Why Use Parametric Equations?

1. Model motion: $t$ often represents time
2. Describe curves that aren't functions (fail vertical line test)
3. Separate horizontal and vertical motion
4. Express direction and speed of motion

Example: Linear Motion

Consider: $x = 2t + 1$ $y = 3t - 2$

As $t$ increases, the point $(x, y)$ traces out a path.

Converting to Rectangular Form

To eliminate the parameter $t$:

1. Solve one equation for $t$
2. Substitute into the other equation
3. Simplify to get $y$ in terms of $x$ (or vice versa)

Example

Given: $x = 2t + 1$ and $y = 3t - 2$

Step 1: Solve for $t$ from the first equation. $x = 2t + 1$ $t = \frac{x - 1}{2}$

Step 2: Substitute into the second equation. $y = 3t - 2 = 3\left(\frac{x - 1}{2}\right) - 2$

$y = \frac{3x - 3}{2} - 2 = \frac{3x - 3 - 4}{2} = \frac{3x - 7}{2}$

Rectangular form: $y = \frac{3}{2}x - \frac{7}{2}$

Parametric Equations for Common Curves

Circle (radius $r$, center at origin)

$x = r\cos(t)$ $y = r\sin(t)$ $0 \leq t \leq 2\pi$

Ellipse

$x = a\cos(t)$ $y = b\sin(t)$

Line Segment

$x = x_1 + t(x_2 - x_1)$ $y = y_1 + t(y_2 - y_1)$ $0 \leq t \leq 1$

Direction and Orientation

The orientation shows the direction of motion as $t$ increases.

• Plot points for increasing values of $t$
• Draw arrows to show direction of travel
• Different parametrizations can trace the same curve with different orientations

Calculus with Parametric Equations

Derivative (slope of tangent line): $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}$

Note: This is NOT $\frac{dy}{dt}$!

Substitute into the second equation:

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

$y = \frac{(x + 1)^2}{4} + 3$

$y = \frac{x^2 + 2x + 1}{4} + 3$

$y = \frac{x^2 + 2x + 1 + 12}{4} = \frac{x^2 + 2x + 13}{4}$

Or: $4y = x^2 + 2x + 13$

Part (b): When $t = 2$:

$x = 2(2) - 1 = 3$ $y = 2^2 + 3 = 7$

Point: $(3, 7)$

Part (c): As $t$ increases from negative to positive:

• When $t = -1$: $x = -3$, $y = 4$
• When $t = 0$: $x = -1$, $y = 3$
• When $t = 1$: $x = 1$, $y = 4$
• When $t = 2$: $x = 3$, $y = 7$

The curve is a parabola opening upward, and motion is from left to right as $t$ increases.

$t = -\frac{60}{2(-16)} = \frac{60}{32} = 1.875$ seconds

Maximum height:

$y = -16(1.875)^2 + 60(1.875) + 5$ $y = -16(3.516) + 112.5 + 5$ $y = -56.25 + 112.5 + 5$ $y = 61.25$ feet

Part (b): The projectile hits the ground when $y = 0$:

$-16t^2 + 60t + 5 = 0$

Using the quadratic formula: $t = \frac{-60 \pm \sqrt{60^2 - 4(-16)(5)}}{2(-16)}$

$t = \frac{-60 \pm \sqrt{3600 + 320}}{-32} = \frac{-60 \pm \sqrt{3920}}{-32}$

$t = \frac{-60 \pm 62.61}{-32}$

$t = \frac{-60 + 62.61}{-32} = -0.082$ (reject, negative time)

$t = \frac{-60 - 62.61}{-32} = \frac{-122.61}{-32} = 3.832$ seconds

Horizontal distance: $x = 40(3.832) = 153.3$ feet

The projectile hits the ground after 3.83 seconds at a distance of 153.3 feet.

$\frac{dx}{dt} = 2$

$\frac{dy}{dt} = 2t$

Step 2: Use the formula for the derivative.

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{2} = t$

Step 3: Evaluate at $t = 3$.

$\frac{dy}{dx}\bigg|_{t=3} = 3$

Answer: $\frac{dy}{dx} = 3$ at $t = 3$

This means the slope of the tangent line at the point where $t = 3$ is $3$.