Derivatives of parametric functions

Two functions, one parameter. A parametric curve hands you $x$ and $y$ separately, each as a function of a third variable $t$. Neither equation alone describes the curve; together, as $t$ sweeps through its range, the point $(x(t), y(t))$ traces the path.

Why not just eliminate $t$? For simple curves — say $x = at^2$, $y = 2at$ — you could solve the first equation for $t$ and substitute into the second to recover an explicit $y(x)$. But for curves built from trigonometric expressions in $t$, this elimination is often messy or impossible in closed form. Differentiating directly through $t$ sidesteps that difficulty entirely.

The governing formula. Provided $f$ and $g$ are differentiable and $\dfrac{dx}{dt} \neq 0$, $y$ can be regarded as a function of $x$ implicitly, through $t$, and

where: $dy/dt$ is the rate of change of $y$ with respect to the parameter, $dx/dt$ is the rate of change of $x$ with respect to the same parameter, and the ratio of the two gives the actual slope of the curve at that point — generally still expressed in terms of $t$, which is the expected form of the answer.

Reading the geometry. As the parameter $t$ increases by a small amount, the point $(x,y)$ shifts by $(dx, dy)$; the slope of the little displacement is exactly $dy/dx$, and writing both displacements through their common rate of change with $t$ is what makes the ratio formula work.

Where this leads next. Once $\dfrac{dy}{dx}$ is expressed in terms of $t$, it is itself a function of $t$, and the syllabus's very next topic asks for the second derivative $\dfrac{d^2y}{dx^2}$ of a parametric curve. That calculation reuses exactly the same idea one level up: differentiate $\dfrac{dy}{dx}$ with respect to $t$, then divide by $\dfrac{dx}{dt}$ again, rather than differentiating the messy ratio directly with respect to $x$. Seeing parametric differentiation as "one clean ratio, repeatable" is what makes that later topic straightforward instead of intimidating.

If $x = at^2$ and $y = 2at$ — the standard parametrisation of a parabola — find $\dfrac{dy}{dx}$, then use it to write the equation of the tangent at $t = 2$, taking $a = 1$.

Step 1. Differentiate $x$ with respect to $t$: $\dfrac{dx}{dt} = 2at$.

Step 2. Differentiate $y$ with respect to $t$: $\dfrac{dy}{dt} = 2a$.

Step 3. Divide to get the slope: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2a}{2at} = \dfrac{1}{t}$ (for $t \neq 0$). This is the standard slope of the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$.

Step 4. Evaluate at $t = 2$, $a = 1$: the point is $(x,y) = (1\cdot4, 2\cdot1\cdot2) = (4,4)$, and the slope is $\dfrac{1}{t} = \dfrac{1}{2}$. The tangent line through $(4,4)$ with slope $\tfrac12$ is $y - 4 = \tfrac12(x-4)$, which simplifies to $y = \tfrac12 x + 2$.

The tangent to the parabola $x=t^2$, $y=2t$ at $t=2$ is $y = \dfrac{1}{2}x + 2$.

Competitive stretch

For the trigonometric pair $x = \cos\theta - \cos2\theta$, $y = \sin\theta - \sin2\theta$, the same three steps give $\dfrac{dx}{d\theta} = -\sin\theta + 2\sin2\theta$ and $\dfrac{dy}{d\theta} = \cos\theta - 2\cos2\theta$, so $\dfrac{dy}{dx} = \dfrac{\cos\theta - 2\cos2\theta}{2\sin2\theta - \sin\theta}$ — left in terms of $\theta$, since there is no simpler closed form in $x$ alone. Notice that no elimination of $\theta$ was attempted here: unlike the parabola, this curve mixes two different multiples of the same angle, and solving either equation for $\theta$ alone is not realistic. This is exactly the situation the formula is built for — it delivers a usable slope regardless of whether the parameter could ever be eliminated.

• Trying to eliminate the parameter first. Solving for $t$ and substituting is usually unnecessary and sometimes impossible in closed form; go straight through $\dfrac{dy/dt}{dx/dt}$.
• Inverting the ratio. Writing $\dfrac{dy}{dx} = \dfrac{dx/dt}{dy/dt}$ instead of $\dfrac{dy/dt}{dx/dt}$ — always differentiate $y$ on top, $x$ on the bottom, both with respect to the same parameter.
• Forgetting the condition $dx/dt \neq 0$. Wherever $dx/dt = 0$, the formula breaks down (a vertical tangent, undefined slope).
• Mixing parameters. Differentiating $x$ with respect to $t$ but $y$ with respect to a different variable in a problem that uses two parameter names.
• Dividing before simplifying trig terms. For curves built from sums or differences of sines and cosines at different multiples of the parameter, applying a sum-to-product identity to $dx/dt$ and $dy/dt$ first — before forming the ratio — usually gives a far cleaner final slope than dividing the raw derivatives directly.

Ask the tutor: If you want to see how a sum-to-product identity can turn a messy trigonometric parametric derivative into a clean one, walk through it with the AI tutor step by step.