7. Parametric Differentiation

Video coming soon4 worked examples

Theory

When a curve is given by $x$ and $y$ in terms of a parameter $t$ , each value of $t$ produces one point on the curve. Eliminating the parameter between the two equations gives the constraint equation — the ordinary relationship between $x$ and $y$ . Several standard forms are worth recognising:

\begin{aligned} x = at^2,\ y = 2at &\ \longrightarrow\ y^2 = 4ax &&\text{parabola} \\ x = r\cos\theta,\ y = r\sin\theta &\ \longrightarrow\ x^2 + y^2 = r^2 &&\text{circle} \\ x = a\cos\theta,\ y = b\sin\theta &\ \longrightarrow\ \tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1 &&\text{ellipse} \\ x = a\sec\theta,\ y = b\tan\theta &\ \longrightarrow\ \tfrac{x^2}{a^2} - \tfrac{y^2}{b^2} = 1 &&\text{hyperbola} \\ x = ct,\ y = \tfrac{c}{t} &\ \longrightarrow\ xy = c^2 &&\text{rectangular hyperbola} \end{aligned}

Usually, though, we differentiate without eliminating anything. The gradient comes from the chain rule:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

The second derivative differentiates $\frac{dy}{dx}$ with respect to $t$ , then divides by $\frac{dx}{dt}$ again:

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{dx/dt}

The Golden Rule: divide the $t$ -derivatives to get $\frac{dy}{dx}$ . For the second derivative, differentiate $\frac{dy}{dx}$ with respect to $t$ and divide by $\frac{dx}{dt}$ once more.

⚠️ Common Examiner Traps

The second derivative is NOT $\frac{d^2y/dt^2}{d^2x/dt^2}$ : you must divide by $\frac{dx}{dt}$ , not by $\frac{d^2x}{dt^2}$ .
Dividing vs differentiating: $\frac{dy}{dx}$ is a quotient of the two $t$ -derivatives.
Chain rule again: the second derivative needs another division by $\frac{dx}{dt}$ .

Worked examples

Example 1

A curve is defined by $x = t^2$ , $y = t^3$ . Find $\dfrac{dy}{dx}$ .

Step 1: Differentiate each with respect to $t$ :

\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2

Step 2: Divide:

\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t

Example 2

For the curve $x = t^2$ , $y = t^3$ , find $\dfrac{d^2y}{dx^2}$ .

Step 1: From the previous result, $\frac{dy}{dx} = \frac{3}{2}t$ . Differentiate this with respect to $t$ :

\frac{d}{dt}\!\left(\frac{3}{2}t\right) = \frac{3}{2}

Step 2: Divide by $\frac{dx}{dt} = 2t$ :

\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}

Example 3

A curve has parametric equations $x = 2\cos\theta$ , $y = 5\sin\theta$ . Find its constraint equation and name the curve.

Step 1: Rearrange each equation to isolate the trigonometric function:

\cos\theta = \frac{x}{2}, \qquad \sin\theta = \frac{y}{5}

Step 2: Eliminate $\theta$ using the identity $\cos^2\theta + \sin^2\theta = 1$ :

\left(\frac{x}{2}\right)^2 + \left(\frac{y}{5}\right)^2 = 1

Step 3: Write it in standard form:

\frac{x^2}{4} + \frac{y^2}{25} = 1

This is an ellipse, centred on the origin, with semi-axes $2$ along $x$ and $5$ along $y$ .

Example 4

A curve is defined by $x = 2t$ , $y = t^2 - 1$ . Find the equation of the tangent at the point where $t = 3$ .

Step 1: Differentiate each equation with respect to $t$ and divide:

\frac{dx}{dt} = 2, \qquad \frac{dy}{dt} = 2t \qquad\implies\qquad \frac{dy}{dx} = \frac{2t}{2} = t

Step 2: Evaluate the gradient at $t = 3$ :

\left.\frac{dy}{dx}\right|_{t=3} = 3

Step 3: Find the coordinates of the point by substituting $t = 3$ into the parametric equations:

x = 2(3) = 6, \qquad y = 3^2 - 1 = 8

Step 4: Use $y - b = m(x - a)$ with $m = 3$ at $(6, 8)$ :

y - 8 = 3(x - 6) \implies y = 3x - 10