Parametric Equations

A2 · P3 · 14 min

A curve can be described by giving $x$ and $y$ separately as functions of a third variable, the parameter, usually $t$ or $\theta$ . Think of $t$ as time and the point $\big(x(t), y(t)\big)$ as a particle moving across the page: each value of $t$ gives one position, and as $t$ runs through its values the particle traces the curve. In P3 you must find the gradient of such a curve using the chain rule, and use it for tangents, normals, and points where the tangent is horizontal or vertical. A parametric differentiation question appears on most papers, typically worth five to seven marks.

Parametric curves

Definition

A curve is given parametrically when its coordinates are written as $x = f(t)$ , $y = g(t)$ for a parameter $t$ taking values in some interval. Each value of $t$ gives exactly one point.

Example: $x = 3\cos\theta$ , $y = 2\sin\theta$ for $0 \le \theta < 2\pi$ . At $\theta = 0$ the point is $(3, 0)$ ; at $\theta = \tfrac{\pi}{2}$ , $(0, 2)$ ; at $\theta = \pi$ , $(-3, 0)$ ; at $\theta = \tfrac{3\pi}{2}$ , $(0, -2)$ . The curve is an ellipse, traced anticlockwise.

(3 cos(t), 2 sin(t))

The ellipse $x = 3\cos\theta$ , $y = 2\sin\theta$ . Its Cartesian equation is $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ .

Parametric form is powerful because it describes curves that are hard or impossible to write as $y = f(x)$ : closed loops, curves that cross themselves, curves with vertical tangents. It also keeps the algebra simple: two easy equations instead of one awkward one.

Differentiating parametric equations

The question is: as $t$ changes, how fast does $y$ change compared with $x$ ? Over a small change $\delta t$ in the parameter, $x$ changes by about $\dfrac{dx}{dt}\,\delta t$ and $y$ by about $\dfrac{dy}{dt}\,\delta t$ . The gradient of the chord is the ratio of those changes:

\frac{\delta y}{\delta x} \approx \frac{\frac{dy}{dt}\,\delta t}{\frac{dx}{dt}\,\delta t} = \frac{dy/dt}{dx/dt}

Formally it is the chain rule, $\dfrac{dy}{dx} = \dfrac{dy}{dt}\times\dfrac{dt}{dx}$ , together with $\dfrac{dt}{dx} = 1\Big/\dfrac{dx}{dt}$ .

Gradient of a parametric curve

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{dy}{dt} \div \frac{dx}{dt}, \qquad \text{provided } \frac{dx}{dt} \ne 0

The answer is in terms of the parameter $t$ .

A sanity check: $x = 2t$ , $y = 6t$ is the straight line $y = 3x$ . The formula gives $\dfrac{6}{2} = 3$ . Correct.

Tangent or normal to a parametric curve

Differentiate: find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ .
Divide: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , simplified, in terms of $t$ .
Substitute the given parameter value into $\dfrac{dy}{dx}$ to get the gradient $m$ (or $-\dfrac{1}{m}$ for the normal).
Substitute the same parameter value into $x = f(t)$ and $y = g(t)$ to get the point.
Write the line: $y - y_1 = m(x - x_1)$ .

If the question gives a point $(x, y)$ instead of a value of $t$ , first find the value of $t$ that gives that point, checking it satisfies both equations.

Horizontal and vertical tangents

From $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ :

Tangent parallel to the $x$ -axis (a stationary point): $\dfrac{dy}{dt} = 0$ with $\dfrac{dx}{dt} \ne 0$ .
Tangent parallel to the $y$ -axis: $\dfrac{dx}{dt} = 0$ with $\dfrac{dy}{dt} \ne 0$ .

Solve for $t$ , then find the coordinates. If both derivatives are zero at the same $t$ , the curve may have a cusp there and the formula gives no information.

Nature of a stationary point

The sign of $\dfrac{dy}{dx}$ tells you whether $y$ increases with $x$ . To test the nature of a stationary point at $t = t_0$ , look at the sign of $\dfrac{dy}{dx}$ for $t$ just below and just above $t_0$ , and check which way $x$ is moving. If $x$ increases with $t$ (that is, $\dfrac{dx}{dt} > 0$ ), then "before" in $t$ is also "to the left" in $x$ , and the usual sign test works. Second derivatives in parametric form are not required in P3.

Converting to a Cartesian equation

Although the syllabus focuses on differentiation, questions sometimes ask for the Cartesian equation, and it helps you recognise the curve. Eliminate the parameter:

Make $t$ the subject of one equation and substitute into the other: $x = 2t + 1$ , $y = t^2$ gives $t = \dfrac{x - 1}{2}$ and $y = \dfrac{(x - 1)^2}{4}$ .
For trigonometric parameters, isolate $\cos\theta$ and $\sin\theta$ and use $\cos^2\theta + \sin^2\theta = 1$ : $x = 3\cos\theta$ , $y = 2\sin\theta$ gives $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ .
Use other identities as needed: $x = \sec\theta$ , $y = \tan\theta$ gives $x^2 - y^2 = 1$ from $\sec^2\theta - \tan^2\theta = 1$ ; $x = \cos 2\theta$ , $y = \sin\theta$ gives $x = 1 - 2y^2$ from $\cos 2\theta = 1 - 2\sin^2\theta$ .

Worked examples

The syllabus curve

A curve has parametric equations $x = t - e^{2t}$ , $y = t + e^{2t}$ . Find the equation of the tangent at the point where $t = 0$ .

Solution

$\dfrac{dx}{dt} = 1 - 2e^{2t}$ , $\dfrac{dy}{dt} = 1 + 2e^{2t}$ , so

\frac{dy}{dx} = \frac{1 + 2e^{2t}}{1 - 2e^{2t}}

At $t = 0$ : $\dfrac{dy}{dx} = \dfrac{3}{-1} = -3$ . The point is $x = 0 - 1 = -1$ , $y = 0 + 1 = 1$ .

Tangent: $y - 1 = -3(x + 1)$ , i.e. $y = -3x - 2$ .

Tangent to an ellipse

The ellipse $x = 3\cos\theta$ , $y = 2\sin\theta$ . Find the equation of the tangent at the point where $\theta = \tfrac{\pi}{4}$ , in the form $ax + by = c$ .

Solution

$\dfrac{dx}{d\theta} = -3\sin\theta$ , $\dfrac{dy}{d\theta} = 2\cos\theta$ , so $\dfrac{dy}{dx} = -\dfrac{2\cos\theta}{3\sin\theta} = -\dfrac{2}{3}\cot\theta$ .

At $\theta = \tfrac{\pi}{4}$ : gradient $-\tfrac{2}{3}$ ; point $\left(\dfrac{3}{\sqrt{2}}, \dfrac{2}{\sqrt{2}}\right) = \left(\dfrac{3\sqrt{2}}{2}, \sqrt{2}\right)$ .

y - \sqrt{2} = -\frac{2}{3}\left(x - \frac{3\sqrt{2}}{2}\right) \quad\Rightarrow\quad 3y - 3\sqrt{2} = -2x + 3\sqrt{2} \quad\Rightarrow\quad 2x + 3y = 6\sqrt{2}

Horizontal and vertical tangents

A curve has parametric equations $x = t^2 + 2t$ , $y = t^3 - 12t$ . Find the coordinates of the points where the tangent is (a) parallel to the $x$ -axis, (b) parallel to the $y$ -axis.

Solution

$\dfrac{dx}{dt} = 2t + 2$ , $\dfrac{dy}{dt} = 3t^2 - 12$ .

(a) $\dfrac{dy}{dt} = 0$ : $t^2 = 4$ , $t = \pm 2$ . Neither makes $\dfrac{dx}{dt}$ zero.

$t = 2$ : $(4 + 4,\ 8 - 24) = (8, -16)$ .
$t = -2$ : $(4 - 4,\ -8 + 24) = (0, 16)$ .

(b) $\dfrac{dx}{dt} = 0$ : $t = -1$ , where $\dfrac{dy}{dt} = -9 \ne 0$ . Point: $(1 - 2,\ -1 + 12) = (-1, 11)$ .

A stationary point and its nature

A curve has parametric equations $x = 1 + \ln t$ , $y = t^2 - 4\ln t$ , for $t > 0$ . Find the exact coordinates of the stationary point and determine its nature.

Solution

$\dfrac{dx}{dt} = \dfrac{1}{t}$ , $\dfrac{dy}{dt} = 2t - \dfrac{4}{t}$ , so

\frac{dy}{dx} = \left(2t - \frac{4}{t}\right) \div \frac{1}{t} = 2t^2 - 4

Zero when $t^2 = 2$ , so $t = \sqrt{2}$ (as $t > 0$ ). Coordinates: $x = 1 + \ln\sqrt{2} = 1 + \tfrac{1}{2}\ln 2$ , $y = 2 - 4\ln\sqrt{2} = 2 - 2\ln 2$ .

Nature: $\dfrac{dx}{dt} = \tfrac{1}{t} > 0$ , so $x$ increases with $t$ . For $t$ slightly less than $\sqrt{2}$ , $\dfrac{dy}{dx} = 2t^2 - 4 < 0$ ; slightly more, $\dfrac{dy}{dx} > 0$ . The gradient goes from negative to positive as $x$ increases: a minimum at $\left(1 + \tfrac{1}{2}\ln 2,\ 2 - 2\ln 2\right)$ .

Exam-hard: where the normal meets the curve again

The curve $C$ has parametric equations $x = t^2$ , $y = 2t$ .

(a) Find the equation of the normal to $C$ at the point $P$ where $t = 2$ .

(b) The normal meets $C$ again at $Q$ . Find the coordinates of $Q$ .

Solution

(a) $\dfrac{dx}{dt} = 2t$ , $\dfrac{dy}{dt} = 2$ , so $\dfrac{dy}{dx} = \dfrac{1}{t}$ . At $t = 2$ : gradient $\tfrac{1}{2}$ , point $P(4, 4)$ . The normal has gradient $-2$ :

y - 4 = -2(x - 4) \quad\Rightarrow\quad y = 12 - 2x

(b) Substitute the parametric equations into the normal:

2t = 12 - 2t^2 \quad\Rightarrow\quad t^2 + t - 6 = 0 \quad\Rightarrow\quad (t + 3)(t - 2) = 0

$t = 2$ is $P$ itself, so $Q$ has $t = -3$ : $Q = (9, -6)$ .

y^2 = 4x y = 12 - 2x

The curve $x = t^2$ , $y = 2t$ (the parabola $y^2 = 4x$ ) and its normal at $P(4, 4)$ , which meets the curve again at $Q(9, -6)$ .

The technique in (b) is the key idea: substituting $x(t)$ and $y(t)$ into the equation of a line gives an equation in $t$ whose roots are the parameters of the intersection points. You already know one root (the point $P$ ), which makes factorising easy.

Exam-hard: a cardioid

The curve with parametric equations $x = 2\cos\theta - \cos 2\theta$ , $y = 2\sin\theta - \sin 2\theta$ , for $0 < \theta < \pi$ , is part of a cardioid.

(a) Show that $\dfrac{dy}{dx} = \dfrac{\cos\theta - \cos 2\theta}{\sin 2\theta - \sin\theta}$ .

(b) Find the exact coordinates of the point where the tangent is parallel to the $x$ -axis.

Solution

(a) $\dfrac{dx}{d\theta} = -2\sin\theta + 2\sin 2\theta$ and $\dfrac{dy}{d\theta} = 2\cos\theta - 2\cos 2\theta$ . Dividing and cancelling the $2$ :

\frac{dy}{dx} = \frac{2(\cos\theta - \cos 2\theta)}{2(\sin 2\theta - \sin\theta)} = \frac{\cos\theta - \cos 2\theta}{\sin 2\theta - \sin\theta}

(b) Need $\cos 2\theta = \cos\theta$ . Using $\cos 2\theta = 2\cos^2\theta - 1$ :

2\cos^2\theta - \cos\theta - 1 = 0 \quad\Rightarrow\quad (2\cos\theta + 1)(\cos\theta - 1) = 0

$\cos\theta = 1$ gives $\theta = 0$ , outside the interval. So $\cos\theta = -\tfrac{1}{2}$ , $\theta = \tfrac{2\pi}{3}$ .

Check the denominator: $\sin\tfrac{4\pi}{3} - \sin\tfrac{2\pi}{3} = -\tfrac{\sqrt{3}}{2} - \tfrac{\sqrt{3}}{2} = -\sqrt{3} \ne 0$ .

$x = 2\left(-\tfrac{1}{2}\right) - \cos\tfrac{4\pi}{3} = -1 + \tfrac{1}{2} = -\tfrac{1}{2}$ , and $y = 2\cdot\tfrac{\sqrt{3}}{2} - \sin\tfrac{4\pi}{3} = \sqrt{3} + \tfrac{\sqrt{3}}{2} = \tfrac{3\sqrt{3}}{2}$ .

The point is $\left(-\tfrac{1}{2}, \tfrac{3\sqrt{3}}{2}\right)$ .

(2 cos(t) - cos(2t), 2 sin(t) - sin(2t)) y = 3 sqrt(3) / 2

The full cardioid. The horizontal tangent $y = \tfrac{3\sqrt{3}}{2}$ touches the top of the curve at $\theta = \tfrac{2\pi}{3}$ .

Common mistakes

Multiplying instead of dividing. $\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$ , not $\dfrac{dy}{dt} \times \dfrac{dx}{dt}$ . Check with a straight line.
Dividing the wrong way up. $\dfrac{dx/dt}{dy/dt}$ is $\dfrac{dx}{dy}$ , the reciprocal of what you want.
Giving the answer as a value of $t$ . "Find the point" means give $(x, y)$ .
Substituting $t$ into the wrong place. The point comes from $x(t)$ and $y(t)$ , the gradient from $\dfrac{dy}{dx}$ . Do not put the $x$ -coordinate into an expression in $t$ .
Forgetting the chain rule in $\dfrac{d}{dt}e^{2t} = 2e^{2t}$ or $\dfrac{d}{d\theta}\sin 2\theta = 2\cos 2\theta$ .
Missing the restriction. If $t > 0$ or $0 < \theta < \pi$ is given, reject parameter values outside it.

Exam tip

Typical marks: one for each of $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ , one for dividing correctly, then marks for the gradient, the point and the line.
"Show that $\dfrac{dy}{dx} = \ldots$ ": simplify with identities as needed until you match the printed form exactly. In trigonometric questions, the identity step usually earns its own mark.
Leave equations of tangents and normals exact, in the form requested. If no form is specified, $y - y_1 = m(x - x_1)$ with exact values is acceptable, but simplifying is safer.
In "meets the curve again" questions, always substitute the parametric equations into the line. Do not try to find the Cartesian equation first unless asked.

Summary

Parametric curve: $x = f(t)$ , $y = g(t)$ ; each $t$ gives one point.
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , in terms of $t$ .
Tangent and normal: gradient and point both come from the same value of $t$ .
Horizontal tangent: $\dfrac{dy}{dt} = 0$ . Vertical tangent: $\dfrac{dx}{dt} = 0$ .
Nature of a stationary point: sign test on $\dfrac{dy}{dx}$ , checking the direction in which $x$ moves.
Cartesian equation: eliminate $t$ by substitution or an identity such as $\cos^2\theta + \sin^2\theta = 1$ .
Intersections with a line: substitute $x(t)$ , $y(t)$ into the line and solve for $t$ .

Practice

Question

A curve has $x = 3t^2$ , $y = 6t$ . Find $\dfrac{dy}{dx}$ in terms of $t$ and the equation of the tangent at $t = 1$ .
A curve has $x = \ln t$ , $y = t^2 - 1$ for $t > 0$ . Find $\dfrac{dy}{dx}$ in terms of $t$ , and the exact coordinates of the point where the gradient is $8$ .
A curve has $x = \cos 2\theta$ , $y = \sin\theta$ . Show that $\dfrac{dy}{dx} = -\dfrac{1}{4\sin\theta}$ , and find the Cartesian equation of the curve.
A curve has $x = e^t$ , $y = te^{-t}$ . Find $\dfrac{dy}{dx}$ and the exact coordinates of the stationary point.
A curve has $x = t + \dfrac{1}{t}$ , $y = t - \dfrac{1}{t}$ for $t > 0$ . Find the point where the tangent is parallel to the $y$ -axis, and show that the Cartesian equation is $x^2 - y^2 = 4$ .
A curve has $x = 2\theta - \sin 2\theta$ , $y = 1 - \cos 2\theta$ , for $0 < \theta < \pi$ . Show that $\dfrac{dy}{dx} = \cot\theta$ , and find the equation of the tangent at $\theta = \tfrac{\pi}{4}$ .
The curve $x = t^3$ , $y = t^2$ . Find the equation of the normal at the point where $t = 1$ , and show that this normal does not meet the curve again.
A curve has $x = e^{2t} - 2t$ , $y = 4e^t$ . Show that $\dfrac{dy}{dx} = \dfrac{2e^t}{e^{2t} - 1}$ , and find the exact value of $t$ at the point where the gradient is $1$ .
A curve has $x = 2\sin^2\theta$ , $y = 3\tan\theta$ , for $0 < \theta < \tfrac{\pi}{2}$ . Show that $\dfrac{dy}{dx} = \dfrac{3}{4\sin\theta\cos^3\theta}$ . Find the equation of the tangent at the point where $\theta = \tfrac{\pi}{4}$ , and the coordinates of the point where it meets the $x$ -axis.

Answers

$\dfrac{dy}{dx} = \dfrac{6}{6t} = \dfrac{1}{t}$ . At $t = 1$ : gradient $1$ , point $(3, 6)$ . Tangent: $y = x + 3$ .
$\dfrac{dy}{dx} = \dfrac{2t}{1/t} = 2t^2$ . $2t^2 = 8 \Rightarrow t = 2$ (as $t > 0$ ). Point: $(\ln 2, 3)$ .
$\dfrac{dx}{d\theta} = -2\sin 2\theta = -4\sin\theta\cos\theta$ , $\dfrac{dy}{d\theta} = \cos\theta$ . So $\dfrac{dy}{dx} = \dfrac{\cos\theta}{-4\sin\theta\cos\theta} = -\dfrac{1}{4\sin\theta}$ . Cartesian: $x = 1 - 2\sin^2\theta = 1 - 2y^2$ .
$\dfrac{dx}{dt} = e^t$ , $\dfrac{dy}{dt} = e^{-t} - te^{-t} = e^{-t}(1 - t)$ . So $\dfrac{dy}{dx} = (1 - t)e^{-2t}$ . Zero at $t = 1$ : the point $\left(e, e^{-1}\right)$ .
$\dfrac{dx}{dt} = 1 - \dfrac{1}{t^2}$ , $\dfrac{dy}{dt} = 1 + \dfrac{1}{t^2}$ . Vertical tangent: $\dfrac{dx}{dt} = 0 \Rightarrow t = 1$ (as $t > 0$ ), where $\dfrac{dy}{dt} = 2 \ne 0$ . Point: $(2, 0)$ . Cartesian: $x^2 - y^2 = \left(t + \tfrac{1}{t}\right)^2 - \left(t - \tfrac{1}{t}\right)^2 = 4$ .
$\dfrac{dx}{d\theta} = 2 - 2\cos 2\theta = 4\sin^2\theta$ , $\dfrac{dy}{d\theta} = 2\sin 2\theta = 4\sin\theta\cos\theta$ . So $\dfrac{dy}{dx} = \dfrac{4\sin\theta\cos\theta}{4\sin^2\theta} = \cot\theta$ . At $\theta = \tfrac{\pi}{4}$ : gradient $1$ , point $\left(\tfrac{\pi}{2} - 1, 1\right)$ . Tangent: $y - 1 = x - \tfrac{\pi}{2} + 1$ , i.e. $y = x + 2 - \tfrac{\pi}{2}$ .
$\dfrac{dy}{dx} = \dfrac{2t}{3t^2} = \dfrac{2}{3t}$ . At $t = 1$ : gradient $\tfrac{2}{3}$ , point $(1, 1)$ ; normal gradient $-\tfrac{3}{2}$ . Normal: $y - 1 = -\tfrac{3}{2}(x - 1)$ , i.e. $3x + 2y = 5$ . Substituting: $3t^3 + 2t^2 - 5 = 0$ , which factorises as $(t - 1)(3t^2 + 5t + 5) = 0$ . The quadratic has discriminant $25 - 60 < 0$ , so $t = 1$ is the only intersection.
$\dfrac{dx}{dt} = 2e^{2t} - 2$ , $\dfrac{dy}{dt} = 4e^t$ , so $\dfrac{dy}{dx} = \dfrac{4e^t}{2e^{2t} - 2} = \dfrac{2e^t}{e^{2t} - 1}$ . Gradient $1$ : with $u = e^t$ , $u^2 - 2u - 1 = 0$ , $u = 1 \pm \sqrt{2}$ . Since $u > 0$ , $u = 1 + \sqrt{2}$ and $t = \ln(1 + \sqrt{2})$ .
$\dfrac{dx}{d\theta} = 4\sin\theta\cos\theta$ , $\dfrac{dy}{d\theta} = 3\sec^2\theta$ , so $\dfrac{dy}{dx} = \dfrac{3}{\cos^2\theta}\cdot\dfrac{1}{4\sin\theta\cos\theta} = \dfrac{3}{4\sin\theta\cos^3\theta}$ . At $\theta = \tfrac{\pi}{4}$ : $\sin\theta\cos^3\theta = \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{2\sqrt{2}} = \tfrac{1}{4}$ , so the gradient is $3$ . Point: $x = 2\cdot\tfrac{1}{2} = 1$ , $y = 3$ . Tangent: $y - 3 = 3(x - 1)$ , i.e. $y = 3x$ . It meets the $x$ -axis at the origin $(0, 0)$ .

Parametric curves

Differentiating parametric equations

Horizontal and vertical tangents

Nature of a stationary point

Converting to a Cartesian equation

Worked examples

Practice

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