The Central Idea
A vector-valued function assigns a vector to each input.
In two dimensions,
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t)\rangle }. \]In three dimensions,
\[ \boxed{ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle }. \]The parameter \(t\) controls where the output point lies.
Scalar-Valued versus Vector-Valued
A scalar-valued function might be
\[ f(t)=t^2. \]Its output is one number.
A vector-valued function might be
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Its output is an ordered pair.
Component Functions
The coordinate functions
\[ x(t), \qquad y(t), \qquad z(t) \]are called the component functions of \(\mathbf{r}(t)\).
Most calculus operations on vector-valued functions are performed componentwise.
Evaluating a Vector-Valued Function
Let
\[ \mathbf{r}(t) = \langle t^2,2t-1\rangle. \]At \(t=3\),
\[ \mathbf{r}(3) = \langle9,5\rangle. \]Thus,
\[ \boxed{ \mathbf{r}(3)=\langle9,5\rangle }. \]Vector-Valued Functions as Curves
As \(t\) varies, the point
\[ \mathbf{r}(t) \]moves through the plane or space.
The set of all output points forms a curve.
The vector-valued function therefore gives a parameterization of that curve.
Example: A Parabola
Consider
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Since
\[ x=t \]and
\[ y=t^2, \]eliminating the parameter gives
\[ \boxed{ y=x^2 }. \]Thus the vector-valued function traces a parabola.
Orientation of a Parameterized Curve
A parameterization contains more information than the geometric curve alone.
It also tells us the direction in which the curve is traced as \(t\) increases.
This direction is called the curve's orientation.
Same Curve, Different Parameterizations
The unit circle can be parameterized by
\[ \mathbf{r}_1(t) = \langle\cos t,\sin t\rangle \]or
\[ \mathbf{r}_2(t) = \langle\cos(2t),\sin(2t)\rangle. \]Both trace the same circle, but \(\mathbf{r}_2\) moves around it at a different parameter rate.
Example: A Circle
Let
\[ \mathbf{r}(t) = \langle a\cos t,a\sin t\rangle. \]Then
\[ x^2+y^2 = a^2\cos^2 t+a^2\sin^2 t. \]Therefore,
\[ \boxed{ x^2+y^2=a^2 }. \]The curve is a circle of radius \(a\).
Example: A Helix
Consider
\[ \mathbf{r}(t) = \langle\cos t,\sin t,t\rangle. \]The first two coordinates trace a circle:
\[ x^2+y^2=1. \]Meanwhile,
\[ z=t \]increases steadily.
The result is a helix winding upward around the \(z\)-axis.
Limits of Vector-Valued Functions
Limits are computed componentwise.
If
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]then
\[ \boxed{ \lim_{t\to a}\mathbf{r}(t) = \left\langle \lim_{t\to a}x(t), \lim_{t\to a}y(t), \lim_{t\to a}z(t) \right\rangle } \]provided the component limits exist.
Continuity
A vector-valued function is continuous at \(t=a\) when each component is continuous there.
Equivalently,
\[ \boxed{ \lim_{t\to a}\mathbf{r}(t) = \mathbf{r}(a) }. \]Derivative of a Vector-Valued Function
The derivative is defined by the vector difference quotient
\[ \boxed{ \mathbf{r}'(t) = \lim_{h\to0} \frac{ \mathbf{r}(t+h)-\mathbf{r}(t) }{h} }. \]When the component derivatives exist, differentiation is componentwise.
Componentwise Derivative Formula
If
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]then
\[ \boxed{ \mathbf{r}'(t) = \langle x'(t),y'(t),z'(t)\rangle }. \]Example: Differentiate a Vector Function
Let
\[ \mathbf{r}(t) = \langle t^2,e^t,\sin t\rangle. \]Then
\[ \boxed{ \mathbf{r}'(t) = \langle2t,e^t,\cos t\rangle }. \]Tangent Vector
The derivative
\[ \mathbf{r}'(t) \]points in the instantaneous direction of motion along the curve.
Therefore, when
\[ \mathbf{r}'(t)\neq\mathbf{0}, \]it is a tangent vector to the curve.
Tangent Line to a Parameterized Curve
At \(t=t_0\), a tangent line can be written
\[ \boxed{ \mathbf{L}(s) = \mathbf{r}(t_0) + s\mathbf{r}'(t_0) }. \]The point is \(\mathbf{r}(t_0)\), and the direction is \(\mathbf{r}'(t_0)\).
Example: Tangent Line
Let
\[ \mathbf{r}(t) = \langle t,t^2,t^3\rangle. \]At \(t=1\),
\[ \mathbf{r}(1) = \langle1,1,1\rangle. \]Also,
\[ \mathbf{r}'(t) = \langle1,2t,3t^2\rangle, \]so
\[ \mathbf{r}'(1) = \langle1,2,3\rangle. \]Complete the Tangent Line
Therefore,
\[ \boxed{ \mathbf{L}(s) = \langle1,1,1\rangle + s\langle1,2,3\rangle }. \]Position, Velocity, and Acceleration
When \(t\) represents time and
\[ \mathbf{r}(t) \]represents position, then
\[ \boxed{ \mathbf{v}(t)=\mathbf{r}'(t) } \]is velocity, and
\[ \boxed{ \mathbf{a}(t)=\mathbf{r}''(t) } \]is acceleration.
Velocity Has Direction and Magnitude
The velocity vector tells both:
- the instantaneous direction of motion;
- the instantaneous rate of change of position.
Its magnitude gives speed.
Speed
Speed is the magnitude of velocity:
\[ \boxed{ \text{speed} = \|\mathbf{v}(t)\| = \|\mathbf{r}'(t)\| }. \]In three dimensions,
\[ \boxed{ \|\mathbf{r}'(t)\| = \sqrt{ (x'(t))^2 + (y'(t))^2 + (z'(t))^2 } }. \]Velocity versus Speed
Velocity is a vector.
Speed is a scalar.
Two objects can have the same speed while moving in different directions, so they can have different velocities.
Example: Circular Motion
Let
\[ \mathbf{r}(t) = \langle a\cos t,a\sin t\rangle. \]Then
\[ \mathbf{v}(t) = \langle-a\sin t,a\cos t\rangle. \]The speed is
\[ \sqrt{ a^2\sin^2 t + a^2\cos^2 t }. \]Therefore,
\[ \boxed{ \text{speed}=a }. \]Acceleration in Circular Motion
Differentiate again:
\[ \mathbf{a}(t) = \langle-a\cos t,-a\sin t\rangle. \]Therefore,
\[ \boxed{ \mathbf{a}(t) = -\mathbf{r}(t) }. \]The acceleration points inward toward the center of the circle.
Unit Tangent Vector
When
\[ \mathbf{r}'(t)\neq\mathbf{0}, \]the unit tangent vector is
\[ \boxed{ \mathbf{T}(t) = \frac{ \mathbf{r}'(t) }{ \|\mathbf{r}'(t)\| } }. \]It records direction without including speed.
Example: Unit Tangent
For
\[ \mathbf{r}(t) = \langle t,t^2\rangle, \]we have
\[ \mathbf{r}'(t) = \langle1,2t\rangle. \]Its magnitude is
\[ \sqrt{1+4t^2}. \]Therefore,
\[ \boxed{ \mathbf{T}(t) = \frac{ \langle1,2t\rangle }{ \sqrt{1+4t^2} } }. \]Arc Length of a Parameterized Curve
The length of a smooth curve
\[ \mathbf{r}(t), \qquad a\le t\le b, \]is
\[ \boxed{ L = \int_a^b \|\mathbf{r}'(t)\| \,dt }. \]Arc length is therefore the accumulated speed along the parameter interval.
Distance Traveled
If
\[ \mathbf{r}(t) \]is the position of a moving object, then the distance traveled from \(t=a\) to \(t=b\) is
\[ \boxed{ \int_a^b \|\mathbf{v}(t)\| \,dt }. \]This is distinct from net displacement.
Displacement
The displacement from time \(a\) to time \(b\) is
\[ \boxed{ \mathbf{r}(b)-\mathbf{r}(a) }. \]Equivalently,
\[ \boxed{ \int_a^b \mathbf{v}(t)\,dt = \mathbf{r}(b)-\mathbf{r}(a) }. \]Distance versus Displacement
Distance traveled is a scalar:
\[ \int_a^b \|\mathbf{v}(t)\|\,dt. \]Displacement is a vector:
\[ \int_a^b \mathbf{v}(t)\,dt. \]An object can travel a positive distance and still have zero displacement if it returns to its starting point.
Integration of Vector-Valued Functions
Integration is performed componentwise.
If
\[ \mathbf{r}(t) = \langle f(t),g(t),h(t)\rangle, \]then
\[ \boxed{ \int \mathbf{r}(t)\,dt = \left\langle \int f(t)\,dt, \int g(t)\,dt, \int h(t)\,dt \right\rangle }. \]Constants of Integration
For an indefinite vector integral, the constant of integration is itself a constant vector:
\[ \boxed{ \mathbf{C} = \langle C_1,C_2,C_3\rangle }. \]Thus,
\[ \int \mathbf{r}(t)\,dt = \mathbf{R}(t)+\mathbf{C}. \]Recovering Position from Velocity
If velocity is known,
\[ \mathbf{v}(t), \]then position can be recovered by integration:
\[ \mathbf{r}(t) = \int \mathbf{v}(t)\,dt. \]An initial position determines the constant vector.
Example: Recover Position
Suppose
\[ \mathbf{v}(t) = \langle2t,3,-\sin t\rangle. \]Integrating,
\[ \mathbf{r}(t) = \langle t^2,3t,\cos t\rangle + \mathbf{C}. \]An initial condition such as
\[ \mathbf{r}(0)=\mathbf{r}_0 \]determines \(\mathbf{C}\).
Product Rules with Vector Functions
If \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are vector-valued functions, then
\[ \boxed{ \frac{d}{dt} \bigl( \mathbf{u}\cdot\mathbf{v} \bigr) = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' }. \]Cross-Product Rule
In three dimensions,
\[ \boxed{ \frac{d}{dt} \bigl( \mathbf{u}\times\mathbf{v} \bigr) = \mathbf{u}'\times\mathbf{v} + \mathbf{u}\times\mathbf{v}' }. \]The order of the cross-product factors matters.
Derivative of the Squared Magnitude
Since
\[ \|\mathbf{r}\|^2 = \mathbf{r}\cdot\mathbf{r}, \]differentiating gives
\[ \boxed{ \frac{d}{dt} \|\mathbf{r}(t)\|^2 = 2\mathbf{r}(t)\cdot\mathbf{r}'(t) }. \]Constant Distance from the Origin
If
\[ \|\mathbf{r}(t)\| \]is constant, then
\[ \frac{d}{dt} \|\mathbf{r}(t)\|^2 = 0. \]Therefore,
\[ \boxed{ \mathbf{r}(t)\cdot\mathbf{r}'(t)=0 }. \]The position vector is perpendicular to the velocity vector.
Circle Geometry Revisited
For circular motion,
\[ \|\mathbf{r}(t)\|=a \]is constant.
Therefore,
\[ \mathbf{r}\cdot\mathbf{v}=0. \]The radius is perpendicular to the tangent direction.
Chain Rule with a Scalar Field
Suppose
\[ f(x,y,z) \]is scalar-valued and
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle. \]Then
\[ \boxed{ \frac{d}{dt} f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t) }. \]Interpretation of the Chain Rule
The rate of change of a scalar field along a path equals the gradient of the field dotted with the path velocity.
This connects:
- vector-valued motion;
- gradients;
- directional derivatives;
- multivariable chain rules.
Example: Temperature along a Path
Suppose
\[ T(x,y)=x^2+y^2 \]and a particle follows
\[ \mathbf{r}(t) = \langle t,2t\rangle. \]Then
\[ \nabla T = \langle2x,2y\rangle \]and
\[ \mathbf{r}'(t) = \langle1,2\rangle. \]Complete the Temperature Rate
Along the path,
\[ \nabla T(\mathbf{r}(t)) = \langle2t,4t\rangle. \]Therefore,
\[ \frac{dT}{dt} = \langle2t,4t\rangle \cdot \langle1,2\rangle. \]Hence,
\[ \boxed{ \frac{dT}{dt}=10t }. \]Reparameterization
A curve can be described using a new parameter.
Suppose
\[ t=\phi(s). \]Then
\[ \mathbf{R}(s) = \mathbf{r}(\phi(s)) \]traces the same geometric path under suitable conditions, but perhaps at a different speed or orientation.
Derivative under Reparameterization
By the chain rule,
\[ \boxed{ \frac{d\mathbf{R}}{ds} = \mathbf{r}'(\phi(s)) \phi'(s) }. \]Thus a change of parameter rescales the tangent vector by the rate at which the old parameter changes.
Regular Curves
A parameterized curve is often called regular when
\[ \boxed{ \mathbf{r}'(t)\neq\mathbf{0} } \]throughout the interval under consideration.
Regularity guarantees a well-defined tangent direction.
Stationary Parameter Values
If
\[ \mathbf{r}'(t_0)=\mathbf{0}, \]the usual tangent direction cannot be read directly from the first derivative at \(t_0\).
The curve may still possess a geometric tangent, but additional analysis may be required.
Vector-Valued Functions and Motion Models
Position functions can model:
- projectile motion;
- planetary or orbital motion;
- vehicle trajectories;
- robot paths;
- particle motion in fields;
- paths through parameter or state space.
Example: Constant Acceleration
If acceleration is constant,
\[ \mathbf{a}(t)=\mathbf{a}_0, \]then integrating once gives
\[ \mathbf{v}(t) = \mathbf{v}_0 + t\mathbf{a}_0. \]Integrating again gives
\[ \boxed{ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}_0 + \frac12t^2\mathbf{a}_0 }. \]Projectile Motion
In an idealized constant-gravity model,
\[ \mathbf{a} = \langle0,-g\rangle. \]If initial position is
\[ \mathbf{r}_0 = \langle x_0,y_0\rangle \]and initial velocity is
\[ \mathbf{v}_0 = \langle v_{0x},v_{0y}\rangle, \]then
\[ \boxed{ \mathbf{r}(t) = \left\langle x_0+v_{0x}t,\, y_0+v_{0y}t-\frac12gt^2 \right\rangle }. \]Velocity and Acceleration Units
If position is measured in meters and time in seconds, then
\[ \mathbf{r} \]has units of meters,
\[ \mathbf{v} \]has units of meters per second, and
\[ \mathbf{a} \]has units of meters per second squared.
Common Error: Differentiating the Magnitude Instead of the Vector
If
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle, \]then
\[ \mathbf{r}'(t) = \langle x'(t),y'(t)\rangle. \]This is not the same as
\[ \frac{d}{dt}\|\mathbf{r}(t)\|. \]Common Error: Confusing Velocity and Speed
Velocity is
\[ \mathbf{v}(t). \]Speed is
\[ \|\mathbf{v}(t)\|. \]Do not drop the direction information unless the problem specifically asks for speed.
Common Error: Forgetting Componentwise Calculus
Derivatives and integrals of vector-valued functions are computed component by component.
Each coordinate follows the ordinary one-variable rules.
Common Error: Using the Position Vector as the Tangent Direction
The tangent direction is determined by
\[ \mathbf{r}'(t), \]not generally by
\[ \mathbf{r}(t). \]Position and tangent direction are different geometric objects.
Common Error: Assuming Every Parameterization Has Unit Speed
A parameter \(t\) does not automatically represent distance traveled.
Unit-speed motion requires
\[ \boxed{ \|\mathbf{r}'(t)\|=1 }. \]Common Error: Confusing Distance Traveled with Displacement
Distance is
\[ \int_a^b \|\mathbf{r}'(t)\|\,dt. \]Displacement is
\[ \mathbf{r}(b)-\mathbf{r}(a). \]They answer different questions.
A Reliable Differentiation Procedure
- Write the vector-valued function in component form.
- Differentiate each component separately.
- Reassemble the derivatives in the original coordinate order.
- Evaluate at the requested parameter value if needed.
- Interpret the result as tangent direction or velocity when appropriate.
A Reliable Motion Procedure
- Identify the position vector \(\mathbf{r}(t)\).
- Differentiate once to obtain velocity.
- Differentiate again to obtain acceleration.
- Take the magnitude of velocity to obtain speed.
- Integrate speed for distance traveled.
- Subtract endpoint positions for displacement.
- Check physical units and initial conditions.
Example: Full Motion Workflow
Let
\[ \mathbf{r}(t) = \langle t^2,2t,t^3\rangle. \]Then
\[ \mathbf{v}(t) = \langle2t,2,3t^2\rangle. \]Differentiate again:
\[ \mathbf{a}(t) = \langle2,0,6t\rangle. \]Speed in the Full Workflow
The speed is
\[ \|\mathbf{v}(t)\| = \sqrt{ 4t^2 + 4 + 9t^4 }. \]Therefore,
\[ \boxed{ \text{speed} = \sqrt{9t^4+4t^2+4} }. \]Practice 1: Evaluate a Vector Function
Let
\[ \mathbf{r}(t) = \langle t^2,3t-1\rangle. \]Find
\[ \mathbf{r}(2). \]Practice 2: Differentiate
Find
\[ \mathbf{r}'(t) \]for
\[ \mathbf{r}(t) = \langle t^3,e^t,\cos t\rangle. \]Practice 3: Tangent Vector
Find a tangent vector to
\[ \mathbf{r}(t) = \langle t,t^2\rangle \]at \(t=2\).
Practice 4: Tangent Line
Find the tangent line to
\[ \mathbf{r}(t) = \langle t,t^2,t^3\rangle \]at \(t=1\).
Practice 5: Velocity and Acceleration
Let
\[ \mathbf{r}(t) = \langle t^2,3t,t^3\rangle. \]Find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\).
Practice 6: Speed
For
\[ \mathbf{r}(t) = \langle3\cos t,3\sin t\rangle, \]find the speed.
Practice 7: Unit Tangent
Find the unit tangent vector for
\[ \mathbf{r}(t) = \langle t,2t\rangle. \]Practice 8: Displacement
Let
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Find the displacement from \(t=0\) to \(t=2\).
Practice 9: Scalar Field along a Path
Let
\[ f(x,y)=x^2+y \]and
\[ \mathbf{r}(t)=\langle t,t^2\rangle. \]Find
\[ \frac{d}{dt}f(\mathbf{r}(t)). \]Practice 10: Identify the Error
A student says that if
\[ \mathbf{v}(t)=\langle3,4\rangle, \]then the speed is
\[ \langle3,4\rangle. \]Explain the error.
Practice Answers
Practice 1
\[ \mathbf{r}(2) = \langle4,5\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}(2)=\langle4,5\rangle }. \]Practice 2
\[ \boxed{ \mathbf{r}'(t) = \langle3t^2,e^t,-\sin t\rangle }. \]Practice 3
We have
\[ \mathbf{r}'(t) = \langle1,2t\rangle. \]At \(t=2\),
\[ \boxed{ \mathbf{r}'(2)=\langle1,4\rangle }. \]Practice 4
At \(t=1\),
\[ \mathbf{r}(1)=\langle1,1,1\rangle \]and
\[ \mathbf{r}'(1)=\langle1,2,3\rangle. \]Therefore,
\[ \boxed{ \mathbf{L}(s) = \langle1,1,1\rangle + s\langle1,2,3\rangle }. \]Practice 5
\[ \boxed{ \mathbf{v}(t) = \langle2t,3,3t^2\rangle } \]and
\[ \boxed{ \mathbf{a}(t) = \langle2,0,6t\rangle }. \]Practice 6
The velocity is
\[ \langle-3\sin t,3\cos t\rangle. \]Its magnitude is
\[ \sqrt{ 9\sin^2 t + 9\cos^2 t } = 3. \]Therefore,
\[ \boxed{3}. \]Practice 7
We have
\[ \mathbf{r}'(t) = \langle1,2\rangle. \]Its magnitude is
\[ \sqrt5. \]Therefore,
\[ \boxed{ \mathbf{T} = \left\langle \frac1{\sqrt5}, \frac2{\sqrt5} \right\rangle }. \]Practice 8
\[ \mathbf{r}(2) = \langle2,4\rangle \]and
\[ \mathbf{r}(0) = \langle0,0\rangle. \]Thus,
\[ \boxed{ \text{displacement} = \langle2,4\rangle }. \]Practice 9
First,
\[ f(\mathbf{r}(t)) = t^2+t^2 = 2t^2. \]Therefore,
\[ \boxed{ \frac{d}{dt} f(\mathbf{r}(t)) = 4t }. \]The same result follows from
\[ \nabla f\cdot\mathbf{r}'(t). \]Practice 10
The vector
\[ \langle3,4\rangle \]is velocity, not speed.
Speed is the magnitude:
\[ \boxed{ \sqrt{3^2+4^2}=5 }. \]What to Remember
Vector-valued functions describe curves, trajectories, and changing vectors.
- A vector-valued function is built from scalar component functions.
- A parameterized curve records both geometric location and orientation.
- Limits, derivatives, and integrals are computed componentwise.
- \(\mathbf{r}'(t)\) is the tangent vector and, in motion problems, the velocity.
- \(\mathbf{r}''(t)\) is acceleration.
- Speed is \(\|\mathbf{r}'(t)\|\).
- Arc length is accumulated speed: \(L=\int\|\mathbf{r}'(t)\|\,dt\).
- The chain rule \(\frac{d}{dt}f(\mathbf{r}(t)) =\nabla f\cdot\mathbf{r}'(t)\) connects paths with scalar fields.
The central structural idea is
\[ \boxed{ \text{parameter} \longrightarrow \text{position vector} \longrightarrow \text{velocity} \longrightarrow \text{acceleration} }. \]Where We Go Next
Vector-valued functions describe how several coordinates change together along a path. The final lesson in this subsection steps back and studies how multivariable outputs respond to changes in many inputs more generally.
The next lesson, Multivariable Sensitivity, brings together partial derivatives, gradients, directional derivatives, linear approximation, Jacobians, and related tools to quantify how small changes in inputs affect outputs and models.