Parametric Functions

Concept Explanation

Unlike traditional functions where $y$ is directly dependent on $x$ (e.g. $y = f(x)$), parametric equations define both $x$ and $y$ independently based on the parameter $t$. This is highly useful for describing trajectories of moving objects, as it records not only the path itself but also when the object occupies each position, and in which direction it is moving.

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Evaluating and Graphing Parametric Curves

To graph a parametric curve manually:

1. Create a table of values: Choose sequential values of the parameter $t$ in its domain.
2. Calculate coordinates: For each $t$, calculate the corresponding values for $x(t)$ and $y(t)$.
3. Plot the points: Plot the resulting coordinate pairs $(x,y)$ in order of increasing $t$.
4. Indicate orientation: Draw arrows along the curve to show the direction of motion as $t$ increases.

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Eliminating the Parameter (Converting to Cartesian)

Sometimes it is useful to rewrite a parametric curve in standard Cartesian coordinates $(x,y)$ to see its geometric shape. To "eliminate the parameter":

• Method 1 (Substitution): Solve one of the equations for $t$ (e.g. $t = \frac{x - x_0}{a}$), and substitute this expression into the other equations. This works well for linear and simple polynomial paths.
• Method 2 (Trig Identities): If the equations involve sine and cosine, isolate the trig functions and use the Pythagorean identity $\cos^2 t + \sin^2 t = 1$.
  • Example: $\frac{x}{r} = \cos t$ and $\frac{y}{r} = \sin t \implies \left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = \cos^2 t + \sin^2 t = 1 \implies x^2 + y^2 = r^2$ (a circle of radius $r$).

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Why We Use It in Multivariable Calculus (Calculus III)

Parametric functions are the fundamental building blocks of spatial geometry and vector calculus:

1. Vector-Valued Functions

A curve in 3D space is described as a single vector-valued function $\vec{r}(t)$ whose component scalar functions are the parametric equations of the curve:

Taking derivatives and integrals of $\vec{r}(t)$ gives velocity, acceleration, and arc lengths.

2. Parameterization for Line Integrals

To evaluate a line integral $\int_C f(x,y,z) \, ds$ or calculate work done by a force field along a path $C$, $\int_C \vec{F} \cdot d\vec{r}$, you must first parameterize the path $C$. Parameterizing expresses $x$, $y$, and $z$ in terms of a single variable $t$, converting a complex multidimensional path integral into a simple single-variable integral in terms of $t$:

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Example Problems

Example Problem 1: Circular Motion & Eliminating Parameter

Graph and find the Cartesian equation for the parametric equations:

Solution

Step 1: Evaluate points for key parameter values Construct a table of values:

$t$	$x(t) = 3\cos t$	$y(t) = 3\sin t$	Point $(x,y)$
$0$	$3\cos(0) = 3$	$3\sin(0) = 0$	$(3,0)$
$\pi/2$	$3\cos(\pi/2) = 0$	$3\sin(\pi/2) = 3$	$(0,3)$
$\pi$	$3\cos(\pi) = -3$	$3\sin(\pi) = 0$	$(-3,0)$
$3\pi/2$	$3\cos(3\pi/2) = 0$	$3\sin(3\pi/2) = -3$	$(0,-3)$
$2\pi$	$3\cos(2\pi) = 3$	$3\sin(2\pi) = 0$	$(3,0)$

Plotting these points in order shows a circle of radius $3$. The orientation is counterclockwise (starting at $(3,0)$, moving up to $(0,3)$, then left to $(-3,0)$, etc.).

Step 2: Eliminate the parameter Isolate the trigonometric terms:

$$\frac{x}{3} = \cos t \quad \text{and} \quad \frac{y}{3} = \sin t$$

Use the identity $\cos^2 t + \sin^2 t = 1$:

$$\left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \implies \frac{x^2}{9} + \frac{y^2}{9} = 1 \implies x^2 + y^2 = 9$$

The Cartesian equation is $x^2 + y^2 = 9$, which represents a circle centered at the origin with radius $3$.

Example Problem 2: 3D Line Segment Parameterization

Find parametric equations for the 3D line segment starting at the point $P(1, 2, -1)$ and ending at the point $Q(3, 5, 2)$ for the domain $0 \leq t \leq 1$.

Solution

Step 1: Find the direction vector ($\vec{v}$) The vector $\vec{v}$ points from the start point $P$ to the end point $Q$:

$$\vec{v} = \overrightarrow{PQ} = \langle Q_x - P_x, Q_y - P_y, Q_z - P_z \rangle$$

$$\vec{v} = \langle 3 - 1, 5 - 2, 2 - (-1) \rangle = \langle 2, 3, 3 \rangle$$

Step 2: Set up the vector equation of the line segment The vector equation is $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ where $\vec{r}_0 = \langle 1, 2, -1 \rangle$ is the starting point:

$$\vec{r}(t) = \langle 1, 2, -1 \rangle + t\langle 2, 3, 3 \rangle$$

Step 3: Extract component parametric equations

$$x(t) = 1 + 2t, \quad y(t) = 2 + 3t, \quad z(t) = -1 + 3t \quad \text{for } 0 \leq t \leq 1$$

Step 4: Verify boundary conditions

• At $t = 0$: $x(0)=1, y(0)=2, z(0)=-1 \implies$ point is $(1, 2, -1)$ (Start point $P$)
• At $t = 1$: $x(1)=3, y(1)=5, z(1)=2 \implies$ point is $(3, 5, 2)$ (End point $Q$)

The parametric equations are $x(t) = 1 + 2t$, $y(t) = 2 + 3t$, $z(t) = -1 + 3t$ for $0 \leq t \leq 1$.

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Connections & References

• Parent Concepts: Calculus Foundations
• Related Concepts: Solving Right Triangles, Line Integrals
• Course Links: Calc-3 Discussion 01