AP Precalculus Unit 4 Notes: Understanding Parametric, Projectile, and Implicit Conic Models

Parametric Functions and Planar Motion

What a parametric function is (and why you use it)

A parametric function describes a relationship between two variables (often x and y) by introducing a third variable called a **parameter** (commonly t). Instead of writing y directly as a function of x, you write both coordinates as functions of the parameter:

x = f(t)

y = g(t)

You should think of t as an “input dial” that generates a point \big(x(t), y(t)\big) on a curve. As you change t, the point moves and traces out a path in the plane.

This matters because many real situations are naturally driven by something like time, rotation angle, or another controlling quantity. For example:

• A moving object has an x-position and a y-position that both depend on time.
• A point moving around a circle is easiest to describe using an angle parameter.
• Some curves fail the “vertical line test” and cannot be written as a single explicit function y=f(x), but they can still be described parametrically.

A common misconception is thinking “parametric” is just a harder way to write the same thing. In reality, parameterization adds information: it tells you how the curve is traced (direction, speed changes, repeats), not just which points are on the curve.

Notation you’ll see (same idea, different packaging)

Parametric descriptions are often written in one of these equivalent ways:

All three represent the same geometric path; the vector form is especially useful when talking about motion.

How to interpret a parametric graph (start point, direction, and repeats)

When you graph a parametric curve, you’re graphing the set of points produced by the pair \big(x(t),y(t)\big) for t in some interval.

To understand what’s happening, you typically do three things:

1. Find the start point: plug in the starting parameter value (like t=0).
2. Find the direction of motion: evaluate a few increasing values of t and see where the point goes.
3. Check for retracing or looping: sometimes different t values give the same point.

A classic example is the unit circle.

Example 1: A circle traced parametrically

Consider:

x=\cos(t)

y=\sin(t)

for 0\le t\le 2\pi.

What it is: each t gives a point on the unit circle.

Why it works: the identity

\cos^2(t)+\sin^2(t)=1

implies

x^2+y^2=1

So every point satisfies the circle equation.

How it traces:

• At t=0, \big(x,y\big)=\big(1,0\big).
• As t increases slightly, \sin(t) becomes positive and \cos(t) decreases, so the point moves counterclockwise.
• By t=2\pi, you return to \big(1,0\big).

A common mistake is to think the curve is “the function” and ignore the interval for t. The interval matters: if you used 0\le t\le \pi, you would only trace the top half of the circle.

Eliminating the parameter (connecting parametric and Cartesian forms)

Often you’re asked to rewrite a parametric curve as a Cartesian equation in x and y only. This is called eliminating the parameter.

The basic strategy:

1. Solve one equation for t (or for an expression involving t).
2. Substitute into the other equation.
3. Simplify to get a relationship between x and y.

Sometimes you can’t solve cleanly for t, but you can still eliminate it using identities (like trig identities).

Example 2: Eliminate the parameter (a parabola)

Given:

x=t+1

y=t^2-4

Solve the first equation for t:

t=x-1

Substitute into y:

y=(x-1)^2-4

So the parametric curve lies on the parabola y=(x-1)^2-4. Notice the parameter still matters for _how_ the parabola is traced. If t is restricted, you only get part of the parabola.

Planar motion: position, displacement, and (average) velocity

When t represents time, parametric functions become a planar motion model.

• Position at time t is the point \big(x(t),y(t)\big).
• Displacement from time t=a to t=b is the change in position:

\langle x(b)-x(a),\ y(b)-y(a)\rangle

• Average velocity vector over [a,b] is displacement divided by elapsed time:

\left\langle \frac{x(b)-x(a)}{b-a},\ \frac{y(b)-y(a)}{b-a}\right\rangle

• Average speed over [a,b] is total distance traveled divided by elapsed time. Be careful: distance traveled is usually not the same as displacement magnitude if the motion curves or reverses.

A common mistake is mixing up:

• Displacement (straight-line change in position)
• Distance traveled (length of the path actually followed)

In many AP Precalculus problems, you’ll be asked for displacement or average velocity because they are accessible from function values.

Example 3: Displacement and average velocity

Suppose a point moves with

x(t)=2t-1

y(t)=t^2

Find displacement and average velocity from t=1 to t=3.

Compute positions:

• At t=1: \big(x(1),y(1)\big)=\big(1,1\big)
• At t=3: \big(x(3),y(3)\big)=\big(5,9\big)

Displacement:

\langle 5-1,\ 9-1\rangle=\langle 4,8\rangle

Average velocity:

\left\langle \frac{4}{3-1},\ \frac{8}{3-1}\right\rangle=\langle 2,4\rangle

Interpretation: on average, the motion changes x by 2 units per unit time and y by 4 units per unit time over that interval.

Exam Focus

• Typical question patterns:
  • Given x(t) and y(t), identify key points (start/end), direction of travel, or whether the curve loops/retraces.
  • Eliminate t to find a Cartesian equation, then interpret what part of the curve is traced based on the t-interval.
  • Compute displacement and average velocity over a time interval using function values.
• Common mistakes:
  • Treating the parametric curve as if it were a single-valued function y=f(x) and forgetting that it can fail the vertical line test.
  • Eliminating t correctly but ignoring restrictions on x or y caused by the allowed t values.
  • Confusing distance traveled with displacement (especially when motion doubles back).

Parametric Functions Modeling Projectile Motion

Why projectile motion is naturally parametric

A projectile is an object launched into the air that then moves under the influence of gravity (and often, in simplified models, without air resistance). The key reason parametrics fit projectile motion so well is that:

• Horizontal motion and vertical motion happen at the same time
• They follow different kinds of functions

In the simplest model (constant gravitational acceleration downward, no horizontal acceleration), the horizontal position is linear in time while the vertical position is quadratic in time.

So instead of forcing a direct equation y=f(x) immediately, you model motion as:

x=x(t)

y=y(t)

This keeps the time information, which is what you usually care about first (how long it’s in the air, when it hits the ground, when it reaches maximum height).

The standard projectile model (what each parameter means)

A common parameterization is:

x(t)=x_0+v_x t

y(t)=y_0+v_y t-\frac{1}{2}gt^2

Where:

• t is time (usually with t\ge 0)
• \big(x_0,y_0\big) is the initial position at t=0
• v_x is constant horizontal velocity component
• v_y is initial vertical velocity component
• g is the positive gravitational acceleration constant (units depend on the system)

Two important ideas make this model powerful:

1. Independence of components: horizontal and vertical positions are computed separately, then paired.
2. Quadratic vertical position: the parabola in y(t) creates a maximum height and eventually a descent.

A common misconception is that the projectile “slows down” horizontally because it’s going up and down. In this simplified model, the horizontal speed is constant (no air resistance), and only the vertical motion accelerates.

Key questions you can answer from the parametric model

Most projectile problems reduce to solving one of these:

1. Time when it hits the ground: solve y(t)=0 (or y(t)=y_{ground}).
2. Maximum height: occurs at the vertex of the quadratic y(t).
3. Range (horizontal distance traveled before landing): evaluate x(t) at the landing time.

Example 1: Finding flight time and range

A projectile is launched from ground level with:

x(t)=20t

y(t)=15t-5t^2

Assume the projectile lands when it returns to the ground, so y(t)=0.

Step 1: Solve for landing time

15t-5t^2=0

Factor:

5t(3-t)=0

So t=0 (launch) or t=3 (landing). Flight time is 3.

Step 2: Find range

x(3)=20\cdot 3=60

So the projectile lands 60 units horizontally from the launch point.

Interpretation: The path in the plane is a parabola, but time is what lets you find when the projectile is at each point.

Maximum height from the vertex idea

Since y(t)=y_0+v_y t-\frac{1}{2}gt^2 is a quadratic in t opening downward, the maximum occurs at the vertex. For a quadratic

y(t)=at^2+bt+c

the vertex time is

t=-\frac{b}{2a}

In projectile form, a=-\frac{1}{2}g and b=v_y, so the time to maximum height is

t=\frac{v_y}{g}

You can then plug that time back into y(t) to get the maximum height.

A common mistake is to try to find maximum height by maximizing y with respect to x directly. It’s much simpler (and more meaningful) to maximize y(t) because t drives the motion.

Example 2: Maximum height

Using the same model:

y(t)=15t-5t^2

Here a=-5 and b=15. Vertex time:

t=-\frac{15}{2(-5)}=\frac{15}{10}=1.5

Maximum height:

y(1.5)=15(1.5)-5(1.5)^2

Compute:

y(1.5)=22.5-5(2.25)=22.5-11.25=11.25

So the maximum height is 11.25 units.

Connecting projectile parametrics to a Cartesian parabola (eliminating time)

You can often eliminate t to show that the path is a parabola in the xy-plane.

From

x=x_0+v_x t

solve for t:

t=\frac{x-x_0}{v_x}

Then substitute into y(t):

y=y_0+v_y\left(\frac{x-x_0}{v_x}\right)-\frac{1}{2}g\left(\frac{x-x_0}{v_x}\right)^2

This is a quadratic in x, confirming the path is parabolic (in this idealized model).

A frequent exam-style trap: you may correctly eliminate t but forget that physically meaningful times are typically t\ge 0. That can restrict which part of the parabola represents the actual motion.

Example 3: Eliminate t for a projectile

Given:

x(t)=20t

y(t)=15t-5t^2

Solve for t from x=20t:

t=\frac{x}{20}

Substitute:

y=15\left(\frac{x}{20}\right)-5\left(\frac{x}{20}\right)^2

Simplify:

y=\frac{3}{4}x-\frac{1}{80}x^2

So the path is a downward-opening parabola in the plane.

Exam Focus

• Typical question patterns:
  • Given x(t) and y(t), solve y(t)=0 (or another height) to find when the projectile lands or reaches a target level.
  • Find maximum height by using the vertex of y(t) and then evaluate x(t) at that time if asked where it happens.
  • Eliminate t to show the trajectory is a parabola and interpret domain restrictions from t\ge 0.
• Common mistakes:
  • Choosing the wrong root when solving y(t)=0 (forgetting t=0 is launch and the other root is landing).
  • Treating negative times as physically meaningful in context.
  • Mixing up “range” (horizontal distance) with total distance traveled along the curved path.

Implicitly Defined Functions and Conic Sections

Explicit versus implicit: what “implicit” really means

An explicitly defined function gives y directly in terms of x, like

y=f(x)

An implicitly defined relationship gives an equation involving x and y together, like

F(x,y)=0

For example, the circle

x^2+y^2=1

defines y implicitly. You _can_ solve for y:

y=\sqrt{1-x^2}

y=-\sqrt{1-x^2}

But notice what happened: solving for y produced **two branches**, not a single function. That’s one major reason implicit equations are useful: they naturally represent curves that are not functions of x.

This connects directly back to parametric functions: parametrics provide a way to describe such curves with a single parameter, often avoiding messy “plus or minus” branches.

Conic sections as implicit equations

Conic sections (circles, ellipses, parabolas, hyperbolas) commonly appear as second-degree equations in x and y. A general second-degree form is

Ax^2+Bxy+Cy^2+Dx+Ey+F=0

In many AP Precalculus contexts, you’ll focus on cases with B=0 (no xy term), which are easier to identify and rewrite by completing the square.

Why conics matter here:

• They are classic examples of implicit relationships.
• Many can be parameterized (especially circles and ellipses), creating a strong link between implicit and parametric representations.
• You’re often asked to recognize a conic, rewrite it in standard form, or interpret geometric features (center, radius, vertices).

Completing the square (the main tool for rewriting conics)

When you see an implicit conic equation, you often want to rewrite it into a standard, recognizable form. The core algebra skill is completing the square.

Example pattern:

x^2+6x+y^2-4y=12

Group x terms and y terms:

\left(x^2+6x\right)+\left(y^2-4y\right)=12

Complete squares inside each group:

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

y^2-4y=(y-2)^2-4

Substitute back:

(x+3)^2-9+(y-2)^2-4=12

Combine constants:

(x+3)^2+(y-2)^2=25

Now you can see it’s a circle centered at (-3,2) with radius 5.

A common mistake is to “complete the square” but forget to balance the equation (you must subtract what you added, or add the same amount to both sides).

The four main conics (standard forms and what to look for)

Here are standard forms you’ll commonly aim for.

Circle

A circle with center \big(h,k\big) and radius r:

(x-h)^2+(y-k)^2=r^2

Key features: center and radius.

Ellipse

An ellipse centered at \big(h,k\big):

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

Key features: center; horizontal semi-axis length a (if it’s under x) and vertical semi-axis length b (if it’s under y). The larger denominator corresponds to the longer direction.

Parabola

Two common orientations:

(y-k)^2=4p(x-h)

(x-h)^2=4p(y-k)

Key features: vertex \big(h,k\big) and opening direction (sign of p).

Hyperbola

Two orientations:

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1

Key features: center and which direction it opens (the positive term).

Parametric descriptions of conics (and how they relate to implicit form)

Parametric equations are a natural way to generate conics.

Circle and ellipse via trig parameterization

Because trig functions satisfy

\cos^2(t)+\sin^2(t)=1

you can build circles and ellipses by scaling and shifting:

Circle of radius r centered at origin:

x=r\cos(t)

y=r\sin(t)

Ellipse centered at origin:

x=a\cos(t)

y=b\sin(t)

Eliminate t by dividing:

\frac{x}{a}=\cos(t)

\frac{y}{b}=\sin(t)

Then use the identity:

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

This gives the ellipse equation.

A misconception to watch for: students sometimes think t is an angle measured in degrees without being told. In most precalculus contexts, trig-based parametrics assume radians.

Parabola via a simple parameterization

Parabolas can be parameterized without trig. For example:

x=t^2

y=2t

Eliminate t:

t=\frac{y}{2}

Substitute into x=t^2:

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

So:

y^2=4x

This is a right-opening parabola. The parametric form traces the parabola in a particular way: as t goes from negative to positive, y increases from negative to positive while x stays nonnegative.

Example 1: Identify and rewrite an implicit conic

Classify and rewrite:

x^2+y^2-4x+6y-12=0

Group and move constant:

x^2-4x+y^2+6y=12

Complete squares:

x^2-4x=(x-2)^2-4

y^2+6y=(y+3)^2-9

Substitute:

(x-2)^2-4+(y+3)^2-9=12

Add constants:

(x-2)^2+(y+3)^2=25

This is a circle with center \big(2,-3\big) and radius 5.

Example 2: From parametric ellipse to implicit equation

Given:

x=4\cos(t)

y=2\sin(t)

Divide:

\frac{x}{4}=\cos(t)

\frac{y}{2}=\sin(t)

Use the identity:

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

So the ellipse is

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

Interpretation: the ellipse is wider in the x direction (semi-axis 4) than in the y direction (semi-axis 2).

Implicit equations and “is it a function?”

A frequent conceptual question is whether an implicit curve defines y as a function of x. The fastest reasoning tool is the **vertical line test**: if some vertical line intersects the curve more than once, then y is not a function of x.

But implicit equations can still define a function on a restricted domain. For the circle

x^2+y^2=1

you can restrict to the top semicircle:

y=\sqrt{1-x^2}

with domain -1\le x\le 1. So the implicit curve represents a full circle, while an explicit function might represent only one branch.

A common mistake is to say “it’s not a function” and stop. On many problems, you’re expected to mention that it can become a function after restricting to a branch or domain.

Exam Focus

• Typical question patterns:
  • Rewrite an implicit quadratic equation by completing the square, then classify it as circle, ellipse, parabola, or hyperbola and state key features (center, radius, vertex).
  • Convert between parametric and implicit forms by eliminating the parameter (especially for circles/ellipses using trig identities).
  • Determine whether an implicit relation defines y as a function of x (often using the vertical line test or by solving for y and noting branches).
• Common mistakes:
  • Completing the square incorrectly (adding a value inside parentheses but not compensating outside).
  • Misidentifying conics by looking only for an x^2 and y^2 term; you must consider signs and coefficients.
  • Eliminating parameters but forgetting that identities like \cos^2(t)+\sin^2(t)=1 are what justify replacing trig expressions with 1.