Parametric equations flashcards

Test yourself one card at a time: say the answer, flip, then mark it. Cards you know drop out; the rest come back until you know them. Your progress stays in this browser.

Parametric equations of a curve
x = f(t), y = g(t): both coordinates given in terms of a third variable, the parameter t. Each value of t gives one point.
Parameter
The third variable (often t or θ) in terms of which x and y are both given; it may represent time or an angle, or nothing physical at all.
Cartesian equation (of a parametric curve)
An equation linking x and y only, found by eliminating the parameter.
Domain of the Cartesian form of a parametric curve
The set of x-values the curve actually reaches, found from x = f(t) over the allowed values of t.
Range of a parametric curve
The set of y-values the curve actually reaches, found from y = g(t) over the allowed values of t.
dy/dx for a parametric curve
dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0. (No — learn it)
Area under a parametric curve
Area = ∫ y (dx/dt) dt, with the limits changed to the values of t at the ends of the region. (No — learn it)
Identity linking sin t and cos t
sin²t + cos²t = 1. (No — learn it)
cos 2t in terms of sin t, and in terms of cos t
cos 2t = 1 − 2 sin²t = 2 cos²t − 1. (No — learn it)
Identity linking tan t and sec t
sec²t = 1 + tan²t (divide sin² + cos² = 1 by cos²). (No — derive it)
Identity linking cot t and cosec t
cosec²t = 1 + cot²t (divide sin² + cos² = 1 by sin²). (No — derive it)
∫ tan t dt
ln |sec t| + c. (Yes)
x = 3 sin t, y = 4 cos 2t: which part of y = 4 − 8x²/9 is the curve?
Only the arc with −3 ≤ x ≤ 3, because −1 ≤ sin t ≤ 1. The rest of the parabola is never reached.
How do you show direction on a sketch of a parametric curve?
Plot points in order of increasing t and add arrows along the curve in that direction.
x = t², y = t³ − 3t: why can this curve cross itself?
Different t-values can give the same point: t = √3 and t = −√3 both give (3, 0), with different gradients.
x = a cos θ, y = b sin θ describes…
an ellipse x²/a² + y²/b² = 1, centre O (a circle if a = b).
x = p + r cos θ, y = q + r sin θ describes…
a circle with centre (p, q) and radius r.
Where is the tangent to a parametric curve horizontal?
Where dy/dt = 0 (and dx/dt ≠ 0).
Where is the tangent to a parametric curve vertical?
Where dx/dt = 0 (and dy/dt ≠ 0).
Why is the area formula ∫ y (dx/dt) dt and not ∫ y dt?
Area is ∫ y dx. Changing the variable to t is a substitution, so dx becomes (dx/dt) dt.
Why can a parametric area integral come out negative?
If x decreases as t increases, dx/dt < 0; with the smaller t as the lower limit the sign flips. Put the t for the left-hand end as the lower limit.
Why must you state a domain with the Cartesian equation?
The parameter's range can restrict x (and y), so the Cartesian equation may describe extra points not on the curve.
Why is dy/dx = (dy/dt) ÷ (dx/dt) true?
Chain rule: dy/dx = (dy/dt) × (dt/dx), and dt/dx = 1 ÷ (dx/dt).
Cartesian equation of x = 3t − 1, y = 2/t?
t = (x + 1)/3, so y = 6/(x + 1).
Cartesian equation of x = 1 + 2 cos t, y = 2 sin t − 3?
cos t = (x − 1)/2, sin t = (y + 3)/2, so (x − 1)² + (y + 3)² = 4: circle, centre (1, −3), radius 2.
x = eᵗ, y = e²ᵗ + 1, t ∈ ℝ: Cartesian equation, domain and range?
y = x² + 1 with x > 0 (since eᵗ > 0), so y > 1.
x = sin t, y = 4 cos 2t: Cartesian equation?
cos 2t = 1 − 2 sin²t, so y = 4 − 8x², for −1 ≤ x ≤ 1.
x = t³, y = t² + t: gradient at t = 1?
dy/dx = (2t + 1)/(3t²) = 3/3 = 1.
x = 4 cos θ, y = 2 sin θ: dy/dx, and its value at θ = π/4?
dy/dx = 2 cos θ/(−4 sin θ) = −½ cot θ; at θ = π/4 it is −½.
Area between x = t² + 2t, y = 3 − t (0 ≤ t ≤ 3) and the axes?
∫₀³ (3 − t)(2t + 2) dt = ∫₀³ (6 + 4t − 2t²) dt = 18.
Area under x = tan t, y = sin 2t for 0 ≤ t ≤ π/4?
∫ sin 2t sec²t dt = ∫ 2 tan t dt = [2 ln sec t]₀^(π/4) = ln 2.
x = 4 cos t, 0 ≤ t ≤ π/2: t-limits for the region from x = 0 to x = 2?
x = 0 at t = π/2 and x = 2 at t = π/3, so the integral runs from π/2 to π/3 (then swap and change sign).
Finding the parameter of a given point
Solve one equation for t, then check each possible value in the other: x = t² = 4 allows t = ±2, but only one gives the right y.
Eliminating θ from x = 2 + 3 cos θ, y = 3 sin θ
Isolate first: cos θ = (x − 2)/3, sin θ = y/3, then (x − 2)² + y² = 9 — not x² + y² = 9.
Limits in a parametric area integral
Convert every x-limit to its t-value and write the change as a line of working; never put x-values into a t-integral.
Tangent equation to a parametric curve
Substitute the parameter value to get a number for the gradient; the final equation must contain only x and y.
y² = … with a restricted parameter
Use the sign of y on the curve (e.g. sec t > 0 for 0 ≤ t < π/2) to choose the positive or negative root.

Practise this topicNotes Questions (28)Test builder