Parametric Equations, Cartesian Conversion & Parametric Calculus
Many curves in mathematics and physics, such as projectiles, orbits, cycloids, and figure-of-eight Lissajous figures, cannot be written as a single function y = f(x) because multiple y-values exist for a single x-value. Parametric equations solve this by expressing both x and y separately in terms of a third variable called the parameter (usually t or θ): x = f(t) and y = g(t). At A-Level, you must eliminate the parameter to find the Cartesian equation (including finding its domain and range), differentiate parametrically to locate tangents and turning points, and integrate parametrically to find areas.
- Eliminating the Parameter: Algebraic Substitution
- When x = f(t) and y = g(t) are algebraic functions, rearrange the simpler equation to make t the subject (e.g., t = h(x)) and substitute this expression for t into the other equation to obtain a Cartesian equation linking x and y directly. Always determine the domain (set of valid x-values) and range (set of valid y-values) of the Cartesian curve by evaluating x = f(t) and y = g(t) over the restricted interval given for the parameter t.
- Eliminating the Parameter: Trigonometric Identities
- When x and y are trigonometric functions of θ (or t), never use arcsin or arccos to isolate θ. Instead, isolate the trig ratios (e.g., sin θ and cos θ) and substitute them into a Pythagorean or double-angle identity: 1. Use sin²θ + cos²θ ≡ 1 (for example, x = x₀ + r cos θ and y = y₀ + r sin θ gives the circle (x - x₀)² + (y - y₀)² = r²); 2. Use sec²θ - tan²θ ≡ 1 or cosec²θ - cot²θ ≡ 1; 3. Use sin(2θ) ≡ 2 sin θ cos θ or cos(2θ) ≡ 1 - 2 sin²θ ≡ 2 cos²θ - 1.
- Intersections of Parametric Curves with Lines and Axes
- To find where a parametric curve x = f(t), y = g(t) crosses the x-axis, set y = g(t) = 0, solve for t, and substitute those t-values into x = f(t). To find where it crosses the y-axis, set x = f(t) = 0, solve for t, and substitute into y = g(t). More generally, to find where the curve intersects a straight line ax + by + c = 0, substitute x = f(t) and y = g(t) directly into the line equation to get a(f(t)) + b(g(t)) + c = 0 and solve for t.
Definite Integral & Trapezium Rule Step-by-Step Visualizer
Choose a function, slide the limits a and b to see the area under the curve update live, and compare Exact Integration vs the Trapezium Rule.
Step-by-Step Working for ∫03 (1x²) dx
- Step 1, Find the antiderivative F(x):F(x) = [(1/3)x³] from x = 0 to x = 3
- Step 2: Substitute upper limit (b = 3) minus lower limit (a = 0):F(3) - F(0) = (9.0000) - (0.0000) = 9.0000
- Step 3, Compare with Trapezium Rule (4 strips, h = 0.750):Area ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)] = 9.2813
Difference from exact integral: 3.13% (increase
nstrips to make the approximation more accurate!).
+ C on indefinite integrals, and use radians whenever integrating trigonometric functions like sin(x) or cos(x)!Key Concepts & Exam Definitions (6)
- Eliminating the Parameter: Algebraic Substitution
- When x = f(t) and y = g(t) are algebraic functions, rearrange the simpler equation to make t the subject (e.g., t = h(x)) and substitute this expression for t into the other equation to obtain a Cartesian equation linking x and y directly. Always determine the domain (set of valid x-values) and range (set of valid y-values) of the Cartesian curve by evaluating x = f(t) and y = g(t) over the restricted interval given for the parameter t.
- Eliminating the Parameter: Trigonometric Identities
- When x and y are trigonometric functions of θ (or t), never use arcsin or arccos to isolate θ. Instead, isolate the trig ratios (e.g., sin θ and cos θ) and substitute them into a Pythagorean or double-angle identity: 1. Use sin²θ + cos²θ ≡ 1 (for example, x = x₀ + r cos θ and y = y₀ + r sin θ gives the circle (x - x₀)² + (y - y₀)² = r²); 2. Use sec²θ - tan²θ ≡ 1 or cosec²θ - cot²θ ≡ 1; 3. Use sin(2θ) ≡ 2 sin θ cos θ or cos(2θ) ≡ 1 - 2 sin²θ ≡ 2 cos²θ - 1.
- Intersections of Parametric Curves with Lines and Axes
- To find where a parametric curve x = f(t), y = g(t) crosses the x-axis, set y = g(t) = 0, solve for t, and substitute those t-values into x = f(t). To find where it crosses the y-axis, set x = f(t) = 0, solve for t, and substitute into y = g(t). More generally, to find where the curve intersects a straight line ax + by + c = 0, substitute x = f(t) and y = g(t) directly into the line equation to get a(f(t)) + b(g(t)) + c = 0 and solve for t.
- Parametric Differentiation: Tangents and Normals
- By the Chain Rule, dy/dx = (dy/dt) ÷ (dx/dt) (provided dx/dt ≠ 0). This allows you to find the gradient of the curve at any parameter value t without ever converting to a Cartesian equation. To find the tangent or normal at a specific parameter value t = t₀, evaluate x₀ = f(t₀), y₀ = g(t₀), and m = (dy/dt) / (dx/dt) at t = t₀, then use y - y₀ = m(x - x₀) for the tangent or y - y₀ = (-1 / m)(x - x₀) for the normal.
- Horizontal vs Vertical Tangents (Turning Points)
- Since dy/dx = (dy/dt) / (dx/dt): 1. A stationary point (horizontal tangent, parallel to the x-axis) occurs where the numerator is zero: dy/dt = 0 (with dx/dt ≠ 0). 2. A vertical tangent (parallel to the y-axis, where dy/dx is undefined) occurs where the denominator is zero: dx/dt = 0 (with dy/dt ≠ 0).
- Parametric Integration for Area Under a Curve
- The Cartesian area between a curve and the x-axis from x = a to x = b is Area = ∫_(x=a)^(x=b) y dx. Since dx = (dx/dt) dt, changing the variable of integration from x to the parameter t gives the Parametric Area Formula: Area = ∫_(t=t₁)^(t=t₂) y(t) · (dx/dt) dt, where t₁ and t₂ are the parameter values corresponding to x = a and x = b. Be careful to convert the x-limits into t-limits and check whether t₁ > t₂ (which may introduce a minus sign that swaps the limits).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
How do you calculate dy/dx for a curve defined parametrically by x = f(t) and y = g(t)?
On a parametric curve, what is the condition for a tangent parallel to the x-axis vs a tangent parallel to the y-axis?
How do you eliminate θ from x = 3 cos θ and y = 2 sin(2θ) for 0 ≤ θ ≤ π?
What is the formula for the area under a parametric curve between parameter values t = t₁ and t = t₂?
If x = t² - 1 and y = 2t for -1 ≤ t ≤ 3, what are the domain and range of the Cartesian function?
How do you find the coordinates where a parametric curve x = t², y = t³ - 4t meets the line y = x - 4?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A curve has parametric equations x = 2 sin t, y = cos(2t) + 3 sin t for 0 ≤ t ≤ 2π. Find a Cartesian equation for the curve in the form y = f(x) and state its domain. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Use the double-angle identity cos(2t) ≡ 1 - 2 sin²t to express y purely in terms of sin t: y = 1 - 2 sin²t + 3 sin t [1 mark]. 2. Rearrange x = 2 sin t to write sin t = x / 2 [1 mark]. 3. Substitute sin t = x / 2 into the expression for y: y = 1 - 2(x / 2)² + 3(x / 2) [1 mark]. 4. Expand and simplify: y = 1 + (3/2)x - (1/2)x² [1 mark]. 5. Since -1 ≤ sin t ≤ 1 for 0 ≤ t ≤ 2π, the domain of x = 2 sin t is -2 ≤ x ≤ 2 [1 mark].
- Q2
A curve C is defined by the parametric equations x = t² + 1, y = 4t - t³ for t ∈ ℝ. Find the coordinates of the stationary points on C, and find the equation of the normal to C at the point where t = 2. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Differentiate x and y with respect to t: dx/dt = 2t and dy/dt = 4 - 3t²; hence dy/dx = (4 - 3t²) / (2t) [1 mark]. 2. At stationary points, dy/dx = 0 → 4 - 3t² = 0 → t = ±2 / √3 [1 mark]. 3. Substitute t = ±2/√3 into (x, y): x = (4/3) + 1 = 7/3, and y = ±8/√3 ∓ 8/(3√3) = ±16√3 / 9; stationary points are (7/3, 16√3 / 9) and (7/3, -16√3 / 9) [1 mark]. 4. At t = 2, find the point and tangent gradient: x = 2² + 1 = 5, y = 4(2) - 2³ = 0, and dy/dx = (4 - 12) / 4 = -2 [1 mark]. 5. Find the normal gradient m_N = -1 / (-2) = +1/2 [1 mark]. 6. Equation of normal through (5, 0): y - 0 = (1/2)(x - 5) → x - 2y - 5 = 0 (or y = (1/2)x - 5/2) [1 mark].
- Q3
A curve is given parametrically by x = 3t - sin t, y = 1 - cos t for 0 ≤ t ≤ π. Show that the area under the curve between x = 0 and x = 3π - 0 is given by ∫₀^π (3 - 4 cos t + cos²t) dt, and evaluate this exact area. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Find dx/dt: dx/dt = 3 - cos t [1 mark]. 2. Convert the x-limits into t-limits: when x = 0, t = 0; when x = 3π, t = π [1 mark]. 3. Substitute y and dx/dt into the parametric area formula Area = ∫ y (dx/dt) dt: Area = ∫₀^π (1 - cos t)(3 - cos t) dt = ∫₀^π (3 - 4 cos t + cos²t) dt [1 mark]. 4. Use the double-angle identity cos²t ≡ (1 + cos 2t) / 2 to rewrite the integrand: 3 - 4 cos t + 1/2 + (1/2) cos(2t) = 7/2 - 4 cos t + (1/2) cos(2t) [1 mark]. 5. Integrate term by term with respect to t: [(7/2)t - 4 sin t + (1/4) sin(2t)]₀^π [1 mark]. 6. Evaluate at the limits t = π and t = 0: [(7π/2 - 0 + 0) - (0)] = 7π / 2 [1 mark].
- Q4
A curve has parametric equations x = 4t / (1 + t²), y = 4t² / (1 + t²) for t ∈ ℝ. Show that all points on the curve satisfy the Cartesian equation x² + (y - 2)² = 4. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Simplify y - 2 in terms of t by putting over the common denominator (1 + t²): y - 2 = [4t² - 2(1 + t²)] / (1 + t²) = (2t² - 2) / (1 + t²) [1 mark]. 2. Square both x and (y - 2): x² = 16t² / (1 + t²)² and (y - 2)² = (4t⁴ - 8t² + 4) / (1 + t²)² [1 mark]. 3. Add x² + (y - 2)² over the common denominator (1 + t²)²: [16t² + 4t⁴ - 8t² + 4] / (1 + t²)² = (4t⁴ + 8t² + 4) / (1 + t²)² [1 mark]. 4. Factorise the numerator as 4(t⁴ + 2t² + 1) = 4(1 + t²)² and cancel (1 + t²)² to obtain x² + (y - 2)² = 4 [1 mark].
Common Exam Questions (4)
Why did I lose a mark when finding the domain of x = t² - 4t for -1 ≤ t ≤ 3?
If you only substitute the endpoints t = -1 (giving x = 5) and t = 3 (giving x = -3), you miss the turning point of the quadratic x(t) at t = 2, where x reaches its true minimum of 2² - 4(2) = -4! Always check for stationary points of x(t) and y(t) inside the t-interval when finding domain and range.
When integrating parametrically, why did my integral come out negative?
As x increases from left to right (a to b), the parameter t might actually decrease (so t₁ at x = a is larger than t₂ at x = b), making dx/dt negative. When you swap the limits so the smaller t-value is at the bottom, multiplying by -1 makes the integrand and final area positive.
How do I find the second derivative d²y/dx² for a parametric curve (if asked in an extension question)?
Never divide d²y/dt² by d²x/dt²! By the chain rule, d²y/dx² = d/dx(dy/dx) = [d/dt(dy/dx)] ÷ (dx/dt). You must differentiate your simplified expression for dy/dx with respect to t, and then divide by dx/dt.
What does a parametric curve look like if t is restricted to 0 ≤ t ≤ π instead of 0 ≤ t ≤ 2π for x = r cos t, y = r sin t?
For 0 ≤ t ≤ π, sin t ≥ 0 so y ≥ 0: the curve is only the upper semicircle rather than the full circle.
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