution' correctly describes all of them at once.

Common error: Sketching only the particular solution and treating it as if it were the differential equation's one and only graph.

Correct: The differential equation itself is satisfied by an entire family of curves, one for every value of the arbitrary constant — the dashed curve above solves the exact same equation as the solid one. The general solution names the whole family; the boundary condition is what selects a single member of it, not what creates the family in the first place.

Marked, line by line

The height, h metres, of a shrub, t years after it was planted, is modelled by the differential equation $\dfrac{dh}{dt} = \dfrac{2h^{\frac32}}{5t^2}$ , $t>0$ . Given that $h=1$ when $t=1$ , (a) solve the differential equation to show that $h = \dfrac{at^2}{(1+bt)^2}$ , where $a$ and $b$ are constants to be found. (6) (b) Hence find, according to the model, the limit of the height of the shrub. (2) — REAL Pearson Edexcel International A-Level question, WMA14 Paper 01A, January 2026, Q3 (Publications Code WMA14_01_2601_MS, Question Paper Log Number P87595A). Independently re-verified against the primary-source question paper and mark scheme PDFs (cover pages checked directly, both `pdftotext -layout` and `-raw` cross-checked against each other and against the rendered PDF pages themselves) and re-derived from scratch with sympy — not carried over from any prior transcription. This is the first genuinely real, cited past-paper question in this lesson; every other marked-solution, worked-chain and chain-drill above is VERIDIAN-original, as their own text already says. One real limitation, stated rather than smoothed over: no examiner report for the January 2026 series exists yet — the series is too recent — so this item verifies real mark-scheme MECHANICS but does not yet contribute a documented candidate MISCONCEPTION; treat it as a real anchor for how this closing step is actually marked, not a trap-taxonomy source.

8 marks available

(a) — 6 marks

01
Separate the variables: $\dfrac{dh}{dt} = \dfrac{2h^{\frac32}}{5t^2} \;\Rightarrow\; \displaystyle\int \dfrac{1}{2h^{\frac32}}\,dh = \int \dfrac{1}{5t^2}\,dt$ .
Correct separation of variables (equivalent forms allowed). There is no need to see the integral signs themselves, but dh and dt must both be present and in the correct positions, or implied by further correct work.
B1
02
Integrate each side: $\displaystyle\int \dfrac{1}{2}h^{-\frac32}\,dh = \int \dfrac{1}{5}t^{-2}\,dt \;\Rightarrow\; -h^{-\frac12} = -\dfrac15 t^{-1}\ (+c)$ .
M1 for reaching a result of the general shape αh^(-1/2) = βt^(-1) (+c) — the correct power on each side, whatever the coefficients. A1 for the coefficients themselves being right: -h^(-1/2) = -(1/5)t^(-1) (+c), or any correct equivalent form (e.g. -10h^(-1/2) = -2t^(-1) (+c)). If no constant of integration is present at all, no further mark in this part can be scored — it has to be carried forward into the next line, not silently dropped.
M1 A1
03
Apply $h=1$ when $t=1$ : $-(1)^{-\frac12} = -\dfrac15(1)^{-1} + c \;\Rightarrow\; -1 = -\dfrac15 + c \;\Rightarrow\; c = -\dfrac45$ .
Uses the given boundary condition, at an appropriate point in the candidate's own integrated equation (however poor its coefficients might be), to find a numerical value for the constant of integration.
M1
04
Rearrange to make h the subject. Form a single fraction on the h-side before inverting — not term-by-term — then square both full sides, not each term separately: $-h^{-\frac12} = -\dfrac15t^{-1} - \dfrac45 \;\Rightarrow\; \dfrac{1}{\sqrt h} = \dfrac{1}{5t} + \dfrac45 = \dfrac{1+4t}{5t} \;\Rightarrow\; \sqrt h = \dfrac{5t}{1+4t} \;\Rightarrow\; h = \dfrac{25t^2}{(1+4t)^2}$ .
Correct algebra from the candidate's own integrated expression to make h the subject. Two genuine sub-steps have to both be present: combining into a single fraction before inverting (inverting each term separately instead is not credited — h stays trapped inside a sum otherwise), and squaring each FULL SIDE of the resulting equation rather than each term of it (a slip that lands on 5t² instead of 25t² is condoned here, since the method — squaring a whole side — is still correct even though the arithmetic isn't).
M1
05
$h = \dfrac{25t^2}{(1+4t)^2}\quad (a=25,\ b=4)$ .
Correct answer only (cao): a=25 and b=4, in the exact printed target form h = at²/(1+bt)². Follows directly from the sympy cross-check: h(1)=1 exactly, and differentiating h(t) with respect to t reproduces 2h^(3/2)/(5t²) exactly (checked symbolically and numerically at six separate t-values, not just at the boundary point) — confirming the closed form genuinely solves the original differential equation, not merely the boundary condition it was built to satisfy.
A1

(b) — 2 marks

101
Divide every term of $h = \dfrac{25t^2}{(1+4t)^2}$ by $t^2$ (equivalently, read off the ratio of the two leading $t^2$ coefficients): $h = \dfrac{25}{\left(\frac1t+4\right)^2}$ . As $t\to\infty$ , $\frac1t \to 0$ , so $h \to \dfrac{25}{4^2} = \dfrac{25}{16}$ .
Recognises the limiting value as (their a)/(their b)², directly from an answer to (a) in the form h = at²/(c+bt)², a>0. This is the genuinely new closing step this real series adds to the topic: it is read straight off the closed rational form — dividing every term by t² and letting t tend to infinity, so every term with a lower power of t vanishes and only the ratio of the two leading coefficients survives — never by differentiating h with respect to t to hunt for a stationary point. That second route is a dead end for this exact question, and checkably so: dh/dt = 2h^(3/2)/(5t²) is strictly positive for every finite t>0 (since h>0 throughout the domain t>0), so this h(t) never actually has a critical point anywhere — it only approaches its limit asymptotically as t grows without bound, which is exactly why the real mark scheme credits the algebraic long-run-behaviour reading and not a calculus-optimisation search for dh/dt=0.
M1
102
Limit of the height $= \dfrac{25}{16}$ m $= 1.5625$ m.
The exact value 25/16 (or 1.5625), following through on the candidate's own a and b from part (a) — condoning interval notation or an inequality (h ≤ 25/16, or 0 < h < 25/16) in place of a bare value. Units are not required, but must be correct if given.
A1ft

Named traps

sphere-formula-used-for-a-cylinder: Confirmed directly on a real cylindrical rates-of-change question: candidates "using the volume of the cylinder as V = (4/3)πr²h" (Oct 2021, Q9) — the volume of a SPHERE, applied to a cylinder, and even then written with an h that the sphere formula does not have. Neither formula is in the exam formula booklet: WMA14-verified-facts.md §3 confirms the P1 mensuration entry gives only a sphere's surface area and a cone's curved surface area, never a volume for any shape. Every solid-volume formula in this topic has to be recognised correctly from memory, from the shape actually described.
two-term-rate-collapsed-to-one: Confirmed on the same question: candidates "using either dV/dt = 0.6π or dV/dt = −0.15πh" (Oct 2021, Q9) — each is exactly half of what should have been a single two-term rate (an inflow AND an outflow), with the other term silently dropped. The marked-solution above is built specifically so this error is checkable: dropping the inflow gives dh/dt = −h/50, which visibly fails to match the printed "show that" target.
constant-integrated-as-a-logarithm: Confirmed on the same question, at the solving stage: candidates "integrating 1/320 to ln(320t)" (Oct 2021, Q9) — a plain constant integrated with respect to time should give (constant)×t, not a logarithm of time. The analogous line in the marked-solution above is $\int \frac{1}{50}\,dt = \frac{t}{50}$ ; the documented wrong version would write $\ln(50t)$ instead.
answer-given-in-the-wrong-time-unit: Confirmed on the same question, at the very last step: candidates "giving the units for the answer 208 as seconds rather than minutes" (Oct 2021, Q9) — the number itself was right, and the mark was lost purely because the stated unit did not match the unit the whole question had been working in from its first line. This is exactly why the marked-solution's part (c) states "minutes" explicitly rather than leaving the reader to assume it.
denominator-mishandled-when-separating: Confirmed on a real separable-equation question: "many candidates who decided to 'move' the 4, ended up with an incorrect starting equation of ∫4/y² dy = ∫1/(4x+5)^(3/2) dx at some point" (Oct 2021, Q2) — a genuine rearrangement error before any integration has even begun, not a calculus mistake. The same report also documents a matching error at the OTHER end of the same question, after both sides had already been correctly integrated and solved for y: converting an equation of the form a/y = b√(4x+5) + c into y = 1/(a·b√(4x+5)) + 1/c — distributing a reciprocal across a sum, which is not a legal algebraic move (1/(p+q) is not 1/p + 1/q). One question, two independently documented ways to lose marks on pure algebra either side of a perfectly good middle section.
fudged-reverse-fit-on-a-show-that-answerfudged-reverse-fit: Confirmed on a real "show that a particular solution equals a printed expression" question: "the given answer here persuaded many candidates to 'adjust' their working following obvious mistakes" (Jan 2021, Q10). Because the target is printed on the page in an ag question, a derivation that quietly changes a wrong intermediate number to make the last line match is a real, examiner-documented pattern — not a hypothetical one — and it is specifically NOT credited even when the final line is correct, because the mark is for the derivation, not for the coincidence of matching text. The warrantCheck attached to the marked-solution's part (a) above is built directly around this trap.
related-quantity-treated-as-fixed-instead-of-substituted: Confirmed on a real cone-based rates-of-change question — surface area rather than volume, but the same connected-rates setup skill spec 5.2 covers: candidates "treating l in the given S formula as a constant" (Jan 2024, Q4), where l (the cone's slant height) was actually a function of the radius via Pythagoras and needed to be substituted as such BEFORE differentiating, not held fixed. The same report records the question as challenging for "a very large majority of students," with "the concept of rates of change... shown to be not well understood by most." The cylinder in this lesson's own worked example is the case where a quantity genuinely IS a constant (the cross-sectional radius, with vertical tank walls) — this trap is what happens when that assumption is applied to a shape where it no longer holds.

Retrieval — with feedback on every choice

Question 1

2 marks

To solve $\dfrac{dy}{dx} = \dfrac{y^2}{2x+5}$ , which is the correctly separated form?

Question 2

3 marks

A cone-shaped pile of sand keeps its shape as it grows, so its radius r and height h are always related by $r = \frac{3}{4}h$ , giving $V = \frac{3\pi}{16}h^3$ . Sand is added so that $\dfrac{dV}{dt}$ is constant. Which is the correct way to find $\dfrac{dh}{dt}$ ?

Question 3

1 mark

A tank problem defines t in minutes throughout its working, and the question specifically asks for the time in minutes. The final calculation gives 208. Which is the correct final answer to state?

Question 4

1 mark

A "show that" part asks you to show that $\dfrac{dh}{dt} = \dfrac{4-h}{50}$ . Partway through your own derivation you reach $\dfrac{dh}{dt} = \dfrac{4-h}{45}$ — close to, but not exactly, the printed answer. What is the safest thing to do?

Reference — not a study method, a lookup

dy/dx = f(x)g(y) → separate: dy/g(y) = f(x)dx, then integrate — one constant, not two.
General solution has +C; a given (x,y) boundary condition turns it into the particular solution.
Connected rates: relate the two quantities first, differentiate that relation, then chain rule: dV/dt = dV/dh × dh/dt.
Booklet gives sphere surface area and cone curved surface area only — every volume formula is memory, not lookup.
Ask: is the geometric quantity actually constant, or is it itself a function of the variable you're differentiating?
State the unit the question was worked in from the start — a number alone is not a complete AO5 answer.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was checked against WMA14-verified-facts.md, itself checked against the primary Pearson documents, not carried over from prior course material. This is the first lesson written for WMA14, a brand-new unit with no other lessons yet registered anywhere in this course — it is not yet linked from any course page or index. The rates-of-change tank scenario (first marked-solution) and the pure-technique worked example (worked-chain) are both VERIDIAN-original — inspired by the real question types the examiner reports document, never a reproduction of any actual Pearson question — so the mark allocations attached to them are modelled on the verified mark-scheme conventions in WMA14-verified-facts.md §4 (M/A/B definitions, dependency, follow-through, ag/"show that" gating) rather than transcribed from a real mark scheme, which for an original question does not exist. One exception, added 2026-09-07: the SECOND marked-solution block (the shrub-height question, "hence find the limit") is a real, cited Pearson question — WMA14/01A, January 2026, Q3, Publications Code WMA14_01_2601_MS — with its real mark scheme transcribed and independently re-verified against the primary-source PDFs; the first marked-solution above it and both worked-chain/chain-drill blocks remain VERIDIAN-original, and all are clearly distinguished in their own question text. Every documented trap named in this lesson is confirmed against exactly three series and four questions — January 2021 Q10, October 2021 Q2, October 2021 Q9, and January 2024 Q4 — no broader or older pattern than that is claimed anywhere above; the January 2026 item above is deliberately excluded from that count, since no examiner report exists yet for that series to confirm a documented misconception from, only real mark-scheme mechanics. WMA14-verified-facts.md §8 flags this research bank as drawing on a narrower base (10 series with at least one document, 4 full examiner reports) than this course's Economics/Business lessons (15 series, 13 reports); this lesson's trap-taxonomy is scoped to what those 4 reports actually say, not presented as more exhaustive than that.

Question 12 marks

To solve $\dfrac{dy}{dx} = \dfrac{y^2}{2x+5}$ , which is the correctly separated form?

$\displaystyle\int y^{-2}\,dy = \int (2x+5)^{-1}\,dx$
Correct. Moving $y^2$ from the right side to divide the left side means dividing by it — i.e. multiplying by its reciprocal, $y^{-2}$ — and moving $(2x+5)$ leaves it exactly where it already was, dividing the right side.
B $\displaystyle\int y^2\,dy = \int (2x+5)\,dx$
$y^2$ has not been reciprocated when it moved across — it was multiplying the right side (via 1/y², buried inside the fraction), so on the left it needs to divide, i.e. appear as $y^{-2}$ , not as $y^2$ itself.
C $\displaystyle\int 2y\,dy = \int (2x+5)\,dx$
This differentiates $y^2$ (giving $2y$ ) rather than simply moving it across the equation. Separating variables is pure rearrangement — no differentiation happens until after both sides have been correctly separated.
D $\displaystyle\int y^{-2}\,dy = \int (2x+5)\,dx$
The left side is correctly reciprocated, but the right side's denominator has not been — $(2x+5)$ was dividing the original right-hand side, so it needs to keep dividing after rearrangement, appearing as $(2x+5)^{-1}$ . A real WMA14 report documents exactly this family of error — mishandling a denominator while separating, rather than a calculus mistake — on a genuine separable-equation question (Oct 2021, Q2).

Traps tested: Reciprocal relationship inverted · Differentiates instead of separates · Denominator not reciprocated when moved

Question 23 marks

Differentiate $V = \dfrac{3\pi}{16}h^3$ with respect to h to get $\dfrac{dV}{dh} = \dfrac{9\pi}{16}h^2$ , then use $\dfrac{dh}{dt} = \dfrac{dV/dt}{dV/dh}$
Correct. V has already been written purely in terms of h (using the fixed ratio between r and h), so differentiating it with respect to h and applying the chain rule connects the given rate to the required one directly.
BSubstitute a fixed numerical value for r into $V = \frac{1}{3}\pi r^2 h$ first, since r is "the radius" and radii are usually treated as constants, then differentiate the result with respect to h
This treats r as fixed when it is genuinely a function of h ( $r = \frac{3}{4}h$ ) — it changes as the pile grows, exactly like it does in real life. A real WMA14 examiner report documents precisely this failure on a cone-based rates-of-change question: treating a quantity that should have been substituted as a function of the other variable as if it were fixed (Jan 2024, Q4). r has to be substituted symbolically, as in option A, before differentiating — not replaced by a single number.
C $\dfrac{dh}{dt} = \dfrac{dV}{dt} \times \dfrac{dV}{dh}$
The chain rule needs the reciprocal of dV/dh here, not dV/dh itself — connecting dV/dt to dh/dt means dividing by dV/dh (or equivalently multiplying by dh/dV), not multiplying by dV/dh.
DSince dV/dt is constant, dh/dt must also be constant
dV/dh depends on h (it's $\frac{9\pi}{16}h^2$ , which grows as h grows), so dh/dt = (dV/dt)/(dV/dh) actually SHRINKS as the pile gets taller, even though dV/dt never changes — the same volume added per second raises the height less and less as the base widens. A constant rate for one quantity does not transfer unchanged to a related quantity.

Traps tested: Treats a varying quantity as fixed · Chain rule inverted · Assumes constant rate transfers unchanged

Question 31 mark

A tank problem defines t in minutes throughout its working, and the question specifically asks for the time in minutes. The final calculation gives 208. Which is the correct final answer to state?

208 minutes
Correct — the unit matches exactly what the question defined and asked for from the very first line.
B208 seconds
This is a real, documented error on an almost identical question: an examiner report records a numerically-correct answer of 208 losing marks for "giving the units for the answer 208 as seconds rather than minutes" (Oct 2021, Q9). The number is right; the unit contradicts every line of working that produced it.
C208 (no unit stated)
A number alone is not a complete AO5 answer on a question that asked for a time. Stating the unit costs one word and confirms the answer is being read in the same terms the question was set in.
D3.47 hours
A correct conversion of a number that was never asked to be converted. The whole problem — the rate, the boundary condition, the integration — was worked entirely in minutes, and the question specifically asks for the time in minutes; switching units at the very last step answers a question that was not set.

Traps tested: Answer restated in a different unit than the working · Units omitted · Unrequested unit conversion at the final step

Question 41 mark

Go back through your own working, line by line, to find the genuine error — never silently change the 45 to a 50 to match the printed answer
Correct. A real WMA14 examiner report on this exact question type records candidates who let a printed target 'persuade' them to adjust flawed working until it matched (Jan 2021, Q10) — examiners are specifically trained to look for this, and it is not credited even when the final line ends up matching.
BChange the 45 to a 50 in your final line, since the method is basically right and the given answer confirms it
This is precisely the documented pattern examiners are trained to catch: a final line that matches the printed target without a genuine derivation reaching it. The mark is for the derivation, not for landing on text that resembles the answer.
CLeave both numbers as they are and move on to the next part, since ag parts are usually only worth one mark anyway
Ag parts on this topic carry real marks — four, in the marked-solution above — and an unresolved mismatch forfeits credit that a short, targeted check could recover. It also does not solve the underlying problem for any later part that depends on this result.
DStart the whole derivation again from a blank page, without looking at the first attempt, to avoid being influenced by it
Slower than it needs to be under real time pressure: the discrepancy (45 vs 50) points at a specific line to re-check, and a targeted check of that one step is far faster than re-deriving four marks' worth of working from nothing.

Traps tested: Fudges working to match the printed answer · Abandons without checking · Inefficient full restart instead of targeted check

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

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Up next

Parametric Equations — Converting to Cartesian Form, and Domain/Range

Eliminating a parameter is one habit, done forward: solve one equation for t, then substitute. Guessing at the answer's shape and adjusting it until it seems to fit — without ever deriving the unknowns by a genuine algebraic identity — is the failure mode to watch for here, not the act of assuming a general form itself: assuming a form and deriving its coefficients by matching, done properly, is a real, valid, full-marks route on this exact spec point. The real record is blunt about how this goes wrong when it is done ungoverned, before any algebra is even attempted: "a small minority assumed the general form of the answer, substituted for x in terms of t and compared this to the expression for y" instead of deriving it, and — on a different question entirely — "a few candidates thought it appropriate to use calculus and scored no marks" on what is a pure substitution question. The domain/range half of this topic is worse still: candidates "confused range with domain and gave answers in terms of x rather than y," and even a script that gets the algebra right often checks only the two ends of the parameter's own domain — missing a minimum or maximum sitting, unannounced, at a point in between, or reaching for a turning point that does not exist at all when the parameter's own domain has no two closed endpoints to check in the first place.

60 min