The Central Idea
Parametric motion writes
\[ x=x(t), \qquad y=y(t), \qquad z=z(t). \]A vector-valued position function combines these coordinate functions into one object:
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle }. \]This notation lets us treat position, displacement, velocity, and acceleration using vector operations.
What Is a Vector-Valued Function?
A vector-valued function assigns a vector to each allowed input.
For example,
\[ \mathbf{r}(t) = \langle t,t^2\rangle \]assigns the vector
\[ \langle t,t^2\rangle \]to each value of \(t\).
In kinematics, this vector usually represents the object's position relative to a chosen origin.
Position Vectors in Two Dimensions
In the plane,
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t)\rangle }. \]At a particular time \(t=t_0\),
\[ \mathbf{r}(t_0) = \langle x(t_0),y(t_0)\rangle. \]This vector points from the origin to the object's location.
Position Vectors in Three Dimensions
In space,
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle }. \]Each component describes motion along one coordinate axis.
The resulting path can be a line, circle, helix, or more complicated space curve.
Example: Evaluating a Position Vector
Let
\[ \mathbf{r}(t) = \langle 2t-1,t^2+3\rangle. \]At \(t=2\),
\[ \mathbf{r}(2) = \langle 2(2)-1,2^2+3\rangle \] \[ = \langle 3,7\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}(2)=\langle3,7\rangle }. \]Component Form
A vector-valued function may also be written using unit vectors.
In two dimensions,
\[ \mathbf{r}(t) = x(t)\mathbf{i}+y(t)\mathbf{j}. \]In three dimensions,
\[ \boxed{ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} }. \]The angle-bracket and unit-vector forms represent the same vector.
The Path of a Position Vector
As \(t\) varies, the endpoint of
\[ \mathbf{r}(t) \]traces a geometric path.
The vector itself changes in magnitude and direction as the object moves.
The position vector should not be confused with the path: the vector points from the origin to the current location, while the path is the collection of locations visited.
Example: A Parabolic Path
Consider
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Since
\[ x=t \]and
\[ y=t^2, \]eliminating \(t\) gives
\[ \boxed{y=x^2}. \]Thus the position vector traces a parabola.
Initial and Final Position Vectors
Suppose motion is studied from \(t=t_i\) to \(t=t_f\).
The initial position vector is
\[ \mathbf{r}_i=\mathbf{r}(t_i), \]and the final position vector is
\[ \mathbf{r}_f=\mathbf{r}(t_f). \]These determine the displacement vector.
Displacement Vector
Displacement is final position minus initial position:
\[ \boxed{ \Delta\mathbf{r} = \mathbf{r}_f-\mathbf{r}_i }. \]Equivalently,
\[ \boxed{ \Delta\mathbf{r} = \mathbf{r}(t_f)-\mathbf{r}(t_i) }. \]Example: Two-Dimensional Displacement
Let
\[ \mathbf{r}(t) = \langle t^2,3t\rangle. \]From \(t=1\) to \(t=4\),
\[ \mathbf{r}(1)=\langle1,3\rangle \]and
\[ \mathbf{r}(4)=\langle16,12\rangle. \]Therefore,
\[ \Delta\mathbf{r} = \langle16,12\rangle-\langle1,3\rangle \] \[ = \boxed{\langle15,9\rangle}. \]Magnitude of Displacement
If
\[ \Delta\mathbf{r} = \langle\Delta x,\Delta y\rangle, \]then its magnitude is
\[ \boxed{ |\Delta\mathbf{r}| = \sqrt{(\Delta x)^2+(\Delta y)^2} }. \]In three dimensions,
\[ \boxed{ |\Delta\mathbf{r}| = \sqrt{ (\Delta x)^2+ (\Delta y)^2+ (\Delta z)^2 } }. \]Average Velocity Vector
Average velocity is displacement divided by elapsed time:
\[ \boxed{ \mathbf{v}_{\text{avg}} = \frac{\Delta\mathbf{r}}{\Delta t} }. \]From \(t_i\) to \(t_f\),
\[ \boxed{ \mathbf{v}_{\text{avg}} = \frac{ \mathbf{r}(t_f)-\mathbf{r}(t_i) }{ t_f-t_i } }. \]Example: Average Velocity Vector
Suppose
\[ \Delta\mathbf{r} = \langle15,9\rangle \]over
\[ \Delta t=3. \]Then
\[ \mathbf{v}_{\text{avg}} = \frac{\langle15,9\rangle}{3} \] \[ = \boxed{\langle5,3\rangle}. \]Instantaneous Velocity Vector
Instantaneous velocity is the derivative of the position vector:
\[ \boxed{ \mathbf{v}(t) = \mathbf{r}'(t) }. \]Differentiation is performed component by component.
If
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]then
\[ \boxed{ \mathbf{v}(t) = \langle x'(t),y'(t),z'(t)\rangle }. \]Example: Velocity from Position
Let
\[ \mathbf{r}(t) = \langle t^2,3t-1\rangle. \]Differentiate each component:
\[ \boxed{ \mathbf{v}(t) = \langle2t,3\rangle }. \]At \(t=4\),
\[ \boxed{ \mathbf{v}(4) = \langle8,3\rangle }. \]Velocity Points Tangent to the Path
When
\[ \mathbf{v}(t)\ne\mathbf{0}, \]the velocity vector points tangent to the path in the direction of motion.
This gives velocity both an algebraic and geometric meaning.
Speed Is the Magnitude of Velocity
Speed is a scalar:
\[ \boxed{ \text{speed} = |\mathbf{v}(t)| }. \]In two dimensions,
\[ \boxed{ |\mathbf{v}(t)| = \sqrt{ [x'(t)]^2+[y'(t)]^2 } }. \]In three dimensions,
\[ \boxed{ |\mathbf{v}(t)| = \sqrt{ [x'(t)]^2+ [y'(t)]^2+ [z'(t)]^2 } }. \]Example: Speed from a Velocity Vector
Suppose
\[ \mathbf{v}(t) = \langle2t,3\rangle. \]Then
\[ |\mathbf{v}(t)| = \sqrt{4t^2+9}. \]At \(t=4\),
\[ |\mathbf{v}(4)| = \sqrt{64+9} \] \[ = \boxed{\sqrt{73}}. \]Acceleration Vector
Acceleration is the derivative of velocity:
\[ \boxed{ \mathbf{a}(t) = \mathbf{v}'(t) }. \]Since velocity is the derivative of position,
\[ \boxed{ \mathbf{a}(t) = \mathbf{r}''(t) }. \]Component Form of Acceleration
If
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]then
\[ \boxed{ \mathbf{a}(t) = \langle x''(t), y''(t), z''(t) \rangle }. \]Example: Acceleration from Position
Let
\[ \mathbf{r}(t) = \langle t^3,2t^2,t\rangle. \]Velocity is
\[ \mathbf{v}(t) = \langle3t^2,4t,1\rangle. \]Acceleration is
\[ \boxed{ \mathbf{a}(t) = \langle6t,4,0\rangle }. \]The Position-Velocity-Acceleration Chain
Vector-valued motion follows the same derivative structure as one-dimensional motion:
\[ \boxed{ \mathbf{r}(t) \xrightarrow{\text{differentiate}} \mathbf{v}(t) \xrightarrow{\text{differentiate}} \mathbf{a}(t) }. \]Each differentiation is performed component by component.
Recovering Velocity from Acceleration
Integration reverses differentiation.
If acceleration is known,
\[ \mathbf{a}(t) = \mathbf{v}'(t), \]then
\[ \mathbf{v}(t) = \int\mathbf{a}(t)\,dt+\mathbf{C}. \]The vector constant \(\mathbf{C}\) is determined by initial velocity.
Example: Recovering Velocity
Suppose
\[ \mathbf{a}(t) = \langle2,6t\rangle. \]Integrating component by component gives
\[ \mathbf{v}(t) = \langle2t,3t^2\rangle + \langle C_1,C_2\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}(t) = \langle2t+C_1,3t^2+C_2\rangle }. \]Using Initial Velocity
Suppose additionally that
\[ \mathbf{v}(0) = \langle4,-1\rangle. \]Then
\[ \langle C_1,C_2\rangle = \langle4,-1\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}(t) = \langle2t+4,3t^2-1\rangle }. \]Recovering Position from Velocity
Since
\[ \mathbf{v}(t) = \mathbf{r}'(t), \]position is recovered by integration:
\[ \boxed{ \mathbf{r}(t) = \int\mathbf{v}(t)\,dt+\mathbf{C} }. \]The constant vector is determined by an initial position.
Example: Recovering Position
Suppose
\[ \mathbf{v}(t) = \langle2t+4,3t^2-1\rangle. \]Integrate:
\[ \mathbf{r}(t) = \langle t^2+4t, t^3-t \rangle + \langle C_1,C_2\rangle. \]If
\[ \mathbf{r}(0) = \langle1,5\rangle, \]then
\[ \boxed{ \mathbf{r}(t) = \langle t^2+4t+1, t^3-t+5 \rangle }. \]Definite-Integral Form
If velocity is known and the initial position at time \(t_0\) is known, then
\[ \boxed{ \mathbf{r}(t) = \mathbf{r}(t_0) + \int_{t_0}^{t}\mathbf{v}(s)\,ds }. \]Similarly,
\[ \boxed{ \mathbf{v}(t) = \mathbf{v}(t_0) + \int_{t_0}^{t}\mathbf{a}(s)\,ds }. \]Distance Traveled
Distance traveled is the accumulation of speed.
From \(t=a\) to \(t=b\),
\[ \boxed{ d = \int_a^b |\mathbf{v}(t)|\,dt }. \]In two dimensions,
\[ \boxed{ d = \int_a^b \sqrt{ [x'(t)]^2+[y'(t)]^2 } \,dt }. \]Distance Is Not Usually the Magnitude of Displacement
The magnitude
\[ |\mathbf{r}(b)-\mathbf{r}(a)| \]gives straight-line separation between the endpoints.
The integral
\[ \int_a^b|\mathbf{v}(t)|\,dt \]gives the length of the path actually traveled.
These are equal only when the motion follows the direct segment without backtracking or detour.
Example: Circular Motion
Let
\[ \mathbf{r}(t) = \langle R\cos t, R\sin t \rangle. \]Then
\[ \mathbf{v}(t) = \langle -R\sin t, R\cos t \rangle. \]Speed is
\[ |\mathbf{v}(t)| = \sqrt{ R^2\sin^2t+ R^2\cos^2t } \] \[ = \boxed{R}. \]Acceleration in Circular Motion
Differentiate velocity:
\[ \mathbf{a}(t) = \langle -R\cos t, -R\sin t \rangle. \]Since
\[ \mathbf{r}(t) = \langle R\cos t, R\sin t \rangle, \]we get
\[ \boxed{ \mathbf{a}(t) = -\mathbf{r}(t) }. \]The acceleration points toward the center.
Velocity and Position in Circular Motion Are Perpendicular
For
\[ \mathbf{r}(t) = \langle R\cos t, R\sin t \rangle \]and
\[ \mathbf{v}(t) = \langle -R\sin t, R\cos t \rangle, \]their dot product is
\[ \mathbf{r}\cdot\mathbf{v} = -R^2\cos t\sin t + R^2\sin t\cos t \] \[ =0. \]Therefore,
\[ \boxed{ \mathbf{r}(t)\perp\mathbf{v}(t) }. \]The Unit Tangent Vector
When
\[ \mathbf{v}(t)\ne\mathbf{0}, \]the unit tangent vector is
\[ \boxed{ \mathbf{T}(t) = \frac{\mathbf{v}(t)} {|\mathbf{v}(t)|} }. \]It points in the direction of motion and has magnitude \(1\).
Example: Unit Tangent Vector
Suppose
\[ \mathbf{v}(t) = \langle3,4\rangle. \]Then
\[ |\mathbf{v}| = 5. \]Therefore,
\[ \boxed{ \mathbf{T} = \left\langle \frac35, \frac45 \right\rangle }. \]Speed and the Unit Tangent Vector
Since
\[ \mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|}, \]we can write velocity as
\[ \boxed{ \mathbf{v} = |\mathbf{v}|\mathbf{T} }. \]This separates velocity into:
- speed, \( |\mathbf{v}| \);
- direction, \(\mathbf{T}\).
When the Velocity Vector Is Zero
If
\[ \mathbf{v}(t_0)=\mathbf{0}, \]then the object is momentarily at rest.
The usual unit tangent expression
\[ \frac{\mathbf{v}}{|\mathbf{v}|} \]is undefined at that instant because the denominator is zero.
Example: Momentary Rest
Let
\[ \mathbf{r}(t) = \langle t^3,t^2\rangle. \]Then
\[ \mathbf{v}(t) = \langle3t^2,2t\rangle. \]At \(t=0\),
\[ \boxed{ \mathbf{v}(0)=\langle0,0\rangle }. \]The object is momentarily at rest there.
Constant Velocity Vector
If
\[ \mathbf{v}(t)=\mathbf{v}_0 \]is constant, then
\[ \mathbf{a}(t)=\mathbf{0}. \]Position changes linearly:
\[ \boxed{ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}_0 }. \]Example: Constant Velocity in the Plane
Suppose
\[ \mathbf{r}_0 = \langle1,2\rangle \]and
\[ \mathbf{v}_0 = \langle3,-1\rangle. \]Then
\[ \boxed{ \mathbf{r}(t) = \langle1+3t,2-t\rangle }. \]The path is a straight line because the direction of velocity never changes.
Constant Acceleration Vector
If acceleration is constant,
\[ \mathbf{a}(t)=\mathbf{a}, \]then
\[ \boxed{ \mathbf{v}(t) = \mathbf{v}_0+t\mathbf{a} }. \]Integrating again,
\[ \boxed{ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}_0 + \frac12t^2\mathbf{a} }. \]Component Form of Constant Acceleration
If
\[ \mathbf{a} = \langle a_x,a_y\rangle, \]then
\[ x(t) = x_0+v_{0x}t+\frac12a_xt^2 \]and
\[ y(t) = y_0+v_{0y}t+\frac12a_yt^2. \]This form will be central in projectile motion.
Example: Constant Vector Acceleration
Let
\[ \mathbf{r}_0=\langle0,0\rangle, \qquad \mathbf{v}_0=\langle4,6\rangle, \qquad \mathbf{a}=\langle0,-2\rangle. \]Then
\[ \mathbf{r}(t) = \langle0,0\rangle + t\langle4,6\rangle + \frac12t^2\langle0,-2\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}(t) = \langle4t,6t-t^2\rangle }. \]Three-Dimensional Helical Motion
Consider
\[ \mathbf{r}(t) = \langle \cos t, \sin t, t \rangle. \]The first two components trace a circle while the third increases linearly.
The resulting path is a helix.
Velocity of the Helix
Differentiate:
\[ \boxed{ \mathbf{v}(t) = \langle -\sin t, \cos t, 1 \rangle }. \]The speed is
\[ |\mathbf{v}| = \sqrt{ \sin^2t+\cos^2t+1 } \] \[ = \boxed{\sqrt2}. \]Acceleration of the Helix
Differentiate velocity:
\[ \boxed{ \mathbf{a}(t) = \langle -\cos t, -\sin t, 0 \rangle }. \]The acceleration has no \(z\)-component in this example.
Different Motions Can Share the Same Path
Consider
\[ \mathbf{r}_1(t) = \langle\cos t,\sin t\rangle \]and
\[ \mathbf{r}_2(t) = \langle\cos 2t,\sin 2t\rangle. \]Both trace the unit circle.
But
\[ |\mathbf{v}_1|=1 \]while
\[ |\mathbf{v}_2|=2. \]The second motion traces the same geometric path twice as rapidly.
Direction Can Also Change under Reparameterization
Compare
\[ \mathbf{r}_1(t) = \langle\cos t,\sin t\rangle \]with
\[ \mathbf{r}_2(t) = \langle\cos t,-\sin t\rangle. \]Both trace the unit circle.
The first moves counterclockwise while the second moves clockwise.
Vector-Valued Functions and Tangent Lines
At time \(t=t_0\), the tangent line to the path can be written using the position and velocity vectors:
\[ \boxed{ \mathbf{L}(s) = \mathbf{r}(t_0) + s\mathbf{v}(t_0) }. \]Here \(s\) is a new parameter along the tangent line.
Example: Tangent Line to a Motion Path
Let
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]At \(t=1\),
\[ \mathbf{r}(1) = \langle1,1\rangle. \]Velocity is
\[ \mathbf{v}(t) = \langle1,2t\rangle, \]so
\[ \mathbf{v}(1) = \langle1,2\rangle. \]Therefore a tangent-line parameterization is
\[ \boxed{ \mathbf{L}(s) = \langle1,1\rangle + s\langle1,2\rangle }. \]Average Velocity Points Along the Chord
The displacement vector
\[ \mathbf{r}(t_2)-\mathbf{r}(t_1) \]points from the initial position to the final position.
Therefore, average velocity points along the chord joining those two points.
Instantaneous velocity instead points along the tangent direction.
Secant Vectors Approach the Tangent Vector
Average velocity over a short interval is
\[ \frac{ \mathbf{r}(t+\Delta t)-\mathbf{r}(t) }{ \Delta t }. \]As
\[ \Delta t\to0, \]this secant-based vector approaches
\[ \boxed{ \mathbf{r}'(t) = \mathbf{v}(t) }. \]Speed and Arc Length
Speed measures the rate at which arc length accumulates.
If \(s(t)\) denotes distance traveled along the path, then
\[ \boxed{ \frac{ds}{dt} = |\mathbf{v}(t)| }. \]This connects the geometry of the path with the kinematics of the motion.
Acceleration Need Not Point Along the Path
Velocity is tangent to the path, but acceleration generally is not.
Acceleration may have:
- a component along the tangent that changes speed;
- a component perpendicular to the tangent that changes direction.
A later lesson on Tangential and Normal Components will formalize this decomposition.
Constant Speed Does Not Imply Zero Acceleration
An object can have
\[ |\mathbf{v}(t)|=\text{constant} \]while
\[ \mathbf{a}(t)\ne\mathbf{0}. \]This happens whenever velocity changes direction without changing magnitude, as in uniform circular motion.
Common Error: Treating \(\mathbf{r}(t)\) as a Scalar
A vector-valued position function has multiple components.
Operations such as differentiation and integration must be applied to each component.
For example,
\[ \frac{d}{dt} \langle t^2,\sin t\rangle = \boxed{ \langle2t,\cos t\rangle }. \]Common Error: Confusing Position Vector with Displacement
The position vector is
\[ \mathbf{r}(t). \]Displacement over an interval is
\[ \mathbf{r}(t_f)-\mathbf{r}(t_i). \]They are not the same quantity unless the initial position happens to be the origin.
Common Error: Confusing Velocity with Speed
Velocity is a vector:
\[ \mathbf{v}(t). \]Speed is its magnitude:
\[ \boxed{ |\mathbf{v}(t)| }. \]Velocity contains directional information that speed does not.
Common Error: Taking the Magnitude before Differentiating
Speed is found from
\[ |\mathbf{r}'(t)|, \]not generally from
\[ \frac{d}{dt}|\mathbf{r}(t)|. \]The first measures how quickly the object moves along its path. The second measures how quickly its distance from the origin changes.
Why \(\frac{d}{dt}|\mathbf{r}|\) Is Different from Speed
Consider uniform circular motion:
\[ \mathbf{r}(t) = \langle R\cos t, R\sin t \rangle. \]The distance from the origin is
\[ |\mathbf{r}(t)|=R, \]so
\[ \frac{d}{dt}|\mathbf{r}(t)|=0. \]But the object is moving with speed
\[ |\mathbf{v}(t)|=R. \]Therefore, the two quantities are clearly different.
Common Error: Forgetting Vector Constants of Integration
If
\[ \mathbf{a}(t) = \langle2,4t\rangle, \]then integrating gives
\[ \mathbf{v}(t) = \langle2t,2t^2\rangle + \langle C_1,C_2\rangle. \]Each component may require its own constant.
Common Error: Assuming Zero Acceleration Means Zero Velocity
If
\[ \mathbf{a}(t)=\mathbf{0}, \]then velocity is constant.
That constant may be nonzero.
Therefore, zero acceleration can describe steady straight-line motion.
Common Error: Ignoring the Parameter Domain
A vector-valued function may be defined only on a restricted interval.
The geometric curve obtained by eliminating the parameter may be larger than the actual path traced over that interval.
A Reliable Procedure for Position, Velocity, and Acceleration
- Write the position vector \(\mathbf{r}(t)\).
- Evaluate \(\mathbf{r}(t)\) for requested positions.
- Differentiate component by component to obtain \(\mathbf{v}(t)\).
- Take \(|\mathbf{v}(t)|\) for speed.
- Differentiate again to obtain \(\mathbf{a}(t)\).
- Evaluate vectors at the requested time.
- Interpret magnitudes and directions geometrically.
A Reliable Procedure for Displacement and Average Velocity
- Find \(\mathbf{r}(t_i)\).
- Find \(\mathbf{r}(t_f)\).
- Subtract to obtain \(\Delta\mathbf{r}\).
- Compute \(\Delta t=t_f-t_i\).
- Divide \(\Delta\mathbf{r}\) by \(\Delta t\).
- Take a magnitude only if a scalar size is requested.
A Reliable Procedure When Acceleration Is Given
- Integrate \(\mathbf{a}(t)\) component by component.
- Add a vector constant of integration.
- Use the initial velocity to determine that constant.
- Integrate \(\mathbf{v}(t)\) component by component.
- Add another vector constant.
- Use the initial position to determine the second constant.
- Check the result by differentiating.
Example: Full Vector-Valued Workflow
Let
\[ \mathbf{r}(t) = \langle t^2-1, 2t^3 \rangle. \]Position at \(t=2\):
\[ \mathbf{r}(2) = \langle3,16\rangle. \]Velocity:
\[ \mathbf{v}(t) = \langle2t,6t^2\rangle. \]At \(t=2\),
\[ \mathbf{v}(2) = \langle4,24\rangle. \]Speed:
\[ |\mathbf{v}(2)| = \sqrt{4^2+24^2} \] \[ = \sqrt{592} \] \[ = \boxed{4\sqrt{37}}. \]Acceleration:
\[ \boxed{ \mathbf{a}(t) = \langle2,12t\rangle }. \]Example: Full Recovery Workflow
Suppose
\[ \mathbf{a}(t) = \langle0,6t\rangle, \] \[ \mathbf{v}(0) = \langle2,-1\rangle, \]and
\[ \mathbf{r}(0) = \langle4,3\rangle. \]Integrate acceleration:
\[ \mathbf{v}(t) = \langle C_1,3t^2+C_2\rangle. \]Using \(\mathbf{v}(0)=\langle2,-1\rangle\),
\[ \boxed{ \mathbf{v}(t) = \langle2,3t^2-1\rangle }. \]Integrate velocity:
\[ \mathbf{r}(t) = \langle2t+D_1,t^3-t+D_2\rangle. \]Using \(\mathbf{r}(0)=\langle4,3\rangle\),
\[ \boxed{ \mathbf{r}(t) = \langle2t+4,t^3-t+3\rangle }. \]Practice 1: Evaluate a Position Vector
Let
\[ \mathbf{r}(t) = \langle3t-1,t^2+2\rangle. \]Find \(\mathbf{r}(2)\).
Practice 2: Displacement
Let
\[ \mathbf{r}(t) = \langle t^2,4t\rangle. \]Find the displacement from \(t=1\) to \(t=3\).
Practice 3: Velocity
Let
\[ \mathbf{r}(t) = \langle t^3,2t^2-1\rangle. \]Find \(\mathbf{v}(t)\).
Practice 4: Speed
If
\[ \mathbf{v} = \langle5,12\rangle, \]find the speed.
Practice 5: Acceleration
Let
\[ \mathbf{r}(t) = \langle t^2,t^3,4t\rangle. \]Find \(\mathbf{a}(t)\).
Practice 6: Unit Tangent Vector
Suppose
\[ \mathbf{v} = \langle8,6\rangle. \]Find the unit tangent vector.
Practice 7: Constant Velocity Motion
An object begins at
\[ \mathbf{r}_0=\langle2,-1\rangle \]with constant velocity
\[ \mathbf{v}_0=\langle3,4\rangle. \]Find \(\mathbf{r}(t)\).
Practice 8: Constant Acceleration Motion
Let
\[ \mathbf{r}_0=\langle0,0\rangle, \qquad \mathbf{v}_0=\langle2,5\rangle, \qquad \mathbf{a}=\langle0,-4\rangle. \]Find \(\mathbf{r}(t)\).
Practice 9: Recover Velocity
Suppose
\[ \mathbf{a}(t) = \langle4t,2\rangle \]and
\[ \mathbf{v}(0) = \langle1,3\rangle. \]Find \(\mathbf{v}(t)\).
Practice 10: Identify the Error
A student says the speed of
\[ \mathbf{r}(t) = \langle3\cos t,3\sin t\rangle \]is
\[ \frac{d}{dt}|\mathbf{r}(t)|=0. \]Explain the error and find the actual speed.
Practice Answers
Practice 1
\[ \mathbf{r}(2) = \langle3(2)-1,2^2+2\rangle \] \[ = \boxed{\langle5,6\rangle}. \]Practice 2
\[ \mathbf{r}(1) = \langle1,4\rangle, \] \[ \mathbf{r}(3) = \langle9,12\rangle. \]Therefore,
\[ \Delta\mathbf{r} = \langle9,12\rangle-\langle1,4\rangle \] \[ = \boxed{\langle8,8\rangle}. \]Practice 3
\[ \mathbf{v}(t) = \mathbf{r}'(t) \] \[ = \boxed{ \langle3t^2,4t\rangle }. \]Practice 4
\[ |\mathbf{v}| = \sqrt{5^2+12^2} \] \[ = \boxed{13}. \]Practice 5
Velocity:
\[ \mathbf{v}(t) = \langle2t,3t^2,4\rangle. \]Acceleration:
\[ \boxed{ \mathbf{a}(t) = \langle2,6t,0\rangle }. \]Practice 6
\[ |\mathbf{v}| = \sqrt{8^2+6^2} = 10. \]Therefore,
\[ \boxed{ \mathbf{T} = \left\langle \frac45, \frac35 \right\rangle }. \]Practice 7
\[ \mathbf{r}(t) = \mathbf{r}_0+t\mathbf{v}_0. \] \[ = \langle2,-1\rangle + t\langle3,4\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}(t) = \langle2+3t,-1+4t\rangle }. \]Practice 8
\[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}_0 + \frac12t^2\mathbf{a}. \] \[ = \langle0,0\rangle + t\langle2,5\rangle + \frac12t^2\langle0,-4\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}(t) = \langle2t,5t-2t^2\rangle }. \]Practice 9
\[ \mathbf{a}(t) = \langle4t,2\rangle. \]Integrate:
\[ \mathbf{v}(t) = \langle2t^2+C_1,2t+C_2\rangle. \]Since
\[ \mathbf{v}(0)=\langle1,3\rangle, \]we have
\[ C_1=1, \qquad C_2=3. \]Therefore,
\[ \boxed{ \mathbf{v}(t) = \langle2t^2+1,2t+3\rangle }. \]Practice 10
The student differentiated the magnitude of the position vector, which measures the rate of change of distance from the origin.
Speed is the magnitude of the velocity vector.
\[ \mathbf{v}(t) = \langle-3\sin t,3\cos t\rangle. \]Therefore,
\[ |\mathbf{v}(t)| = \sqrt{ 9\sin^2t+9\cos^2t } \] \[ = \boxed{3}. \]What to Remember
A vector-valued position function is
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle }. \]Displacement is
\[ \boxed{ \Delta\mathbf{r} = \mathbf{r}(t_f)-\mathbf{r}(t_i) }. \]Velocity is
\[ \boxed{ \mathbf{v}(t) = \mathbf{r}'(t) }. \]Speed is
\[ \boxed{ |\mathbf{v}(t)| }. \]Acceleration is
\[ \boxed{ \mathbf{a}(t) = \mathbf{r}''(t) }. \]The unit tangent vector is
\[ \boxed{ \mathbf{T}(t) = \frac{\mathbf{v}(t)} {|\mathbf{v}(t)|} }. \]The main structural relationships are:
- position is vector-valued;
- velocity is the derivative of position;
- acceleration is the derivative of velocity;
- speed is the magnitude of velocity;
- velocity points tangent to the path;
- integration recovers velocity and position from their rates of change.
Where We Go Next
Vector-valued position functions describe the motion of one object relative to a chosen coordinate system. Many important problems instead compare the motion of two objects or describe one object's motion as observed from another moving frame.
The next lesson, Relative Motion, develops differences of position and velocity vectors to analyze pursuit, separation, closing speed, and motion from moving viewpoints.