OxfordAQA International AS and A-level Mathematics: Parametric equations and exponential and logarithmic functions (9660) – Revision Notes

These notes condense the study guide for the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, sections P2.3 (Cartesian and parametric equations) and P2.5 (eˣ, ln x, growth and decay). This content sits in Unit P2 (Pure Maths), part of the International A-level only. It applies to International AS exams from May/June 2018 and International A-level exams from May/June 2019.

Other pages: the practice questions, the course page, the printable checklist and the International AS page on logarithm laws and a^x = b. To find weak spots first, try a free 10-minute diagnostic.

Key definitions

Term	Meaning
Parameter	A third variable (t or θ) that both x and y depend on
Parametric equations	x = f(t), y = g(t); each value of the parameter gives one point
Cartesian equation	An equation in x and y only
e	An irrational constant, about 2.71828
ln x	logₑx, the inverse of eˣ
Growth / decay	N = Aeᵏᵗ with k > 0 (growth); N = Ae⁻ᵏᵗ with k > 0 (decay)

The four curves named in the specification

Parametric form	Cartesian form	Shape and restriction
x = t², y = 2t	y² = 4x	Parabola, all t
x = a cosθ, y = b sinθ	x²/a² + y²/b² = 1	Ellipse, circle if a = b
x = 1/t, y = 3t	xy = 3	Rectangular hyperbola, t ≠ 0
x = t + 1/t, y = t − 1/t	x² − y² = 4	Hyperbola; t > 0 gives only x ≥ 2

Method: parametric to Cartesian

1. Look for the easier equation to rearrange for the parameter.
2. Substitute into the other equation and simplify.
3. If both equations are trigonometric, use sin²θ + cos²θ = 1, or a double-angle identity such as cos2θ = 2cos²θ − 1.
4. If the equations have forms like t + 1/t, add and subtract them, then multiply the results.
5. Write down any restriction on x or y from the range of cosθ, sinθ, or the sign of t.

Reminder. x = 3 sinθ, y = 1 − cos2θ: since cos2θ = 1 − 2sin²θ, y = 2sin²θ = 2(x/3)² = 2x²/9, with −3 ≤ x ≤ 3.

Method: Cartesian to parametric

• y = f(x): put x = t, y = f(t).
• Circle (x − p)² + (y − q)² = r²: x = p + r cosθ, y = q + r sinθ.
• Ellipse x²/a² + y²/b² = 1: x = a cosθ, y = b sinθ.
• Product curve xy = c: x = t, y = c/t.

Several parametrisations are correct. Check yours by substituting back.

Method: curve meets a line

Substitute the parametric x and y into the line equation. Solve for the parameter, then find the points using the parametric equations. Check one point in the Cartesian equation.

Reminder. x = 1/t, y = 3t meets y = 5 − x. Then 3t = 5 − 1/t, so 3t² − 5t + 1 = 0. The roots are t = (5 ± √13)/6, not integers, so keep the exact form.

eˣ and ln x

Feature	y = eˣ	y = ln x
Inputs allowed	all real x	x > 0 only
Outputs produced	y > 0 only	all real y
Passes through	(0, 1)	(1, 0)
Other key point	(1, e)	(e, 1)
Asymptote	y = 0	x = 0
Relationship	Inverse of ln x	Inverse of eˣ

Reflect either graph in the line y = x to get the other.

Inverse identities:

ln(eˣ) = x  (all x)         e^(ln x) = x  (x > 0)
ln a + ln b = ln(ab)        ln a − ln b = ln(a/b)
k ln a = ln(aᵏ)             ln 1 = 0,  ln e = 1

Transforming the graphs

Equation	Effect	Asymptote	Axis crossing
y = eˣ + c	Translate up c	y = c	(0, 1 + c)
y = eˣ⁻ᵃ	Translate right a	y = 0	(0, e⁻ᵃ)
y = e⁻ˣ	Reflect in y-axis	y = 0	(0, 1)
y = A eˣ	Stretch, factor A	y = 0	(0, A)
y = ln(x − a)	Translate right a	x = a	(a + 1, 0)
y = ln x + c	Translate up c	x = 0	(e⁻ᶜ, 0)

Method: solving equations with e and ln

1. Isolate the exponential or the logarithm.
2. Take ln of both sides (for eˣ) or convert to index form (for ln x).
3. For e²ˣ + a eˣ + b = 0, substitute u = eˣ and solve the quadratic.
4. Reject values where eˣ ≤ 0 or where a logarithm would have a non-positive argument.
5. Give exact answers (ln 3, (e³ − 1)/2) unless a decimal is asked for.

Reminder. Solve e²ˣ − 7eˣ + 10 = 0. Then (eˣ − 2)(eˣ − 5) = 0, so x = ln 2 or x = ln 5.

Inverse function reminder. For f(x) = 3eˣ + 2, put y = 3eˣ + 2, so x = ln[(y − 2)/3] and f⁻¹(x) = ln[(x − 2)/3], with domain x > 2.

Growth and decay formulas

Situation	Model	Useful results
Growth	N = Aeᵏᵗ, k > 0	A = value at t = 0; doubling time = ln 2/k
Decay	N = Ae⁻ᵏᵗ, k > 0	Half-life = ln 2/k
Decay to a limit	T = L + Be⁻ᵏᵗ	T → L as t → ∞; initial T = L + B

Method in steps for a model problem:

1. Use the first condition to find A (often given by t = 0).
2. Use a second condition to form an equation in k: e^(kt) = ratio, so k = (ln ratio)/t.
3. Keep k as an exact logarithm until the last step.
4. For a time, take ln to solve e^(kt) = value.
5. Interpret the answer with units, and say if the model is unrealistic (for example, unlimited growth).

Reminder. N = 50e^(0.2t). N = 200 when e^(0.2t) = 4, so t = ln 4/0.2 ≈ 6.93.

Must-know distinctions

• ln(a + b) is not ln a + ln b. Logs turn products into sums, not sums into sums.
• eˣ is not x to a power. The variable is in the index; the graph is never zero or negative.
• Growth k versus decay k. In N = Ae⁻ᵏᵗ the sign is already in the formula, so k is positive.
• Asymptote versus intercept. For y = 2eˣ − 5 the asymptote is y = −5 and the intercept is (0, −3).
• Domain of the parameter versus the curve. x = 2cosθ gives only −2 ≤ x ≤ 2, whatever the Cartesian equation allows.

Quick self-test

1. Find the Cartesian equation of x = 3t, y = t² − 2.
2. Find the Cartesian equation of x = 4cosθ, y = 7sinθ.
3. Find the Cartesian equation of x = 1/t, y = 3t.
4. Show that x = t + 1/t, y = t − 1/t satisfy x² − y² = 4.
5. Write y = x³ − x in parametric form.
6. State the asymptote and the y-intercept of y = eˣ − 4.
7. State the set of inputs and the set of outputs of y = ln x.
8. Solve ln x = −2.
9. Solve e³ˣ = 7, giving x to 3 significant figures.
10. Find f⁻¹(x) when f(x) = ln(x − 1), x > 1.
11. A substance has half-life 5 years and decays as m = m₀e⁻ᵏᵗ. Find k to 3 significant figures.
12. N = 50e^(0.2t). Find when N = 200, to 3 significant figures.

Answers

1. t = x/3, so y = x²/9 − 2.
2. x²/16 + y²/49 = 1.
3. y = 3/x (that is, xy = 3).
4. (x + y)(x − y) = (2t)(2/t) = 4, so x² − y² = 4.
5. For example x = t, y = t³ − t.
6. Asymptote y = −4; y-intercept (0, −3).
7. Inputs x > 0; outputs all real numbers.
8. x = e⁻² (about 0.135).
9. 3x = ln 7, so x = 0.649.
10. y = ln(x − 1) gives x = eʸ + 1, so f⁻¹(x) = eˣ + 1.
11. e⁻⁵ᵏ = 1/2, so k = ln 2/5 = 0.139.
12. e^(0.2t) = 4, so t = ln 4/0.2 = 6.93.

Where marks are usually lost

• Leaving out the restriction on x (for example −2 ≤ x ≤ 2) after eliminating a trig parameter.
• Writing x² + y² = 1 for an ellipse, or forgetting to divide by a² and b².
• Not squaring both fractions: (x/5)² is x²/25, not x²/5.
• Splitting ln(a + b) into two logarithms.
• Keeping a root with ln of a negative number, such as x = −1 in ln x + ln(x − 2) = ln 3.
• Stating the range of f⁻¹ as the domain, or the domain of f⁻¹ as all real numbers.
• Rounding k early and then getting a final time that is out in the third significant figure.
• Giving a time with no units, or with no comment when the question asks for one.
• Showing too little working on a “show that” question: write every line from the start to the target.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Sections P2.3 (Cartesian and parametric equations of curves and conversion between the two forms) and P2.5 (the function eˣ and its graph, ln x as the inverse of eˣ, exponential growth and decay including exponential functions as models).