Defining and Differentiating Parametric Equations

AP Calculus BC · Parametric, Polar & Vector-Valued Functions

Defining and Differentiating Parametric Equations

Both coordinates answer to t, so the slope is the ratio of the two rates.

key idea$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\qquad (dx/dt\neq0)$

What it says

Parametric equations

\[x=f(t),\qquad y=g(t)\]

Rather than tying $y$ directly to $x$, both coordinates are given as functions of a third variable $t$ — the parameter. As $t$ runs through its interval the point $\left(f(t),g(t)\right)$ traces a curve, in a definite direction and at a definite pace. A parametrisation therefore carries strictly more information than the curve's shape alone.

The parametric derivative

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

Both coordinates respond to $t$, so the slope is the ratio of the two rates. This is the chain rule rearranged: $\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$, solved for $\dfrac{dy}{dx}$. It is valid wherever $\dfrac{dx}{dt}\neq0$.

The curve $x=t^{2}-2$, $y=t^{3}-3t$. Small arrowheads show the direction of travel as $t$ increases, and the amber vector is the tangent at $t=1$. The curve crosses itself — something no graph of a function can do, and a clear sign that $y$ is not a function of $x$ here.

Reading the notation

Parameter $t$The driving variable. Often time, but it may be an angle or any other quantity.

OrientationThe direction the curve is traced as $t$ increases — shown by arrows, and lost entirely if you eliminate the parameter.

$\dfrac{dx}{dt}$How fast the point moves horizontally. Zero here means a vertical tangent.

$\dfrac{dy}{dt}$How fast it moves vertically. Zero here means a horizontal tangent.

Eliminating the parameterSolving one equation for $t$ and substituting. It recovers the shape but discards both the direction and the speed.

At a point where…	$\dfrac{dx}{dt}$	$\dfrac{dy}{dt}$	The tangent is
horizontal tangent	$\neq0$	$=0$	horizontal
vertical tangent	$=0$	$\neq0$	vertical — $\dfrac{dy}{dx}$ undefined
both vanish	$=0$	$=0$	indeterminate — a possible cusp
neither vanishes	$\neq0$	$\neq0$	slanted, slope $\dfrac{dy/dt}{dx/dt}$

Row 3 is the case that needs care: when both derivatives vanish the ratio is $\tfrac00$ and the curve may have a cusp, a corner where the tangent direction genuinely fails to exist.

$\dfrac{dy}{dt}=0$horizontal tangent

$\dfrac{dx}{dt}=0$vertical tangent

Both $=0$check for a cusp

Worked examples

EX 1

The slope at a parameter value

ratio of rates

For $x=t^{2}-2$ and $y=t^{3}-3t$, find $\dfrac{dy}{dx}$ at $t=1$.

Differentiate each coordinate

$\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=3t^{2}-3$.

Form the ratio

$\dfrac{dy}{dx}=\dfrac{3t^{2}-3}{2t}$.

Substitute

At $t=1$: $\dfrac{3-3}{2}=0$.

Interpret

The tangent is horizontal there.

Locate the point

$x=1-2=-1$ and $y=1-3=-2$, so the horizontal tangent occurs at $(-1,-2)$.

EX 2

Tangent line at a point

slope plus a point

Find the equation of the tangent line to $x=\cos t$, $y=\sin 2t$ at $t=\dfrac{\pi}{4}$.

Differentiate

$\dfrac{dx}{dt}=-\sin t$ and $\dfrac{dy}{dt}=2\cos 2t$.

Evaluate the derivatives

At $t=\dfrac{\pi}{4}$: $\dfrac{dx}{dt}=-\dfrac{\sqrt2}{2}$ and $\dfrac{dy}{dt}=2\cos\dfrac{\pi}{2}=0$.

Form the slope

$\dfrac{dy}{dx}=\dfrac{0}{-\sqrt2/2}=0$.

Find the point

$x=\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}$, $y=\sin\dfrac{\pi}{2}=1$.

Write the line

$y=1$ — a horizontal tangent at $\left(\dfrac{\sqrt2}{2},1\right)$.

EX 3

Vertical tangents

the denominator vanishes

For $x=t^{2}-4t$ and $y=t^{3}$, find every point with a vertical tangent.

Set the horizontal rate to zero

$\dfrac{dx}{dt}=2t-4=0$ gives $t=2$.

Check the vertical rate there

$\dfrac{dy}{dt}=3t^{2}=12\neq0$, so this is a genuine vertical tangent, not a cusp.

Find the point

$x=4-8=-4$ and $y=8$.

State the answer

A vertical tangent at $(-4,8)$.

Note the necessary check

Had $\dfrac{dy}{dt}$ also been zero, the ratio would be $\tfrac00$ and further work would be required to classify the point.

EX 4

Eliminating the parameter

what gets lost

Eliminate the parameter from $x=t+1$, $y=t^{2}$ for $t\ge0$, and say what information disappears.

Solve for $t$

$t=x-1$.

Substitute

$y=(x-1)^{2}$ — a familiar parabola.

Track the domain

$t\ge0$ forces $x\ge1$, so only the right half of the parabola is traced.

Identify what is gone

The orientation: the Cartesian equation cannot say that the point moves rightward and upward as $t$ grows.

Note the practical consequence

Two different parametrisations can produce the same Cartesian curve at different speeds or in opposite directions. Eliminating $t$ throws that distinction away.

Boundary cases

$\dfrac{dy}{dx}$ is a ratio, not a quotient rule

Divide $\dfrac{dy}{dt}$ by $\dfrac{dx}{dt}$. There is no product or quotient rule anywhere in this formula.

$\dfrac{dx}{dt}=0$ makes the slope undefined

That is a vertical tangent, not an error. Report it as such rather than trying to force a numerical slope.

Both rates zero needs investigation

$\tfrac00$ signals a possible cusp. Examine the behaviour of the ratio as $t$ approaches the value from each side.

Eliminating the parameter discards direction

The Cartesian equation keeps the shape and loses the motion. Answer questions about orientation or speed from the parametric form.

Course map. Topic 9.1 · LO CHA-3.G — calculate derivatives of parametric functions. BC-only.