$x = f(t), \quad y = g(t)$

💡 Key Idea: The parameter $t$ often represents time, and the equations trace out a path or curve!

Why Use Parametric Equations?

Advantages:

1. Can describe curves that aren't functions

   • Example: A circle fails the vertical line test
   • But parametrically: $x = \cos t$, $y = \sin t$ works perfectly!

2. Natural for motion problems

   • $t$ = time
   • Position: $(x(t), y(t))$
   • Describes the path of a moving object

3. More flexibility

   • Can go backwards, loop, cross itself
   • Can control speed along the curve

Example 1: A Simple Line

Consider: $x = t$, $y = 2t + 1$ for $t \in \mathbb{R}$

Step 1: Make a table

Step 2: Eliminate the parameter

From $x = t$, we have $t = x$

Substitute into $y = 2t + 1$: $y = 2x + 1$

This is a line with slope 2!

Direction: As $t$ increases, we move in the direction of increasing $x$ (left to right).

Example 2: A Circle

Consider: $x = 3\cos t$, $y = 3\sin t$ for $0 \leq t \leq 2\pi$

Step 1: Eliminate the parameter

We know: $\cos^2 t + \sin^2 t = 1$

From the equations: $\frac{x}{3} = \cos t, \quad \frac{y}{3} = \sin t$

Square and add: $\left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 = \cos^2 t + \sin^2 t = 1$

$\frac{x^2}{9} + \frac{y^2}{9} = 1$

$x^2 + y^2 = 9$

This is a circle with center $(0, 0)$ and radius 3!

Direction: As $t$ goes from 0 to $2\pi$:

• At $t = 0$: $(3, 0)$ (starting point)
• At $t = \pi/2$: $(0, 3)$ (top)
• At $t = \pi$: $(-3, 0)$ (left)
• At $t = 3\pi/2$: $(0, -3)$ (bottom)

Counterclockwise motion!

Example 3: An Ellipse

Consider: $x = 4\cos t$, $y = 2\sin t$ for $0 \leq t \leq 2\pi$

Eliminate the parameter:

$\frac{x}{4} = \cos t, \quad \frac{y}{2} = \sin t$

$\left(\frac{x}{4}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$

$\frac{x^2}{16} + \frac{y^2}{4} = 1$

This is an ellipse with semi-major axis 4 (horizontal) and semi-minor axis 2 (vertical)!

Direction of Motion

The orientation of the curve matters!

To find direction:

1. Plug in a few values of $t$
2. See which way the point moves as $t$ increases
3. Draw arrows on the curve

Example: For $x = \cos t$, $y = \sin t$:

• $t = 0 \to \pi/2$: Point moves from $(1,0)$ to $(0,1)$
• This is counterclockwise

Eliminating the Parameter

Common Techniques:

1. Solve for $t$ from one equation, substitute into other

Example: $x = t + 1$, $y = t^2$

From first: $t = x - 1$

Substitute: $y = (x-1)^2$

2. Use trigonometric identities

For $x = a\cos t$, $y = b\sin t$:

$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1$

3. Use relationships between expressions

Example: $x = t^2$, $y = t^3$

Note: $y = t \cdot t^2 = t \cdot x$

But $t = \pm\sqrt{x}$, so $y = \pm x\sqrt{x} = \pm x^{3/2}$

Example 4: A Parabola

Consider: $x = t^2$, $y = 2t$ for $t \geq 0$

Step 1: Solve for $t$

From $y = 2t$: $t = \frac{y}{2}$

Step 2: Substitute

$x = t^2 = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4}$

$y^2 = 4x$

This is a parabola opening to the right!

Restriction: Since $t \geq 0$, we have $y = 2t \geq 0$, so only the upper half!

Domain Restrictions

Pay attention to the range of $t$!

Example: $x = \cos t$, $y = \sin t$ for $0 \leq t \leq \pi$

This traces only the upper semicircle (from $(1,0)$ to $(-1,0)$), not the full circle!

Parametric vs Cartesian

Same Curve, Different Parametrizations

Example: The line $y = x$ can be represented as:

1. $x = t, y = t$
2. $x = 2t, y = 2t$
3. $x = t^3, y = t^3$

All trace the same line, but at different "speeds"!

Common Parametric Curves

Line through $(x_0, y_0)$ with slope $m$:

$x = x_0 + at, \quad y = y_0 + bt$

where $\frac{b}{a} = m$ (slope)

Circle with center $(h, k)$ and radius $r$:

$x = h + r\cos t, \quad y = k + r\sin t$

for $0 \leq t \leq 2\pi$

Ellipse with center $(h, k)$:

$x = h + a\cos t, \quad y = k + b\sin t$

where $a$ = horizontal radius, $b$ = vertical radius

Example 5: A Cycloid

The path traced by a point on the rim of a rolling wheel!

$x = r(t - \sin t), \quad y = r(1 - \cos t)$

where $r$ is the radius of the wheel.

Famous curve in physics - fastest descent path (brachistochrone)!

Finding Intersections

To find where a parametric curve intersects itself or another curve:

Step 1: Set $x(t_1) = x(t_2)$ and $y(t_1) = y(t_2)$

Step 2: Solve for $t_1$ and $t_2$ (with $t_1 \neq t_2$)

Step 3: Those $t$ values give the intersection point(s)

⚠️ Common Mistakes

Mistake 1: Forgetting Domain Restrictions

When eliminating $t$, the Cartesian equation might represent more than the parametric curve!

Always note restrictions on $t$, $x$, or $y$.

Mistake 2: Losing Direction Information

The Cartesian equation doesn't tell you which way the curve is traced!

Always check a few points to determine orientation.

Mistake 3: Assuming One-to-One

A parametric curve can cross itself!

Different values of $t$ can give the same point $(x, y)$.

Applications of Parametric Equations

1. Projectile motion:

   • $x = (v_0\cos\theta)t$
   • $y = (v_0\sin\theta)t - \frac{1}{2}gt^2$

2. Planetary orbits: Elliptical paths

3. Computer graphics: Bezier curves for design

4. Engineering: Cam profiles, gear shapes

📝 Practice Strategy

1. Make a table of values for small integer $t$ values
2. Plot points and connect smoothly
3. Draw arrows showing direction as $t$ increases
4. Eliminate parameter to identify the curve type
5. Check domain - does the parametric curve trace all or part of the Cartesian curve?
6. Identify key features: starting point, ending point, orientation
7. Look for symmetry or special structure

$x = (y-1)^2 - 2(y-1)$

$= y^2 - 2y + 1 - 2y + 2$

$= y^2 - 4y + 3$

Or solving for $y$: $y^2 - 4y + (3 - x) = 0$

Part (b): When $t = 3$:

$x = 3^2 - 2(3) = 9 - 6 = 3$

$y = 3 + 1 = 4$

Point: $(3, 4)$

Step 2: Apply $\cos^2 t + \sin^2 t = 1$

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

$\frac{x^2}{4} + \frac{y^2}{9} = 1$

Step 3: Identify the curve

This is an ellipse with:

• Center: $(0, 0)$
• Horizontal semi-axis: $a = 2$
• Vertical semi-axis: $b = 3$

Step 4: Find direction

Check a few points:

• $t = 0$: $(2, 0)$
• $t = \pi/2$: $(0, 3)$
• $t = \pi$: $(-2, 0)$
• $t = 3\pi/2$: $(0, -3)$

Direction: Counterclockwise, starting at $(2, 0)$.

Answer: Ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$, traced counterclockwise once.

(Note: This requires $x > 0$)

Step 2: Substitute into $y$ equation

$y = e^{2t} + 1 = (e^t)^2 + 1$

Since $e^t = x$: $y = x^2 + 1$

Step 3: Identify restrictions

Since $x = e^t$ and $e^t > 0$ for all $t$:

• Domain: $x > 0$
• This is only the right half of the parabola $y = x^2 + 1$

Step 4: Additional restriction on $y$

When $x > 0$: $x^2 > 0$, so $y = x^2 + 1 > 1$

Therefore: $y > 1$

Answer: Parabola $y = x^2 + 1$ for $x > 0$ (equivalently, $y > 1$).

The curve is only the right branch of the parabola.