Calculus with Parametric Curves

You never have to solve for y in terms of x: on a parametric curve the slope is $dy/dx$ $= (dy/dt)/(dx/dt)$ — differentiate each coordinate in t and divide — and arc length is $\int \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$. All the calculus happens directly in the parameter.

By the end you'll be able to compute a parametric curve's slope with $(dy/dt)/(dx/dt)$, locate its horizontal ($dy/dt = 0$) and vertical ($dx/dt = 0$) tangents, and find its arc length with $\int \sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$.

Predict: at $t = \pm 1$, where $dx/dt = 0$, is the tangent vertical or horizontal? Slide t onto those spots to check.

This is the curve $x = t^3 - 3t$, $y = 3t^2 - 9$ from the previous lesson — same moving point, same direction of travel (gray arrows) — but now it carries its own calculus. The dashed tangent line at the moving point has slope $(dy/dt)/(dx/dt)$, the blue trail shows the arc length swept out so far, and the orange dots mark where a derivative in t vanishes. Hover or focus any dot for exact values.

Parameter t = 0.60

dx/dt = -1.920 · dy/dt = 3.600 · slope dy/dx = -1.875 · length traced from t = -2.2: L = 19.264

x = t³ − 3t, y = 3t² − 9 — tangent from the velocity, arc length from the trail

path traced so far rest of the curve tangent at t dy/dt = 0 or dx/dt = 0

For a parametric curve, $dy/dx = (dy/dt)/(dx/dt)$ gives the slope, and arc length is $\int\sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$ — calculus done directly in the parameter, with no need to eliminate t.

Visual

Picture the moving point from the last lesson. Its velocity has a horizontal part ($dx/dt$) and a vertical part ($dy/dt$), and the tangent line simply points along that velocity — so its slope is rise over run, $(dy/dt)/(dx/dt)$. When $dy/dt = 0$ the point is moving purely sideways: horizontal tangent. When $dx/dt = 0$ it's moving purely up or down: vertical tangent — the situation a plain function $y = f(x)$ can never represent, but which is completely natural here. Arc length is the same picture accumulated: in each instant the point moves along a tiny hypotenuse $\sqrt{dx^2 + dy^2}$, and adding those up gives the blue trail's length in the demo.

Formal

For $x = f(t)$, $y = g(t)$, the chain rule gives $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$, so $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \ne 0. $$ Horizontal tangents occur where $dy/dt = 0$ (and $dx/dt \ne 0$); vertical tangents where $dx/dt = 0$ (and $dy/dt \ne 0$). For arc length, the infinitesimal displacement is $ds = \sqrt{dx^2 + dy^2} = \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$, so over $t \in [\alpha, \beta]$, traversed exactly once, $$ L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt. $$ This generalizes the Cartesian formula $\int\sqrt{1 + (dy/dx)^2}\,dx$ — and if the parametrization retraces the curve, the integral counts the repeated distance, not just the geometric length.

Worked example — slope and special tangents

Let $x = t^3 - 3t$, $y = 3t^2 - 9$ (the demo curve). Differentiate each piece in t: $dx/dt = 3t^2 - 3 = 3(t^2 - 1)$ and $dy/dt = 6t$. So $$ \frac{dy}{dx} = \frac{6t}{3(t^2-1)} = \frac{2t}{t^2 - 1}. $$ Horizontal tangent where $dy/dt = 6t = 0 \Rightarrow t = 0$, at the point $(0, -9)$. Vertical tangents where $dx/dt = 3(t^2 - 1) = 0 \Rightarrow t = \pm 1$, at the points $(2, -6)$ and $(-2, -6)$ — exactly the three orange dots in the demo.

Worked example — arc length of a circle

Find the arc length of $x = 3\sin t$, $y = 3\cos t$ for $t \in [0, 2\pi]$ — a circle of radius 3, traced once. Here $dx/dt = 3\cos t$ and $dy/dt = -3\sin t$, so $(dx/dt)^2 + (dy/dt)^2 = 9\cos^2 t + 9\sin^2 t = 9$, and $$ L = \int_0^{2\pi} \sqrt{9}\,dt = \int_0^{2\pi} 3\,dt = 6\pi, $$ matching the geometric circumference $2\pi r = 6\pi$. The formula checks out.

Your turn

Same idea, different numbers: find the arc length of $x = 4\cos t$, $y = 4\sin t$ for $t \in [0, \pi]$ (half of a radius-4 circle). First, $dx/dt = -4\sin t$ and $dy/dt = 4\cos t$, so $(dx/dt)^2 + (dy/dt)^2 = $ ____ and $L = \int_0^{\pi}$ ____ $\,dt = $ ____

Reveal the answer

$(dx/dt)^2 + (dy/dt)^2 = 16\sin^2 t + 16\cos^2 t = 16$, so the integrand is $\sqrt{16} = 4$ and $L = \int_0^{\pi} 4\,dt = $ $4\pi$ — half the circumference $2\pi(4) = 8\pi$, as the half-interval $[0, \pi]$ suggests.

More info — concavity needs a second chain-rule step

It's tempting to get $d^2y/dx^2$ by differentiating $dy/dx$ with respect to t — but that gives a derivative in the wrong variable. Concavity needs the derivative of the slope with respect to x, so you divide by $dx/dt$ once more: $$ \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}. $$ It's the same move as the first-derivative formula, applied to $dy/dx$ instead of y. The first Paul's Online Math Notes link under Dive deeper works this second-derivative formula through full examples.

Check your understanding

Question 1 of 4

For $x = t^2$, $y = t^3 - 3t$, what is $dy/dx$ at $t = 2$?

Question 2 of 4

At $t = t_0$ a parametric curve has $dx/dt = 0$ and $dy/dt = 5$. What does the tangent look like there?

Question 3 of 4

Find the arc length of $x = 5\cos t$, $y = 5\sin t$ for $t \in [0, 2\pi]$.

Question 4 of 4

$x = 3\cos t$, $y = 3\sin t$ traces a circle of radius 3 once as t runs over $[0, 2\pi]$. If you instead integrate the arc-length formula over $t \in [0, 4\pi]$, what do you get?

Recap

$dy/dx = (dy/dt)/(dx/dt)$: differentiate each coordinate in t, then divide — no need to solve for y in terms of x.
Horizontal tangent where $dy/dt = 0$ (and $dx/dt \ne 0$); vertical tangent where $dx/dt = 0$ (and $dy/dt \ne 0$) — vertical tangents are natural for parametric curves.
Arc length: $L = \int_{\alpha}^{\beta} \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$, valid when the curve is traced exactly once over $[\alpha, \beta]$.
The integrand is the moving point's speed, so the integral measures distance traveled — retracing the curve double-counts the length.

Dive deeper

Paul's Online Math Notes — Tangents with Parametric Equations Derive dy/dx = (dy/dt)/(dx/dt) and locate horizontal and vertical tangents.
Paul's Online Math Notes — Arc Length with Parametric Equations Verify the parametric arc-length formula against the circumference of a circle.

Sources

Calculus with Parametric Curves — Slope and Arc Length