Lesson 10.7 : Parametric Equations: Graphs

New Concept

Parametric equations define a curve using a third variable, the parameter $t$. Instead of $y=f(x)$, we have $x(t)$ and $y(t)$, which lets us track an object's position over time and determine its direction of motion.

What’s next

Now, let's start graphing. You'll work through interactive examples, plotting points from a table to see how these equations create curves with direction.

Property

To sketch a graph by plotting points from a pair of parametric equations:

1. Construct a table with three columns: $t$, $x(t)$, and $y(t)$.
2. Evaluate $x$ and $y$ for values of $t$ over the interval for which the functions are defined.
3. Plot the resulting pairs $(x, y)$. The arrows indicate direction according to increasing values of $t$.

Examples

• To sketch $x(t) = t^2 - 3$ and $y(t) = t + 1$, plotting for $t=-1, 0, 1$ gives points $(-2, 0)$, $(-3, 1)$, and $(-2, 2)$, which begin to trace a parabola.
• For $x(t) = 5 - t$ and $y(t) = \frac{t^2}{2}$, points for $t=0, 2, 4$ are $(5, 0)$, $(3, 2)$, and $(1, 8)$. The curve opens upwards and moves to the left.
• Given $x(t) = 2t + 1$ and $y(t) = \sqrt{t}$ for $t \ge 0$, the points for $t=1, 4, 9$ are $(3, 1)$, $(9, 2)$, and $(19, 3)$, tracing the shape of a square root function.

Explanation

This method traces the curve's path step-by-step. Since the parameter $t$ often represents time, plotting points reveals not just the shape of the path, but also the direction of motion along it, which a standard equation doesn't show.

Property

Parametric equations involving sine and cosine often describe an ellipse or a circle. For equations of the form $x = a \operatorname{cos} t$ and $y = b \operatorname{sin} t$, the graph is an ellipse centered at the origin. The calculations use angle measures in radians for $t$. The ellipse is mapped in a counterclockwise direction as $t$ increases from $0$ to $2\pi$.

Examples

• The equations $x = 6 \operatorname{cos} t$ and $y = 2 \operatorname{sin} t$ trace an ellipse. At $t=0$, the point is $(6,0)$. At $t=\frac{\pi}{2}$, the point is $(0,2)$.
• For $x = 5 \operatorname{cos} t$ and $y = 5 \operatorname{sin} t$, the horizontal and vertical radii are equal, so the graph is a circle with a radius of 5.
• The equations $x = 3 \operatorname{sin} t$ and $y = 4 \operatorname{cos} t$ also trace an ellipse. However, at $t=0$, the starting point is $(0,4)$, and the curve traces in a clockwise direction.

Explanation

Think of this as stretching or shrinking a unit circle. The value of $a$ controls the horizontal radius, and $b$ controls the vertical radius. The parameter $t$ acts as an angle, sweeping around the origin to draw the shape.

Property

To convert parametric equations to a single rectangular equation, eliminate the parameter $t$. There are two common methods:

1. Solve one of the parametric equations for $t$ and substitute the resulting expression into the other equation.
2. For trigonometric forms, isolate the sine and cosine terms and apply the Pythagorean Identity: $\operatorname{cos}^2 t + \operatorname{sin}^2 t = 1$.

Examples

• Given $x = t + 3$ and $y = t^2 + 1$, we solve for $t$ to get $t = x - 3$. Substituting into the second equation gives $y = (x-3)^2 + 1$, which is a parabola.
• For $x = 6 \operatorname{cos} t$ and $y = 2 \operatorname{sin} t$, we write $\frac{x}{6} = \operatorname{cos} t$ and $\frac{y}{2} = \operatorname{sin} t$. Using the Pythagorean identity, we get $(\frac{x}{6})^2 + (\frac{y}{2})^2 = 1$.
• Given $x = e^{3t}$ and $y = e^t$, we can write $x = (e^t)^3$. Since $y=e^t$, we substitute to get the rectangular equation $x = y^3$.

Explanation

Converting to rectangular form helps us identify familiar shapes like parabolas, ellipses, or lines. However, this process removes the parameter $t$, so we lose the information about the curve's direction and the timing of the motion.

Property

The path of an object propelled at an inclination of $\theta$ to the horizontal, with initial speed $v_0$, and at a height $h$ above the horizontal, is given by:
Horizontal distance: $x = (v_0 \operatorname{cos} \theta)t$
Vertical distance: $y = -\frac{1}{2}gt^2 + (v_0 \operatorname{sin} \theta)t + h$
Where $g$ is the acceleration due to gravity, use $g = 32$ ft/s$^2$ or $g = 9.8$ m/s$^2$.

Examples

• A golf ball is hit with $v_0 = 150$ ft/s, at an angle $\theta=30^\circ$ from the ground ($h=0$). The path is modeled by $x = (150 \operatorname{cos} 30^\circ)t$ and $y = -16t^2 + (150 \operatorname{sin} 30^\circ)t$.
• For the golf ball above, its position at $t=1$ second is $x = 150(\frac{\sqrt{3}}{2}) \approx 129.9$ feet and $y = -16(1)^2 + 150(0.5) = 59$ feet.
• To find how long the golf ball is in the air, set its vertical position to zero: $-16t^2 + 75t = 0$. Factoring gives $t(-16t + 75) = 0$, so it lands when $t = \frac{75}{16} \approx 4.69$ seconds.

Explanation

These equations simplify a projectile's flight by separating it into two components. Horizontal motion is steady and constant, while vertical motion is affected by gravity's downward pull. The parameter $t$ links both, tracking the object's position over time.