When x and y are both driven by a third variable, the chain rule links their two separate rates of change into one gradient.
9 min read4 worked examples
The parametric gradient formula
A curve given parametrically has both $x$ and $y$ expressed in terms of a third variable, the parameter $t$: $x=f(t)$, $y=g(t)$. As $t$ varies, the point $(x,y)$ traces out the curve.
ProofDeriving dy/dx for a parametric curve
1
By the chain rule, $\dfrac{dy}{dt}=\dfrac{dy}{dx}\times\dfrac{dx}{dt}$.
y depends on t indirectly, through x — exactly the chain rule's setup.
2
Divide both sides by $\dfrac{dx}{dt}$ (valid wherever $\dfrac{dx}{dt}\ne0$): $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.
Rearrange to isolate the Cartesian gradient dy/dx.
▪ proven
$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$
In practice: differentiate $x$ and $y$ separately with respect to $t$, then divide the two results. The gradient at a specific point on the curve is found by first finding the value of $t$ at that point, then substituting into $\dfrac{dy}{dx}$.
Tangents and normals to parametric curves
Once the gradient at a particular value of $t$ has been found, the tangent and normal follow exactly as usual: find the corresponding $(x,y)$ coordinates by substituting $t$ into $x=f(t)$ and $y=g(t)$, then apply $y-y_1=m(x-x_1)$ with $m$ for the tangent or $-\dfrac{1}{m}$ for the normal.
Two ways to use the parameter
The parameter $t$ plays two roles: it locates the point on the curve (via $x=f(t)$, $y=g(t)$) and it drives the gradient (via $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$). A single value of $t$ pins down both a specific point and the gradient there — no need to eliminate $t$ and go back to Cartesian form.
Worked example 1
Finding dy/dx for a parametric curve
A curve is defined by $x=t^2$, $y=2t^3$. Find $\dfrac{dy}{dx}$ in terms of $t$.
MethodDifferentiate x and y separately, then divide
1
$\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=6t^2$.
Differentiate each parametric equation with respect to t.
Apply y − y₁ = m(x − x₁) with the gradient and point just found.
▪ solved
Worked example 4
Normal to a parametric curve, expressed exactly · Challenge
A curve is defined by $x=t+\dfrac{1}{t}$, $y=t-\dfrac{1}{t}$ for $t>0$. Find the equation of the normal at $t=1$, giving your answer in the form $ax+by+c=0$.
MethodFind dy/dx, take the negative reciprocal, then form the normal
1
$\dfrac{dx}{dt}=1-\dfrac{1}{t^2}$ and $\dfrac{dy}{dt}=1+\dfrac{1}{t^2}$.
Differentiate each parametric equation, treating 1/t as t⁻¹.
2
At $t=1$: $\dfrac{dx}{dt}=1-1=0$ and $\dfrac{dy}{dt}=1+1=2$.
Substitute t = 1 into each derivative.
3
Since $\dfrac{dx}{dt}=0$ at $t=1$, the tangent is vertical — so $\dfrac{dy}{dx}$ is undefined there, and the normal must be horizontal.
Dividing by dx/dt = 0 is undefined; a vertical tangent has a horizontal normal, reasoned directly rather than via −1/m.
4
The point at $t=1$: $x=1+1=2$, $y=1-1=0$.
Substitute t = 1 into the parametric equations.
5
The normal is the horizontal line through $(2,0)$: $y=0$.
A horizontal line through the point has equation y = 0.
6
In the required form: $0x+1y+0=0$, i.e. $y=0$.
Already in the form ax + by + c = 0 with a = 0, b = 1, c = 0.
▪ solved
Look deeper
Just as in implicit differentiation, a zero in the denominator of the parametric gradient formula is a genuine special case, not an error — it signals a vertical tangent (undefined gradient), whose normal is horizontal. Always check whether $\dfrac{dx}{dt}=0$ before dividing.
Common mistake
Inverting the parametric gradient formula — writing $\dfrac{dy}{dx}=\dfrac{dx/dt}{dy/dt}$ instead of $\dfrac{dy/dt}{dx/dt}$. Also common: substituting the wrong value into $\dfrac{dy}{dx}$ — remember the formula gives the gradient in terms of the parameter $t$, not $x$, so points must be identified by their $t$-value, not read off an $x$-coordinate.
Check yourself
A curve has parametric equations $x=t^2-1$, $y=t^3-t$. Find $\dfrac{dy}{dx}$ in terms of $t$, and evaluate it at $t=3$. (Hint: differentiate each equation with respect to t, then divide.)
Method
1
$\dfrac{dx}{dt}=2t$, $\dfrac{dy}{dt}=3t^2-1$.
Differentiate each equation.
2
$\dfrac{dy}{dx}=\dfrac{3t^2-1}{2t}$.
Apply the parametric gradient formula.
3
At $t=3$: $\dfrac{dy}{dx}=\dfrac{3(9)-1}{6}=\dfrac{26}{6}=\dfrac{13}{3}$.