Defining and Differentiating Vector-Valued Functions

3. Visual Connection

5. Concept Development

BC-only topic. A vector-valued function assigns a vector to each input:

When \(\mathbf r(t)\) represents position, its components are the same parametric equations studied earlier. The vector \(\mathbf r(t)\) points from the origin to the point \((x(t),y(t))\), while the ordered points trace the particle's path as \(t\) changes.

Derivative definition

The derivative is the limit of average vector change:

Vector limits are evaluated componentwise, so whenever the component derivatives exist,

A vector-valued function is differentiable exactly where all of its component functions are differentiable.

Position, velocity, acceleration, and speed

Velocity and acceleration are vectors. Speed is the scalar magnitude

The signs of the velocity components describe instantaneous direction: \(x'>0\) means right, \(x'<0\) means left, \(y'>0\) means up, and \(y'<0\) means down. A particle is at rest only when every velocity component is zero at the same time.

Tangent vectors

If \(\mathbf r'(t_0)\ne\mathbf0\), the derivative is tangent to the path and points in the direction of increasing \(t\). A vector equation of the tangent line is

The corresponding unit tangent vector is

Multiplying a tangent vector by any nonzero scalar changes its length or direction but not the geometric tangent line.

Differentiation rules

Apply familiar scalar rules to each component. For a scalar function \(f\), constant \(c\), and vector functions \(\mathbf r,\mathbf u\),

Product, quotient, and chain rules may therefore appear inside individual components. A vector has no ordinary scalar quotient, so divide components only when multiplication by a scalar reciprocal is defined.

Geometry of constant distance

If \(\|\mathbf r(t)\|=R\) is constant, then \(\mathbf r(t)\cdot\mathbf r(t)=R^2\). Differentiating gives

Thus the position and velocity vectors are perpendicular, matching the geometry of motion tangent to a circle centered at the origin.

Units and representations

If position is measured in meters and \(t\) in seconds, velocity components use meters per second, acceleration components use meters per second squared, and speed uses meters per second. From a graph, \(\mathbf r'\) is tangent; from a table, its components are instantaneous coordinate rates; from formulas, differentiate componentwise.

Scope boundary

Topic 9.4 defines and differentiates vector-valued functions. Topic 9.5 reverses the process by integration, while Topic 9.6 combines position, velocity, acceleration, displacement, and total distance in motion problems.

6. Detailed Worked Example and Error Check

Example 1: Differentiate position componentwise

Let \(\mathbf r(t)=\langle t^2-1,2t\rangle\). Then

At \(t=1\), position is \((0,2)\), velocity is \(\langle2,2\rangle\), and speed is \(\sqrt{2^2+2^2}=2\sqrt2\).

Example 2: Connect the limit definition to components

For \(\mathbf r(t)=\langle3t+4,t^2-4t+3\rangle\), the difference quotient separates into two components. Taking each limit gives

Example 3: Uniform circular motion

Let \(\mathbf r(t)=\langle4\cos t,4\sin t\rangle\). Then

Speed is constantly \(4\). Acceleration points toward the origin, while velocity is perpendicular to position and tangent to the circle.

Example 4: Product and chain rules inside components

For \(\mathbf r(t)=\langle te^t,\sin(t^2)\rangle\),

Each component uses its appropriate scalar differentiation rule.

Example 5: Unit tangent vector

Let \(\mathbf r(t)=\langle t^2,2t\rangle\). At \(t=1\),

Example 6: Find when a particle is at rest

Suppose \(\mathbf r(t)=\langle t^3-3t,t^2-2t\rangle\). Its velocity is

The second component is zero only at \(t=1\), and the first is also zero there. The particle is at rest at \(t=1\), at position \((-2,-1)\). Its acceleration then is \(\mathbf a(1)=\langle6,2\rangle\).

Example 7: Tangent line in vector and scalar form

For \(\mathbf r(t)=\langle t^2,t^3\rangle\) at \(t=1\), the point is \((1,1)\) and the tangent vector is \(\langle2,3\rangle\). Thus

Because the horizontal component is nonzero, the same line is \(y-1=\tfrac32(x-1)\).

Example 8: Interpret tabular vector data

Suppose a table gives \(\mathbf r(2)=\langle3,-1\rangle\), \(\mathbf r'(2)=\langle-2,5\rangle\), and \(\mathbf r''(2)=\langle4,1\rangle\). At \(t=2\), velocity is \(\langle-2,5\rangle\), speed is \(\sqrt{29}\), and acceleration is \(\langle4,1\rangle\). The particle is moving left and up.

Common errors

• Differentiating only one component.
• Confusing the point \((x,y)\) with the velocity vector \(\langle x',y'\rangle\).
• Reporting velocity as speed instead of taking its magnitude.
• Taking the magnitude componentwise rather than using the square root of the sum of squares.
• Declaring the particle at rest when only one velocity component is zero.
• Using acceleration as the tangent direction; velocity is tangent to the path.
• Normalizing by \(\|\mathbf r\|\) instead of \(\|\mathbf r'\|\) when finding \(\mathbf T\).
• Dropping units or treating vector and scalar quantities as interchangeable.