13.3 Parametric Equations, Elimination of Parameters & Projectile Motion

13.3 Parametric Equations, Elimination of Parameters & Projectile Motion

Parametric Equations and Curve Orientation

In standard Cartesian representations, a planar curve is expressed as an explicit relation $y = f(x)$ or an implicit relation $F(x, y) = 0$. While effective for geometric analysis, Cartesian forms fail to capture dynamic motion: they reveal where a particle travels, but convey nothing about when it reaches a location, how fast it travels, or its direction of transit.

A parametric system resolves this by expressing both coordinate variables $x$ and $y$ as independent functions of a third variable $t$, termed the parameter: $x = f(t), \quad y = g(t) \quad \text{for } t \in I$ Here, $I \subseteq \mathbb{R}$ is the parameter interval. In physical applications, $t$ typically represents elapsed time, while in geometric constructions, $t$ may represent an angle of rotation or arc length.

As $t$ increases continuously from its lower bound to its upper bound across $I$, the point $(x(t), y(t))$ traces out a continuous collection of points termed a plane curve. Crucially, parametric curves possess an intrinsic orientation (or direction of motion), typically marked on graphs with directional arrows showing the path traced as $t$ increases.

Instantaneous Rate of Change

By the chain rule of differential calculus, the Cartesian tangent slope $\frac{dy}{dx}$ along a smooth parametric curve is given by the ratio of the parametric derivatives: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad \text{provided } f'(t) \neq 0$ If $f'(t) = 0$ and $g'(t) \neq 0$, the curve has a vertical tangent line. If both $f'(t) = 0$ and $g'(t) = 0$, the parameter value represents a singular point or cusp where orientation may abruptly reverse.

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Techniques for Eliminating the Parameter

Eliminating the parameter converts a parametric system into an equivalent rectangular equation in $x$ and $y$. The chosen algebraic strategy depends on the structural form of the parametric functions.

1. Direct Algebraic Substitution

When one or both parametric equations are polynomial, radical, or rational, isolate $t$ in the simpler equation and substitute that expression into the remaining equation:

• Linear System: If $x = 3t + 2$ and $y = -2t + 5$, isolate $t = \frac{x - 2}{3}$. Substituting into $y$ yields the linear equation $y = -2\left(\frac{x - 2}{3}\right) + 5 = -\frac{2}{3}x + \frac{19}{3}$.
• Quadratic / Radical System: If $x = \sqrt{t}$ and $y = 2t - 1$ for $t \ge 0$, then $t = x^2$. Substituting yields $y = 2x^2 - 1$ with the domain restriction $x \ge 0$.

2. Trigonometric Identity Elimination

When parametric equations involve circular or hyperbolic trigonometric functions, direct algebraic isolation of $t$ using inverse functions is cumbersome and introduces extraneous branch restrictions. Instead, isolate the trigonometric components and eliminate $t$ using fundamental Pythagorean identities:

• Circles and Ellipses: Given $x = h + a\cos t$ and $y = k + b\sin t$: $\cos t = \frac{x - h}{a}, \quad \sin t = \frac{y - k}{b}$ Applying $\cos^2 t + \sin^2 t = 1$ yields: $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ If $a = b$, this represents a circle of radius $a$; if $a \neq b$, it represents an ellipse centered at $(h, k)$.
• Hyperbolas: Given $x = h + a\sec t$ and $y = k + b\tan t$: $\sec t = \frac{x - h}{a}, \quad \tan t = \frac{y - k}{b}$ Applying $\sec^2 t - \tan^2 t = 1$ yields: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$

Domain and Range Restrictions

A critical competency on the FTCE Mathematics 6-12 exam is recognizing that eliminating the parameter can artificially expand the curve. The rectangular equation represents the entire geometric locus, but the parametric curve may cover only a subset or segment of that locus:

• Consider $x = \cos t$ and $y = \cos(2t)$ for $t \in \mathbb{R}$.
• Using the double-angle identity: $y = 2\cos^2 t - 1 = 2x^2 - 1$.
• While $y = 2x^2 - 1$ is an unbounded parabola opening upward across all $x \in \mathbb{R}$, the original parametric definition is constrained by the range of cosine: $-1 \le x \le 1$.
• Furthermore, since $\cos(2t) \in [-1, 1]$, the range is strictly $y \in [-1, 1]$.
• Therefore, the parametric curve is merely a bounded parabolic arc, retraced indefinitely as $t$ varies.

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Parameter Elimination Techniques Reference

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Modeling Projectile Motion in the Plane

In classical Newtonian mechanics, an ideal projectile launched into the air undergoes uniform acceleration due to gravity acting purely in the downward vertical direction, while experiencing zero acceleration horizontally (neglecting air resistance).

Let a projectile be launched from an initial coordinate $(x_0, y_0)$ with an initial launch speed $v_0$ at an angle of elevation $\theta$ above the horizontal. The initial velocity decomposes into orthogonal components: $v_{0x} = v_0\cos\theta, \quad v_{0y} = v_0\sin\theta$

Integrating the acceleration vector $\vec{a}(t) = \langle 0, -g \rangle$ twice with respect to time yields the standard kinematic parametric equations: $x(t) = (v_0\cos\theta)t + x_0$ $y(t) = -\frac{1}{2}gt^2 + (v_0\sin\theta)t + y_0$ where $g$ represents the acceleration due to gravity:

• In SI metric units: $g = 9.8\text{ m/s}^2$ (so $\frac{1}{2}g = 4.9$).
• In US customary units: $g = 32\text{ ft/s}^2$ (so $\frac{1}{2}g = 16$).

Derivation of Key Projectile Metrics (for $y_0 = 0$ on Level Ground)

1. Time to Maximum Height ($t_{\text{apex}}$): Vertical velocity vanishes at the peak: $v_y(t) = y'(t) = v_0\sin\theta - gt = 0$, giving: $t_{\text{apex}} = \frac{v_0\sin\theta}{g}$
2. Maximum Vertical Height ($H_{\max}$): Substitute $t_{\text{apex}}$ into $y(t)$: $H_{\max} = y(t_{\text{apex}}) = -\frac{1}{2}g\left(\frac{v_0\sin\theta}{g}\right)^2 + (v_0\sin\theta)\left(\frac{v_0\sin\theta}{g}\right) = \frac{v_0^2\sin^2\theta}{2g}$
3. Total Flight Duration ($t_{\text{flight}}$): Setting $y(t) = 0$ yields $t(-0.5gt + v_0\sin\theta) = 0$. The non-zero landing time is: $t_{\text{flight}} = 2t_{\text{apex}} = \frac{2v_0\sin\theta}{g}$
4. Horizontal Range ($R$): Substitute $t_{\text{flight}}$ into $x(t)$: $R = x(t_{\text{flight}}) = (v_0\cos\theta)\left(\frac{2v_0\sin\theta}{g}\right) = \frac{v_0^2(2\sin\theta\cos\theta)}{g} = \frac{v_0^2\sin(2\theta)}{g}$ Because $\sin(2\theta)$ achieves its maximum value of $1$ when $2\theta = 90^\circ$, the maximum horizontal range on flat ground occurs at a launch angle of $\theta = 45^\circ$.

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Worked Exemplar: Elevated Projectile Trajectory and Barrier Clearance

Problem: A soccer ball is kicked from ground level ($x_0 = 0, y_0 = 0$) with an initial velocity of $v_0 = 20\text{ m/s}$ at an angle of elevation of $\theta = 30^\circ$. A defensive wall of height $2.4\text{ meters}$ stands $25\text{ meters}$ downfield from the launch point. Use $g = 9.8\text{ m/s}^2$ and neglect air resistance.

1. Formulate the parametric equations modeling the trajectory.
2. Determine the time $t$ when the ball reaches the defensive wall.
3. Determine whether the ball clears the defensive wall, and calculate its vertical clearance margin.
4. Calculate the total flight time and the horizontal landing distance on level ground.

Step 1: Formulate Parametric Equations

Calculate the horizontal and vertical components of the initial velocity: $v_{0x} = 20\cos 30^\circ = 20\left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3} \approx 17.3205\text{ m/s}$ $v_{0y} = 20\sin 30^\circ = 20\left(\frac{1}{2}\right) = 10\text{ m/s}$

Substitute into the parametric kinematic model with $g = 9.8\text{ m/s}^2$: $x(t) = (10\sqrt{3})t$ $y(t) = -\frac{1}{2}(9.8)t^2 + 10t = -4.9t^2 + 10t$

Step 2: Time to Reach the Defensive Wall

The wall is located at horizontal distance $x = 25\text{ meters}$: $10\sqrt{3}t = 25 \implies t = \frac{25}{10\sqrt{3}} = \frac{5}{2\sqrt{3}} = \frac{5\sqrt{3}}{6}\text{ seconds}$ Numerically: $t \approx \frac{5(1.73205)}{6} = \frac{8.6603}{6} \approx 1.4434\text{ seconds}$

Step 3: Vertical Height and Barrier Clearance

Evaluate the vertical position function $y(t)$ at $t = \frac{5\sqrt{3}}{6}$: $y\left(\frac{5\sqrt{3}}{6}\right) = -4.9\left(\frac{5\sqrt{3}}{6}\right)^2 + 10\left(\frac{5\sqrt{3}}{6}\right)$ $y = -4.9\left(\frac{75}{36}\right) + \frac{50\sqrt{3}}{6} = -4.9\left(\frac{25}{12}\right) + \frac{25\sqrt{3}}{3}$ $y = -\frac{122.5}{12} + \frac{100\sqrt{3}}{12} = \frac{100\sqrt{3} - 122.5}{12}$ Substituting $\sqrt{3} \approx 1.73205$: $y \approx \frac{100(1.73205) - 122.5}{12} = \frac{173.205 - 122.5}{12} = \frac{50.705}{12} \approx 4.225\text{ meters}$

Comparing with the wall height of $2.4\text{ meters}$: $\text{Clearance Margin} = 4.225 - 2.400 = 1.825 \approx 1.83\text{ meters}$ The ball easily clears the defensive wall by approximately $1.83\text{ meters}$.

Step 4: Total Flight Duration and Horizontal Range

The ball strikes the ground when $y(t) = 0$: $-4.9t^2 + 10t = 0 \implies t(-4.9t + 10) = 0$ The non-zero impact time is: $t_{\text{flight}} = \frac{10}{4.9} = \frac{100}{49} \approx 2.0408\text{ seconds}$

Now calculate the horizontal landing range $R$: $R = x\left(\frac{100}{49}\right) = (10\sqrt{3})\left(\frac{100}{49}\right) = \frac{1000\sqrt{3}}{49} \approx \frac{1732.05}{49} \approx 35.35\text{ meters}$

Alternatively, verify using the range formula: $R = \frac{v_0^2\sin(2\theta)}{g} = \frac{20^2\sin(60^\circ)}{9.8} = \frac{400(\sqrt{3}/2)}{9.8} = \frac{200\sqrt{3}}{9.8} = \frac{1000\sqrt{3}}{49} \approx 35.35\text{ meters}$ The results match identically.