Topics

Parametric differentiation

Covers AS91578

What a parametric equation is

Instead of $y$ being given directly in terms of $x$ , both are given in terms of a third variable called a parameter, usually $t$ or $\theta$ :

$x = f(t) \qquad y = g(t)$

As $t$ varies, the point $(x, y)$ traces out a curve.
Parametric form can describe curves that fail the vertical line test — circles, ellipses and loops — which no single $y = f(x)$ equation can.
Worked example of the idea: $x = t^2$ , $y = 2t$ traces a parabola opening to the right. At $t = 1$ the point is $(1, 2)$ ; at $t = 2$ it is $(4, 4)$ .

The first derivative

The chain rule links the three variables:

$\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\;\dfrac{dy}{dt}\;}{\dfrac{dx}{dt}}$

Differentiate each equation with respect to $t$ , then divide.
Worked through for $x = t^2$ , $y = 2t$ :
- $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 2$
- $\dfrac{dy}{dx} = \dfrac{2}{2t} = \dfrac{1}{t}$
Note that the answer is in terms of $t$ , not $x$ . That is normal and usually what is wanted — to find the gradient at a particular point, substitute the value of $t$ .

The second derivative — where marks are lost

$\dfrac{d^2y}{dx^2}$ is NOT $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ . That is the single most common error in this topic, and the 2025 report names it explicitly.
The correct rule differentiates $\dfrac{dy}{dx}$ with respect to $x$ , which needs the chain rule again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx} = \frac{\;\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\;}{\dfrac{dx}{dt}}$

The procedure:
- Find $\dfrac{dy}{dx}$ in terms of $t$ .
- Differentiate that expression with respect to $t$ .
- Divide by $\dfrac{dx}{dt}$ .
Worked through, continuing $x = t^2$ , $y = 2t$ with $\dfrac{dy}{dx} = t^{-1}$ :
- $\dfrac{d}{dt}\left(t^{-1}\right) = -t^{-2}$
- Divide by $\dfrac{dx}{dt} = 2t$ : $\dfrac{d^2y}{dx^2} = \dfrac{-t^{-2}}{2t} = -\dfrac{1}{2t^3}$
The 2025 report: candidates "struggled to find the correct second derivative for the parametric functions ... by omitting to multiply by $\frac{dt}{dx}$ , having differentiated with respect to the parameter. This resulted in an answer supported by an invalid method."

Tangents and normals to parametric curves

The method is unchanged, except that the point and the gradient both come from a value of $t$ :
- Substitute the given $t$ into $x = f(t)$ and $y = g(t)$ to get the point.
- Substitute the same $t$ into $\dfrac{dy}{dx}$ to get the gradient.
- Use $y - y_1 = m(x - x_1)$ .
For the normal, use the perpendicular gradient $-\dfrac{1}{m}$ .

Stationary points on parametric curves

$\dfrac{dy}{dx} = 0$ requires the numerator to vanish:

$\frac{dy}{dt} = 0 \quad \text{(with } \frac{dx}{dt} \neq 0\text{)}$

A vertical tangent occurs where $\dfrac{dx}{dt} = 0$ while $\dfrac{dy}{dt} \neq 0$ — the gradient is undefined there.
Note that both conditions are read off the two separate $t$ -derivatives, which is quicker than forming $\frac{dy}{dx}$ first.

Common parametric curves

Circle of radius $r$ : $x = r\cos t$ , $y = r\sin t$
Ellipse: $x = a\cos t$ , $y = b\sin t$
Parabola: $x = at^2$ , $y = 2at$
These reappear in the conic sections internal (AS91573), where the same $\frac{dy}{dx}$ calculation is used to find tangents.

Tips and Tricks

The second derivative is the trap on this topic. Write the formula out before you start: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}$ Then follow it mechanically. The 2025 report confirms this exact omission cost candidates the question, because the method itself was invalid — not merely the arithmetic.

Differentiate $\frac{dy}{dx}$ , not $\frac{dy}{dt}$ , in the second step. You must start from the expression you already found for the gradient.

Answers in terms of $t$ are fine unless the question asks for $x$ . Do not waste time trying to eliminate the parameter — that is usually harder than the question intended.

For a tangent, get both the point and the gradient from the same value of $t$ . Mixing $t$ -values is a quiet error that produces a plausible-looking wrong line.

The report notes candidates "were confused when the problem connected their differentiation and co-ordinate geometry knowledge". Parametric tangent questions are exactly that combination, so practise finishing the job — writing the line's equation in gradient-intercept or general form, fully evaluated, not left with substituted values.

Worked ExampleFirst and second derivatives of a parametric curve

A curve is given by $x = 3\cos t$ and $y = 2\sin t$ . Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ in terms of $t$ , and find the equation of the tangent at $t = \dfrac{\pi}{4}$ .

Step 1 — Differentiate each equation with respect to $t$

$\frac{dx}{dt} = -3\sin t \qquad \frac{dy}{dt} = 2\cos t$

Note the minus sign on $\frac{dx}{dt}$ — it comes from differentiating cosine, and it survives into every later step.

Step 2 — Form the first derivative

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t}{-3\sin t} = -\frac{2}{3}\cot t$

Step 3 — Set up the second derivative correctly

Write the rule out before computing anything:

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

Step 4 — Differentiate $\dfrac{dy}{dx}$ with respect to $t$

Using $\dfrac{d}{dt}(\cot t) = -\csc^2 t$ :

$\frac{d}{dt}\left(-\frac{2}{3}\cot t\right) = -\frac{2}{3} \times \left(-\csc^2 t\right) = \frac{2}{3}\csc^2 t$

Step 5 — Divide by $\dfrac{dx}{dt}$

This is the step candidates omit:

$\frac{d^2y}{dx^2} = \frac{\frac{2}{3}\csc^2 t}{-3\sin t} = -\frac{2}{9}\,\frac{\csc^2 t}{\sin t}$

Writing $\csc^2 t = \dfrac{1}{\sin^2 t}$ :

$\frac{d^2y}{dx^2} = -\frac{2}{9\sin^3 t}$

Step 6 — Find the point at $t = \dfrac{\pi}{4}$

Substitute into the original parametric equations, using $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ :

$x = 3 \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \qquad y = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}$

Step 7 — Find the gradient at the same $t$

$m = -\frac{2}{3}\cot\frac{\pi}{4} = -\frac{2}{3} \times 1 = -\frac{2}{3}$

Step 8 — Write the equation of the tangent

Using $y - y_1 = m(x - x_1)$ :

$y - \sqrt{2} = -\frac{2}{3}\left(x - \frac{3\sqrt{2}}{2}\right)$

Expand and simplify:

$y - \sqrt{2} = -\frac{2}{3}x + \sqrt{2}$

$y = -\frac{2}{3}x + 2\sqrt{2}$

Step 9 — Check

The curve is the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ . Substituting the point: $\frac{9/2}{9} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$ ✓ the point is on the curve.

Parametric differentiation

Covers AS91578

What a parametric equation is

Instead of $y$ being given directly in terms of $x$ , both are given in terms of a third variable called a parameter, usually $t$ or $\theta$ :

$x = f(t) \qquad y = g(t)$

As $t$ varies, the point $(x, y)$ traces out a curve.
Parametric form can describe curves that fail the vertical line test — circles, ellipses and loops — which no single $y = f(x)$ equation can.
Worked example of the idea: $x = t^2$ , $y = 2t$ traces a parabola opening to the right. At $t = 1$ the point is $(1, 2)$ ; at $t = 2$ it is $(4, 4)$ .

The first derivative

The chain rule links the three variables:

$\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\;\dfrac{dy}{dt}\;}{\dfrac{dx}{dt}}$

Differentiate each equation with respect to $t$ , then divide.
Worked through for $x = t^2$ , $y = 2t$ :
- $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 2$
- $\dfrac{dy}{dx} = \dfrac{2}{2t} = \dfrac{1}{t}$
Note that the answer is in terms of $t$ , not $x$ . That is normal and usually what is wanted — to find the gradient at a particular point, substitute the value of $t$ .

The second derivative — where marks are lost

$\dfrac{d^2y}{dx^2}$ is NOT $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ . That is the single most common error in this topic, and the 2025 report names it explicitly.
The correct rule differentiates $\dfrac{dy}{dx}$ with respect to $x$ , which needs the chain rule again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx} = \frac{\;\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\;}{\dfrac{dx}{dt}}$

The procedure:
- Find $\dfrac{dy}{dx}$ in terms of $t$ .
- Differentiate that expression with respect to $t$ .
- Divide by $\dfrac{dx}{dt}$ .
Worked through, continuing $x = t^2$ , $y = 2t$ with $\dfrac{dy}{dx} = t^{-1}$ :
- $\dfrac{d}{dt}\left(t^{-1}\right) = -t^{-2}$
- Divide by $\dfrac{dx}{dt} = 2t$ : $\dfrac{d^2y}{dx^2} = \dfrac{-t^{-2}}{2t} = -\dfrac{1}{2t^3}$
The 2025 report: candidates "struggled to find the correct second derivative for the parametric functions ... by omitting to multiply by $\frac{dt}{dx}$ , having differentiated with respect to the parameter. This resulted in an answer supported by an invalid method."

Tangents and normals to parametric curves

The method is unchanged, except that the point and the gradient both come from a value of $t$ :
- Substitute the given $t$ into $x = f(t)$ and $y = g(t)$ to get the point.
- Substitute the same $t$ into $\dfrac{dy}{dx}$ to get the gradient.
- Use $y - y_1 = m(x - x_1)$ .
For the normal, use the perpendicular gradient $-\dfrac{1}{m}$ .

Stationary points on parametric curves

$\dfrac{dy}{dx} = 0$ requires the numerator to vanish:

$\frac{dy}{dt} = 0 \quad \text{(with } \frac{dx}{dt} \neq 0\text{)}$

A vertical tangent occurs where $\dfrac{dx}{dt} = 0$ while $\dfrac{dy}{dt} \neq 0$ — the gradient is undefined there.
Note that both conditions are read off the two separate $t$ -derivatives, which is quicker than forming $\frac{dy}{dx}$ first.

Common parametric curves

Circle of radius $r$ : $x = r\cos t$ , $y = r\sin t$
Ellipse: $x = a\cos t$ , $y = b\sin t$
Parabola: $x = at^2$ , $y = 2at$
These reappear in the conic sections internal (AS91573), where the same $\frac{dy}{dx}$ calculation is used to find tangents.

Tips and Tricks

Differentiate $\frac{dy}{dx}$ , not $\frac{dy}{dt}$ , in the second step. You must start from the expression you already found for the gradient.

Answers in terms of $t$ are fine unless the question asks for $x$ . Do not waste time trying to eliminate the parameter — that is usually harder than the question intended.

For a tangent, get both the point and the gradient from the same value of $t$ . Mixing $t$ -values is a quiet error that produces a plausible-looking wrong line.

Worked ExampleFirst and second derivatives of a parametric curve

A curve is given by $x = 3\cos t$ and $y = 2\sin t$ . Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ in terms of $t$ , and find the equation of the tangent at $t = \dfrac{\pi}{4}$ .

Step 1 — Differentiate each equation with respect to $t$

$\frac{dx}{dt} = -3\sin t \qquad \frac{dy}{dt} = 2\cos t$

Note the minus sign on $\frac{dx}{dt}$ — it comes from differentiating cosine, and it survives into every later step.

Step 2 — Form the first derivative

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t}{-3\sin t} = -\frac{2}{3}\cot t$

Step 3 — Set up the second derivative correctly

Write the rule out before computing anything:

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

Step 4 — Differentiate $\dfrac{dy}{dx}$ with respect to $t$

Using $\dfrac{d}{dt}(\cot t) = -\csc^2 t$ :

$\frac{d}{dt}\left(-\frac{2}{3}\cot t\right) = -\frac{2}{3} \times \left(-\csc^2 t\right) = \frac{2}{3}\csc^2 t$

Step 5 — Divide by $\dfrac{dx}{dt}$

This is the step candidates omit:

$\frac{d^2y}{dx^2} = \frac{\frac{2}{3}\csc^2 t}{-3\sin t} = -\frac{2}{9}\,\frac{\csc^2 t}{\sin t}$

Writing $\csc^2 t = \dfrac{1}{\sin^2 t}$ :

$\frac{d^2y}{dx^2} = -\frac{2}{9\sin^3 t}$

Step 6 — Find the point at $t = \dfrac{\pi}{4}$

Substitute into the original parametric equations, using $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ :

$x = 3 \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \qquad y = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}$

Step 7 — Find the gradient at the same $t$

$m = -\frac{2}{3}\cot\frac{\pi}{4} = -\frac{2}{3} \times 1 = -\frac{2}{3}$

Step 8 — Write the equation of the tangent

Using $y - y_1 = m(x - x_1)$ :

$y - \sqrt{2} = -\frac{2}{3}\left(x - \frac{3\sqrt{2}}{2}\right)$

Expand and simplify:

$y - \sqrt{2} = -\frac{2}{3}x + \sqrt{2}$

$y = -\frac{2}{3}x + 2\sqrt{2}$

Step 9 — Check

The curve is the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ . Substituting the point: $\frac{9/2}{9} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$ ✓ the point is on the curve.

Parametric differentiation

What a parametric equation is

The first derivative

The second derivative — where marks are lost

Tangents and normals to parametric curves

Stationary points on parametric curves

Common parametric curves

Step 1 — Differentiate each equation with respect to ttt

Step 2 — Form the first derivative

Step 3 — Set up the second derivative correctly

Step 4 — Differentiate dydx\dfrac{dy}{dx}dxdy​ with respect to ttt

Step 5 — Divide by dxdt\dfrac{dx}{dt}dtdx​

Step 6 — Find the point at t=π4t = \dfrac{\pi}{4}t=4π​

Step 7 — Find the gradient at the same ttt

Step 8 — Write the equation of the tangent

Step 9 — Check

Parametric differentiation

What a parametric equation is

The first derivative

The second derivative — where marks are lost

Tangents and normals to parametric curves

Stationary points on parametric curves

Common parametric curves

Step 1 — Differentiate each equation with respect to ttt

Step 2 — Form the first derivative

Step 3 — Set up the second derivative correctly

Step 4 — Differentiate dydx\dfrac{dy}{dx}dxdy​ with respect to ttt

Step 5 — Divide by dxdt\dfrac{dx}{dt}dtdx​

Step 6 — Find the point at t=π4t = \dfrac{\pi}{4}t=4π​

Step 7 — Find the gradient at the same ttt

Step 8 — Write the equation of the tangent

Step 9 — Check

Step 1 — Differentiate each equation with respect to $t$

Step 4 — Differentiate $\dfrac{dy}{dx}$ with respect to $t$

Step 5 — Divide by $\dfrac{dx}{dt}$

Step 6 — Find the point at $t = \dfrac{\pi}{4}$

Step 7 — Find the gradient at the same $t$

Step 1 — Differentiate each equation with respect to $t$

Step 4 — Differentiate $\dfrac{dy}{dx}$ with respect to $t$

Step 5 — Divide by $\dfrac{dx}{dt}$

Step 6 — Find the point at $t = \dfrac{\pi}{4}$

Step 7 — Find the gradient at the same $t$