The Central Idea
In ordinary coordinate geometry, a curve is often written in the form
\[ y=f(x). \]In motion, however, both \(x\) and \(y\) may change with time.
Parametric motion writes
\[ \boxed{ x=x(t), \qquad y=y(t) }. \]The shared parameter \(t\) tells us which position the object occupies at each moment.
What Is a Parameter?
A parameter is an auxiliary variable used to generate coordinates or other quantities.
In motion problems, the parameter is usually time:
\[ t. \]But a parameter does not have to represent time. Other symbols such as \(s\), \(u\), or \(\theta\) may also be used.
Parametric Equations
A pair of equations such as
\[ x=2t+1, \qquad y=t^2 \]gives a point \((x,y)\) for each allowed value of \(t\).
The resulting ordered pairs trace a path in the coordinate plane.
Reading a Parametric Position
Suppose
\[ x(t)=2t+1, \qquad y(t)=t^2. \]At \(t=0\),
\[ x(0)=1, \qquad y(0)=0. \]Therefore the object is at
\[ \boxed{(1,0)}. \]At \(t=2\),
\[ x(2)=5, \qquad y(2)=4, \]so the object is at
\[ \boxed{(5,4)}. \]A Parametric Table
For
\[ x=t, \qquad y=t^2, \]selected values give:
| \(t\) | \(x\) | \(y\) | Point |
|---|---|---|---|
| \(-2\) | \(-2\) | \(4\) | \((-2,4)\) |
| \(-1\) | \(-1\) | \(1\) | \((-1,1)\) |
| \(0\) | \(0\) | \(0\) | \((0,0)\) |
| \(1\) | \(1\) | \(1\) | \((1,1)\) |
| \(2\) | \(2\) | \(4\) | \((2,4)\) |
These points lie on the parabola \(y=x^2\).
The Path and the Motion Are Different Ideas
Parametric equations contain two kinds of information.
- The path is the geometric set of points traced.
- The parameterization tells how those points are traversed.
Two different parameterizations can trace the same path at different rates or in different directions.
Example: Same Path, Different Parameterization
Consider
\[ x=t, \qquad y=t^2. \]Also consider
\[ x=2t, \qquad y=4t^2. \]In the second system,
\[ t=\frac{x}{2}. \]Therefore,
\[ y=4\left(\frac{x}{2}\right)^2=x^2. \]Both parameterizations trace the same parabola:
\[ \boxed{y=x^2}. \]Eliminating the Parameter
Eliminating the parameter means removing \(t\) to obtain a direct relationship between \(x\) and \(y\).
Suppose
\[ x=3t+2, \qquad y=t^2. \]Solve the first equation for \(t\):
\[ t=\frac{x-2}{3}. \]Substitute into \(y=t^2\):
\[ y= \left(\frac{x-2}{3}\right)^2. \]Thus the path is
\[ \boxed{ y=\frac{(x-2)^2}{9} }. \]Elimination May Reveal the Path but Lose Timing
Converting
\[ x=x(t), \qquad y=y(t) \]into a Cartesian equation such as
\[ y=f(x) \]can reveal the geometric curve.
But the Cartesian equation usually does not preserve:
- how fast the object moves;
- which direction it travels;
- when it reaches a point;
- whether it passes through a point more than once.
Direction of Motion
The direction in which a parametric curve is traced depends on how the point changes as \(t\) increases.
For
\[ x=t, \qquad y=t^2, \]increasing \(t\) moves from the upper-left branch of the parabola toward the origin, then upward along the upper-right branch.
Reversing the Parameter Reverses the Motion
Compare
\[ x=t, \qquad y=t^2 \]with
\[ x=-t, \qquad y=t^2. \]Both trace
\[ y=x^2, \]but they travel along the parabola in opposite parameter directions.
Parameter Domains Matter
A parameterization may be restricted to a particular interval.
For example,
\[ x=t, \qquad y=t^2, \qquad 0\le t\le2 \]traces only the portion of \(y=x^2\) from
\[ (0,0) \]to
\[ (2,4). \]The full Cartesian parabola is larger than the actual path traversed by this restricted motion.
Parametric Motion on a Line
A straight-line path can be parameterized in many ways.
For example,
\[ x=1+2t, \qquad y=3-t. \]Solve the first equation for \(t\):
\[ t=\frac{x-1}{2}. \]Substitute into the second:
\[ y=3-\frac{x-1}{2}. \]Therefore,
\[ \boxed{ y=\frac72-\frac12x }. \]Parametric Motion on a Circle
A standard circle parameterization is
\[ \boxed{ x=r\cos t, \qquad y=r\sin t }. \]Using
\[ \cos^2t+\sin^2t=1, \]we obtain
\[ \frac{x^2}{r^2} + \frac{y^2}{r^2} = 1. \]Thus
\[ \boxed{x^2+y^2=r^2}. \]Direction around the Circle
For
\[ x=r\cos t, \qquad y=r\sin t, \]the point begins at
\[ (r,0) \]when \(t=0\).
For small positive \(t\), \(y>0\), so the motion proceeds counterclockwise.
Clockwise Circular Motion
The parameterization
\[ x=r\cos t, \qquad y=-r\sin t \]traces the same circle
\[ x^2+y^2=r^2 \]but in the clockwise direction as \(t\) increases.
Shifted Circular Motion
A circle of radius \(r\) centered at \((h,k)\) can be parameterized by
\[ \boxed{ x=h+r\cos t, \qquad y=k+r\sin t }. \]Eliminating the parameter gives
\[ \boxed{ (x-h)^2+(y-k)^2=r^2 }. \]Parametric Motion on an Ellipse
An ellipse centered at the origin can be parameterized by
\[ x=a\cos t, \qquad y=b\sin t. \]Then
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2t+\sin^2t \] \[ =1. \]Therefore,
\[ \boxed{ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 }. \]Time as the Parameter
In kinematics, \(t\) usually represents time.
Then
\[ x=x(t), \qquad y=y(t) \]means that both horizontal and vertical coordinates are determined at every instant.
The equations describe a moving point rather than merely a static curve.
Position Vector Form
Parametric motion in the plane can be written compactly as
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t)\rangle }. \]In three dimensions,
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle }. \]The next lesson will develop this vector-valued viewpoint more fully.
Velocity from Parametric Equations
If
\[ x=x(t), \qquad y=y(t), \]then the velocity components are
\[ \boxed{ v_x=\frac{dx}{dt}, \qquad v_y=\frac{dy}{dt} }. \]The velocity vector is
\[ \boxed{ \mathbf{v}(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle }. \]Example: Velocity from Parametric Motion
Let
\[ x(t)=t^2+1, \qquad y(t)=3t-2. \]Then
\[ \frac{dx}{dt}=2t \]and
\[ \frac{dy}{dt}=3. \]Therefore,
\[ \boxed{ \mathbf{v}(t)=\langle 2t,3\rangle }. \]Speed from Parametric Motion
Speed is the magnitude of the velocity vector.
In two dimensions,
\[ \boxed{ \text{speed} = \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } }. \]This formula combines the horizontal and vertical rates into one scalar measure of motion.
Example: Speed from Parametric Equations
Continue with
\[ \mathbf{v}(t)=\langle 2t,3\rangle. \]Then
\[ |\mathbf{v}(t)| = \sqrt{(2t)^2+3^2} \] \[ = \sqrt{4t^2+9}. \]Therefore,
\[ \boxed{ \text{speed} = \sqrt{4t^2+9} }. \]Acceleration from Parametric Equations
Differentiate the velocity components:
\[ \boxed{ a_x=\frac{d^2x}{dt^2}, \qquad a_y=\frac{d^2y}{dt^2} }. \]Therefore,
\[ \boxed{ \mathbf{a}(t) = \left\langle \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right\rangle }. \]Example: Acceleration from Parametric Motion
Let
\[ x(t)=t^3, \qquad y(t)=2t^2. \]Velocity:
\[ \mathbf{v}(t) = \langle 3t^2,4t\rangle. \]Acceleration:
\[ \boxed{ \mathbf{a}(t) = \langle 6t,4\rangle }. \]Slope of a Parametric Curve
If \(x\) and \(y\) both depend on \(t\), then
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \]provided
\[ \frac{dx}{dt}\ne0. \]Thus
\[ \boxed{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }. \]Why the Parametric Slope Formula Works
Informally, if
\[ \frac{dy}{dt} \]measures vertical change per unit time and
\[ \frac{dx}{dt} \]measures horizontal change per unit time, then dividing the two gives
\[ \frac{\text{vertical change}}{\text{horizontal change}}. \]That is the ordinary slope of the tangent line.
Example: Tangent Slope
Let
\[ x=t^2+1, \qquad y=t^3. \]Then
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2. \]Therefore,
\[ \frac{dy}{dx} = \frac{3t^2}{2t} \] \[ = \frac{3t}{2} \]for \(t\ne0\).
At \(t=2\),
\[ \boxed{ \frac{dy}{dx}=3 }. \]Velocity Is Tangent to the Path
The vector
\[ \mathbf{v}(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle \]points tangent to the path when the velocity is nonzero.
This gives a geometric interpretation of velocity: it points in the instantaneous direction of motion.
Horizontal Tangents
A parametric curve may have a horizontal tangent when
\[ \frac{dy}{dt}=0 \]while
\[ \frac{dx}{dt}\ne0. \]Then
\[ \frac{dy}{dx}=0. \]Vertical Tangents
A vertical tangent may occur when
\[ \frac{dx}{dt}=0 \]while
\[ \frac{dy}{dt}\ne0. \]In that case, the ordinary slope ratio is undefined, corresponding to a vertical tangent direction.
Example: Horizontal and Vertical Tangents
Let
\[ x=t^2, \qquad y=t^3-3t. \]Then
\[ \frac{dx}{dt}=2t \]and
\[ \frac{dy}{dt}=3t^2-3. \]Horizontal tangents occur when
\[ 3t^2-3=0, \]so
\[ t=\pm1. \]A vertical tangent candidate occurs when
\[ 2t=0, \]so
\[ t=0. \]Motion Can Pass through the Same Point More Than Once
Parametric motion can revisit a location.
For example,
\[ x=\cos t, \qquad y=\sin t \]returns to the same point whenever \(t\) increases by \(2\pi\).
Thus one geometric point can correspond to many parameter values.
Self-Intersecting Paths
More complicated parametric curves can cross themselves.
At a self-intersection, different parameter values may produce the same position but different velocity vectors.
This is another reason the parameter contains motion information that the geometric path alone does not.
Example: A Repeated Position
Let
\[ x=\cos t, \qquad y=\sin t. \]At
\[ t=0, \]the position is
\[ (1,0). \]At
\[ t=2\pi, \]the position is again
\[ (1,0). \]The object has completed one full circuit even though its net displacement is zero.
Uniform Circular Parameterization
Let
\[ x(t)=r\cos(\omega t), \qquad y(t)=r\sin(\omega t), \]where \(r\) and \(\omega\) are constants.
This traces a circle of radius \(r\).
The parameter \(\omega\) controls how rapidly the angle changes with time.
Velocity in Uniform Circular Parameterization
Differentiate:
\[ v_x = -r\omega\sin(\omega t) \]and
\[ v_y = r\omega\cos(\omega t). \]Therefore,
\[ \mathbf{v}(t) = \langle -r\omega\sin(\omega t), r\omega\cos(\omega t) \rangle. \]Speed in Uniform Circular Parameterization
The speed is
\[ |\mathbf{v}| = \sqrt{ r^2\omega^2\sin^2(\omega t) + r^2\omega^2\cos^2(\omega t) }. \]Factor:
\[ |\mathbf{v}| = r|\omega| \sqrt{ \sin^2(\omega t)+\cos^2(\omega t) }. \]Therefore,
\[ \boxed{ |\mathbf{v}|=r|\omega| }. \]The speed is constant.
Acceleration in Uniform Circular Parameterization
Differentiate velocity:
\[ \mathbf{a}(t) = \langle -r\omega^2\cos(\omega t), -r\omega^2\sin(\omega t) \rangle. \]Since
\[ \mathbf{r}(t) = \langle r\cos(\omega t), r\sin(\omega t) \rangle, \]we obtain
\[ \boxed{ \mathbf{a}(t) = -\omega^2\mathbf{r}(t) }. \]The acceleration points inward toward the center.
Parametric Motion and Projectile Motion
A projectile can be modeled parametrically by separating horizontal and vertical coordinates.
A common idealized model is
\[ x(t)=x_0+v_{0x}t \]and
\[ y(t)=y_0+v_{0y}t-\frac12gt^2. \]These equations will be developed more fully in the later lesson on Projectile Motion.
Three-Dimensional Parametric Motion
Motion through space can be described by
\[ x=x(t), \qquad y=y(t), \qquad z=z(t). \]Equivalently,
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle. \]The resulting path may be a line, circle, helix, or more complicated space curve.
Example: Helical Motion
Consider
\[ x(t)=\cos t, \qquad y(t)=\sin t, \qquad z(t)=t. \]The \(x\)- and \(y\)-coordinates move around the unit circle while the \(z\)-coordinate steadily increases.
The resulting path is a helix.
Velocity of a Helix
For
\[ \mathbf{r}(t) = \langle \cos t,\sin t,t \rangle, \]velocity is
\[ \boxed{ \mathbf{v}(t) = \langle -\sin t,\cos t,1 \rangle }. \]Its speed is
\[ \sqrt{ \sin^2t+\cos^2t+1 } \] \[ = \boxed{\sqrt2}. \]Acceleration of a Helix
Differentiate again:
\[ \boxed{ \mathbf{a}(t) = \langle -\cos t,-\sin t,0 \rangle }. \]The acceleration has no vertical component in this example.
Parametric Motion Can Stop Momentarily
A moving point is momentarily at rest when all velocity components are zero.
In two dimensions, this requires
\[ \frac{dx}{dt}=0 \]and
\[ \frac{dy}{dt}=0 \]at the same parameter value.
Example: Momentary Rest
Let
\[ x=t^3, \qquad y=t^2. \]Then
\[ \frac{dx}{dt}=3t^2 \]and
\[ \frac{dy}{dt}=2t. \]At
\[ t=0, \]both components are zero.
Therefore,
\[ \boxed{ \mathbf{v}(0)=\mathbf{0} }. \]The Parameter Can Affect Speed without Changing the Path
Compare
\[ x=\cos t, \qquad y=\sin t \]with
\[ x=\cos(3t), \qquad y=\sin(3t). \]Both trace the unit circle.
But the second completes the circle three times as rapidly with respect to \(t\).
Nonuniform Parameterization
A parameterization does not have to move at constant speed.
Consider
\[ x=\cos(t^2), \qquad y=\sin(t^2). \]The path is still the unit circle because
\[ x^2+y^2=1. \]But the angular parameter \(t^2\) changes at a rate \(2t\), so the motion around the circle is not uniform.
Parametric Distance Traveled
Since speed is
\[ \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 }, \]distance traveled from \(t=a\) to \(t=b\) is
\[ \boxed{ d = \int_a^b \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \,dt }. \]This is the arc length of the path traced over the given parameter interval.
Displacement in Parametric Motion
Displacement depends only on the initial and final positions.
From \(t=a\) to \(t=b\),
\[ \boxed{ \Delta\mathbf{r} = \langle x(b)-x(a), y(b)-y(a) \rangle }. \]Distance traveled generally depends on the entire path between those times.
Example: Displacement versus Distance
For
\[ x=\cos t, \qquad y=\sin t, \qquad 0\le t\le2\pi, \]the initial and final positions are both
\[ (1,0). \]Therefore,
\[ \boxed{ \Delta\mathbf{r}=\mathbf{0} }. \]But the distance traveled is the circumference of the unit circle:
\[ \boxed{d=2\pi}. \]Common Error: Treating \(t\) as a Spatial Coordinate
In
\[ x=x(t), \qquad y=y(t), \]the parameter \(t\) usually controls the coordinates.
It is not ordinarily plotted as a third spatial coordinate unless the model explicitly defines it that way.
Common Error: Eliminating the Parameter and Forgetting the Domain
Suppose
\[ x=t, \qquad y=t^2, \qquad 0\le t\le1. \]Eliminating \(t\) gives
\[ y=x^2. \]But the motion traces only the part with
\[ 0\le x\le1. \]The full parabola is not the actual motion path over the restricted domain.
Common Error: Assuming the Cartesian Curve Gives Direction
The equation
\[ x^2+y^2=1 \]describes a unit circle.
It does not tell whether an object moves clockwise, counterclockwise, or changes direction.
That information comes from the parameterization.
Common Error: Dividing by \(dx/dt\) When It Is Zero
The formula
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]requires care when
\[ \frac{dx}{dt}=0. \]Such points may correspond to vertical tangents, singular behavior, or special cases that require separate analysis.
Common Error: Confusing Path Speed with Coordinate Speed
The quantity
\[ \frac{dx}{dt} \]is only the horizontal velocity component.
The total speed in the plane is
\[ \boxed{ \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } }. \]Common Error: Assuming the Parameterization Is Unique
A single path can have infinitely many parameterizations.
Changing the parameterization can alter:
- speed;
- direction;
- starting point;
- number of times the path is traversed.
The geometric path can remain unchanged.
A Reliable Procedure for Reading Parametric Motion
- Identify the parameter and its allowed domain.
- Evaluate \(x(t)\), \(y(t)\), and \(z(t)\) as needed.
- Form ordered pairs or triples to locate the object.
- Track how those positions change as the parameter increases.
- Determine the direction in which the path is traced.
- Eliminate the parameter when the underlying geometric path is needed.
- Preserve the original parameter information when interpreting motion.
A Reliable Procedure for Velocity and Acceleration
- Start with the coordinate functions.
- Differentiate each coordinate once to obtain velocity components.
- Combine the components into \(\mathbf{v}(t)\).
- Take the magnitude of \(\mathbf{v}(t)\) for speed.
- Differentiate the velocity components to obtain acceleration.
- Evaluate at the requested parameter value.
- Interpret the resulting vectors geometrically.
A Reliable Procedure for Parametric Tangents
- Compute \(dx/dt\).
- Compute \(dy/dt\).
- Form \((dy/dt)/(dx/dt)\) when \(dx/dt\ne0\).
- Set \(dy/dt=0\) to search for horizontal tangents.
- Set \(dx/dt=0\) to search for vertical tangent candidates.
- Check the other derivative before classifying the tangent.
Example: Full Parametric Workflow
Let
\[ x(t)=t^2-1, \qquad y(t)=2t+3. \]At \(t=2\),
\[ x(2)=3, \qquad y(2)=7. \]Therefore the position is
\[ \boxed{(3,7)}. \]Velocity:
\[ \mathbf{v}(t) = \langle 2t,2\rangle. \]At \(t=2\),
\[ \boxed{ \mathbf{v}(2)=\langle4,2\rangle }. \]Speed:
\[ |\mathbf{v}(2)| = \sqrt{4^2+2^2} \] \[ = \boxed{2\sqrt5}. \]Acceleration:
\[ \boxed{ \mathbf{a}(t)=\langle2,0\rangle }. \]Example: Full Elimination Workflow
Let
\[ x=2t-1, \qquad y=t^2+2. \]Solve for \(t\):
\[ t=\frac{x+1}{2}. \]Substitute into \(y\):
\[ y= \left(\frac{x+1}{2}\right)^2+2. \]Therefore the Cartesian path is
\[ \boxed{ y=\frac{(x+1)^2}{4}+2 }. \]Practice 1: Evaluate a Position
Let
\[ x(t)=3t-2, \qquad y(t)=t^2+1. \]Find the position at \(t=2\).
Practice 2: Eliminate the Parameter
Given
\[ x=t+1, \qquad y=t^2, \]eliminate \(t\).
Practice 3: Circular Path
Given
\[ x=4\cos t, \qquad y=4\sin t, \]identify the Cartesian equation of the path.
Practice 4: Velocity Vector
Let
\[ x=t^2, \qquad y=5t-1. \]Find \(\mathbf{v}(t)\).
Practice 5: Speed
For
\[ \mathbf{v}(t)=\langle3,4\rangle, \]find the speed.
Practice 6: Acceleration
Let
\[ x=t^3, \qquad y=t^2+2t. \]Find \(\mathbf{a}(t)\).
Practice 7: Tangent Slope
Let
\[ x=t^2+1, \qquad y=t^3. \]Find \(dy/dx\) at \(t=1\).
Practice 8: Direction of Circular Motion
Consider
\[ x=\cos t, \qquad y=-\sin t. \]Is the unit circle traced clockwise or counterclockwise as \(t\) increases from \(0\)?
Practice 9: Three-Dimensional Motion
Let
\[ \mathbf{r}(t) = \langle t,\;t^2,\;2t+1\rangle. \]Find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\).
Practice 10: Identify the Error
A student converts
\[ x=t, \qquad y=t^2, \qquad 0\le t\le1 \]into \(y=x^2\) and says the object travels along the entire parabola.
Explain the error.
Practice Answers
Practice 1
\[ x(2)=3(2)-2=4, \] \[ y(2)=2^2+1=5. \]Therefore,
\[ \boxed{(4,5)}. \]Practice 2
\[ x=t+1 \]gives
\[ t=x-1. \]Substitute:
\[ y=(x-1)^2. \]Therefore,
\[ \boxed{y=(x-1)^2}. \]Practice 3
\[ \frac{x}{4}=\cos t, \qquad \frac{y}{4}=\sin t. \]Squaring and adding,
\[ \frac{x^2}{16} + \frac{y^2}{16} = 1. \]Therefore,
\[ \boxed{x^2+y^2=16}. \]Practice 4
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=5. \]Therefore,
\[ \boxed{ \mathbf{v}(t)=\langle2t,5\rangle }. \]Practice 5
\[ |\mathbf{v}| = \sqrt{3^2+4^2} \] \[ = \boxed{5}. \]Practice 6
Velocity:
\[ \mathbf{v}(t) = \langle3t^2,2t+2\rangle. \]Acceleration:
\[ \boxed{ \mathbf{a}(t) = \langle6t,2\rangle }. \]Practice 7
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2. \]Therefore,
\[ \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}. \]At \(t=1\),
\[ \boxed{ \frac{dy}{dx}=\frac32 }. \]Practice 8
At \(t=0\), the object begins at \((1,0)\).
For small positive \(t\), \(y=-\sin t<0\), so the motion initially moves downward from the rightmost point of the circle.
Therefore, the circle is traced clockwise.
Practice 9
\[ \mathbf{r}(t) = \langle t,t^2,2t+1\rangle. \] \[ \mathbf{v}(t) = \langle1,2t,2\rangle. \] \[ \mathbf{a}(t) = \langle0,2,0\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}(t)=\langle1,2t,2\rangle, \qquad \mathbf{a}(t)=\langle0,2,0\rangle }. \]Practice 10
Eliminating the parameter correctly gives
\[ y=x^2. \]But the parameter restriction
\[ 0\le t\le1 \]implies
\[ 0\le x\le1. \]Therefore, only the corresponding segment of the parabola is traced, not the entire curve.
What to Remember
Parametric motion describes coordinates using a shared parameter:
\[ \boxed{ x=x(t), \qquad y=y(t) }. \]Velocity is
\[ \boxed{ \mathbf{v}(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle }. \]Speed is
\[ \boxed{ |\mathbf{v}(t)| = \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } }. \]Acceleration is
\[ \boxed{ \mathbf{a}(t) = \left\langle \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right\rangle }. \]Tangent slope is
\[ \boxed{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} } \]when \(dx/dt\ne0\).
The most important conceptual distinction is:
- the path tells which points are traced;
- the parameterization tells when and how they are traced;
- eliminating the parameter can reveal the path but may lose motion information;
- the parameter domain determines which part of the path is actually traversed.
Where We Go Next
Parametric equations describe each coordinate of a moving object. A more compact and powerful formulation packages those coordinates into a single vector-valued function.
The next lesson, Vector-Valued Position Functions, develops \(\mathbf{r}(t)\) directly and uses vector operations to organize position, displacement, velocity, acceleration, and motion through space.