A Level · Cambridge (CIE) · 9709

Handle horizontal vertical and indeterminate tangent cases

Use coordinate rates or implicit derivative coefficients to recognise regular vertical and horizontal tangents, and investigate a zero-over-zero ratio separately.

Section 3.4 · Differentiation

Pure Mathematics 2 & 3 · Differentiation · 2.4 / 3.4

Exam scope and assessment notes

Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.

Before you begin

Find parametric and implicit first derivatives.
Represent horizontal and vertical lines and recognise an undefined ratio.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What does a derivative ratio 0/0 tell you by itself?

What you will learn

Find tangent and normal equations when a regular curve has a horizontal or vertical tangent.
Distinguish a non-zero-over-zero gradient from an indeterminate zero-over-zero parametrisation and inspect the original relation.

A regular curve direction can be vertical without a finite gradient

For a differentiable parametric curve, if dx/dt ≠ 0 and dy/dt = 0, the tangent is horizontal and its normal vertical. If dx/dt = 0 and dy/dt ≠ 0, the local direction is vertical: the tangent is x = x₀ and the normal y = y₀. Do not call the gradient zero just because its denominator is zero, or force a vertical tangent into y = mx + c. A non-zero coordinate-rate vector supplies a definite local direction even when dy/dx has no finite value.

On x = cos t, y = sin t, at t = 0 the point is (1, 0) and the rates are 0 and 1. Thus the tangent is x = 1 and the normal y = 0. At t = π/2 the point is (0, 1) and the rates are −1 and 0, giving tangent y = 1 and normal x = 0. The same geometry follows from x² + y² = 1: its formula y′ = −x/y is finite at the top point but needs a separate vertical-tangent check at y = 0.

Zero over zero does not decide a curve's tangent

Implicit differentiation can give a relation A(x, y) + B(x, y)y′ = 0. If B ≠ 0, divide to find y′. If B = 0 and A ≠ 0 at a regular curve point, the finite y-as-a-function-of-x slope fails and the tangent is vertical. If both coefficients vanish, the first derivative relation alone is inconclusive: inspect the original relation or its local branches. Likewise, when both parametric coordinate rates are zero, their 0/0 ratio does not automatically mean a zero slope, no tangent or a unique vertical tangent.

For x = t³ and y = 2t³, both rates vanish at t = 0. Eliminating the parameter gives y = 2x for every real x, so the curve is a straight line with tangent y = 2x and normal y = −x/2 at the origin. The chosen parametrisation stops changing to first order at zero, but the underlying curve still has gradient 2. Recovering the original geometric relation supplies evidence that the indeterminate ratio alone could not provide.

Pause and explain

At a regular parametric point, dx/dt = 0 and dy/dt = 2. What is the tangent direction?

Put the idea to work

Worked example

For x = cos t, y = sin t, find tangent and normal equations at t = 0 and t = π/2 without dividing by a zero rate.

Show the worked solution

At t = 0 the point is (1, 0), with dx/dt = 0 and dy/dt = 1; its tangent is vertical x = 1.
The normal there is horizontal y = 0.
At t = π/2 the point is (0, 1), with rates −1 and 0; its tangent is horizontal y = 1 and its normal vertical x = 0.

Answer At t = 0: tangent x = 1, normal y = 0. At t = π/2: tangent y = 1, normal x = 0.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

For x² + y² = 25, find tangent and normal equations at (5, 0) and explain why y′ = −x/y cannot give a finite gradient there.

Give me a hint

Verify the point and inspect the original circle geometry or its non-zero implicit x coefficient.

Compare my reasoning

The point satisfies the circle equation; its derivative relation is 2x + 2y y′ = 0.
At (5, 0), the y′ coefficient is zero while the x contribution is 10, so the regular tangent is vertical x = 5.
The perpendicular normal is horizontal y = 0; the invalid ratio −5/0 should not be replaced by zero.

Tangent x = 5; normal y = 0; no finite dy/dx at that point.

Look for these in your work

Distinguish an original valid point from an undefined finite slope formula.
Use vertical and horizontal line equations instead of dividing by zero.

2 · Independent

For x = t² and y = t, find tangent and normal equations at t = 0 and check by eliminating t.

Give me a hint

The vertical coordinate rate stays non-zero even where the horizontal rate vanishes.

Compare my reasoning

At zero the point is (0, 0), with dx/dt = 2t = 0 and dy/dt = 1.
The tangent is vertical x = 0 and the normal horizontal y = 0.
Eliminating t gives x = y², whose sideways parabola has the same vertical tangent at its origin.

Tangent x = 0; normal y = 0; Cartesian check x = y².

Look for these in your work

Use the non-zero local direction vector despite a zero horizontal rate.
Confirm the tangent using the original geometric relation.

3 · Transfer

For x = t³ and y = 2t³, find the tangent and normal at t = 0. Explain why substituting into (dy/dt)/(dx/dt) is insufficient.

Give me a hint

Both coordinate rates vanish; eliminate the parameter to identify the underlying curve.

Compare my reasoning

The rates are 3t² and 6t², so direct substitution gives 0/0 rather than a defined gradient.
The original coordinates satisfy y = 2x for every real x, proving that the curve is a line with gradient 2 through the origin.
Its tangent is y = 2x and its perpendicular normal is y = −x/2; the explicit relation resolves the indeterminate parametrisation.

Tangent y = 2x; normal y = −x/2; the 0/0 parameter ratio alone is inconclusive.

Look for these in your work

Recognise the indeterminate ratio without assigning it a value.
Use a verified equivalent curve relation to find the tangent and normal.

Common mistakes

A zero denominator with a non-zero numerator is not a zero gradient.
Use x = constant for a vertical line instead of assigning it a finite gradient.
A 0/0 ratio is inconclusive; use the original relation or nearby curve behaviour for a separate argument.

Recall without your notes

What can you explain now?

For x = t³, y = 2t³ at t = 0, does the 0/0 derivative ratio prove that there is no tangent?

Compare with the explanation

No; eliminating t gives y = 2x, so the tangent at the origin is y = 2x.

The parametrisation's zero rates do not determine the underlying curve's tangent; its explicit line relation does.

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Make it stick

Compare horizontal, vertical and indeterminate derivative cases. Give a valid curve point where the finite slope formula fails and explain which separate geometric evidence resolves the tangent.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Mathematics

Check the official syllabus 2026–2027

Course scope and assessment

312 sequenced lessons provide 936 guided, independent and application problems covering the 160 objectives in sections 1.1–1.8, 2.1–2.6, 3.1–3.9, 4.1–4.5, 5.1–5.5 and 6.1–6.5. The map lists all 38 sections across six alternative components; all eight Pure Mathematics 1 sections and all six Pure Mathematics 2 sections, from Paper 2 Algebra through Numerical solution of equations, currently have full teaching sequences. Paper 3 Algebra, section 3.1, links the six shared foundation lessons plus six lessons on partial fractions and rational binomial expansions; Paper 3 logarithms, trigonometry and numerical equations use the shared lessons, with the Paper 1 graph foundation linked into trigonometry. Paper 3 differentiation combines the shared methods with its inverse-tangent derivative lesson. Paper 3 Integration links four shared standard and identity lessons plus seven advanced lessons, covering all six section 3.5 requirements. Eight Vectors lessons cover all six section 3.7 requirements, with 2D and 3D line and solid applications. Nine Differential equations lessons cover all four section 3.8 requirements, including Paper 3 integration techniques, initial conditions, equilibrium solutions and conditional model interpretations. Eleven Complex numbers lessons cover all eight section 3.9 requirements, with seven Argand diagrams and explicit argument and boundary checks. All nine Pure Mathematics 3 sections now have teaching linked to every mapped requirement. Nine Forces and equilibrium lessons cover all seven section 4.1 requirements, with six force diagrams, signed incline-friction intervals, reaction-dependent pull and push limits and separate-body third-law checks. Nine Kinematics lessons cover the four official section 4.2 requirements, with the graph requirement mapped to three relationships. Six motion diagrams support directed quantities, graph slopes and signed areas, Paper 1 time-calculus methods, reversal distances, constant-acceleration stages and shared-clock particle meetings. Six Momentum lessons cover both section 4.3 requirements, including signed system totals, conservation assumptions, outgoing velocities, coalescence, unknown masses and successive impacts. Four diagrams show system momentum through the stipulated events, and supplied kinetic-energy checks distinguish conservation of momentum from changes in kinetic energy. Twelve Newton’s laws lessons cover all four section 4.4 requirements, with vertical support and release stages, rough horizontal and incline motion, inferred parameters, hanging and surface pulley systems, rigid tow-bar thrust and slack-string transitions. Seven force diagrams support the separate-particle balances. Nine Energy, work and power lessons cover all five section 4.5 requirements, with angle-dependent work, kinetic and reference potential energy, signed net-work balances, smooth curved slides, rough and driven slopes, ideal connected-system energy and instantaneous power on hills. Five diagrams distinguish the energy inputs, losses and valid positive-speed force model. All five Mechanics sections now have teaching and staged practice linked to their mapped requirements. Thirteen Representation of data lessons cover all five section 5.1 requirements, including choice and limitations of displays, ordered and back-to-back stems, raw summaries and box plots, unequal-width density histograms, cumulative-frequency interpolation, descriptive moment calculations, grouped estimates, coding, combining two datasets and correcting totals. Six accessible diagrams retain repeated observations and distinguish numerical spread, frequency area and estimated readings. Ten Permutations and combinations lessons cover both official section 5.2 requirements, with selection and ordered roles, repeated objects, adjacency blocks, non-adjacency gaps, leading-zero cases, labelled multi-row seating and mixed restrictions. Six accessible diagrams show outcome trees, ordered versus unordered choices, repeated copies, blocks, gaps and distinct row seats. Eleven Probability lessons cover all four section 5.3 requirements, with equiprobable elementary outcomes, selection-based probabilities, complements and disjoint regions, independence tests, restricted sample spaces, two-way tables, replacement and removal trees, reversed conditions and unknown parameters. Seven accessible diagrams show dice outcomes, random subsets, event regions, a conditioned grid, a count table and two probability trees. Twelve Discrete random variables lessons cover all three official section 5.4 requirements, with complete probability tables, moment and parameter checks, binomial assumptions, exact probabilities and moments, geometric first-success probabilities, tails, expectation and integer reliability thresholds. Six accessible diagrams distinguish outcome grouping, weighted moments, fixed trials, discrete probability bars, waiting conventions and integer tail deadlines. Twelve Normal distribution lessons cover all three official section 5.5 requirements, with continuous-area interpretation, MF19 cumulative tables, full standardisation, tail and interval regions, conditional probabilities, percentiles, one- and two-parameter inference, binomial suitability, continuity corrections and approximate integer thresholds. Seven accessible diagrams show model areas, table reading, aligned scales, intervals, quantiles, matched binomial and normal shapes and whole integer cells. All five Statistics 1 sections now have teaching and staged practice linked to their mapped requirements. Fourteen Poisson distribution lessons cover all five official section 6.1 requirements, with homogeneous event-count assumptions, exposure scaling, exact masses and tails, conditional count events, moments, parameter inference, probability duration targets, disjoint and overlapping windows, rare-count binomial approximations, matched normal approximations and continuity corrections. Nine accessible diagrams show event exposures, unit conversions, discrete bars, complete tail complements, population moments, shared windows, approximate count models and integer cells. Exact checks distinguish approximate and exact close-target decisions. Fourteen Linear combinations lessons cover the six listed results in official section 6.2, with affine moments, signed scales and constant cases, expectation without independence, independent squared variance contributions, fixed and random fees, independent copies versus reused values, affine normal events, independent normal sums and differences, weighted and conditional areas, coefficient inference, exact independent Poisson sums and overlapping exposure limits. Nine accessible diagrams distinguish scale changes, variance contributions, component costs, copy mechanisms, aligned normal scales, summed areas, Poisson count pairs and unsupported distribution claims. Fourteen Continuous random variables lessons cover both official section 6.3 requirements, with density validity and units, support-aware intervals, complements and conditional regions, first and second moments, percentiles by direct density areas, symmetry checks, parameter inference, infinite-support power and exponential integrals, divergent-moment limits and combined applications. Nine accessible diagrams distinguish density height from probability, normalisation, negative-density failures, central and conditional areas, mean versus median, percentile locations and partial infinite tails. Explicit cumulative distribution function knowledge is not required. Eighteen Sampling and estimation lessons cover all eight official section 6.4 requirements, with population and frame definitions, random-number selection, sampling-quality limits, independent sample-mean moments, normal means, CLT conditions, SE scaling, raw, summarised and coded unbiased estimates, known-variance and large-sample mean intervals, repeated-sampling interpretation, precision planning, proportion intervals, sparse-outcome limits and contextual applications. Ten accessible diagrams distinguish selection frames, number rejection, repeated sample means, population versus mean density, CLT models, variance divisors, interval targets, illustrative coverage, integer size checks and proportion margins. Eighteen Hypothesis tests lessons cover all five official section 6.5 requirements, with pre-specified alternatives, inclusive count tails, explicit equal-tail conventions, integer critical regions and actual significance, exposure-aware Poisson tests, continuity-corrected count approximations, normal and large-sample mean tests, confidence-interval connections, error classifications, fixed-region binomial, Poisson and normal error probabilities and sample-size power comparisons. Nine accessible diagrams show the testing procedure, exact count regions, unequal Poisson tails, discrete significance jumps, corrected count boundaries, matching mean intervals and alternative error distributions. All shared lessons are counted once in the course total. All 38 sections now link teaching and practice. Eight earlier overview lessons remain available, but do not count as full section coverage. This is a content-availability measure, not full programme completion or exam readiness. Written practice is self-checked, not officially marked. Check the 2028–2030 syllabus separately if sitting an examination from 2028; this teaching map uses 2026–2027 requirements.