Vector-Valued Functions Vector-Valued Functions

Introduction

Parametric functions are of the form:

(x,y) = \langle f(t), g(t) \rangle
A parametric function outputs lists of values. In this case:
(x,y)
.

These lists of values are called vectors. The vectors we have seen so far were all 2-dimensional, but vectors can be of any dimension. For example, a vector in 3-dimensional space is of the form:
(x,y,z)
A vector-valued function takes a variable as input and outputs a vector. An example of a 3-dimensional vector-valued function is:
\mathbf{r}(t) = \langle f(t),\, g(t),\, h(t) \rangle
The input of this function is time,
t
, and the output is the 3-dimensional vector,
(x(t),y(t),z(t))

Differentiating Vector-Valued Functions

The derivative of a vector-valued function is found by differentiating each component by the input:

\mathbf{r}'(t) = \left\langle \frac{df}{dt},\, \frac{dg}{dt},\, \frac{dh}{dt} \right\rangle
df/dt
is the rate that the
x
-position changes with respect to
t
.
dg/dt
is the rate that the
y
-position changes with respect to
t
.
dh/dt
is the rate that the
z
-position changes with respect to
t
.

Geometric Interpretation

The derivative of a vector-valued function,

\mathbf{r}'(t)
, is tangent to the curve at
\mathbf{r}(t)
.

For example:
r(t)
: The position of a particle in 3D space.
r'(t)
: The velocity of that particle.

Another Example:
\begin{align*} \mathbf{r}(t) &= \langle \cos(t),\, \sin(t),\, t \rangle \\ \mathbf{r}'(t) &= \langle -\sin(t),\, \cos(t),\, 1 \rangle \end{align*}
The z-component of the derivative shows constant upward speed, while the x and y components trace circular motion.
Click to Rotate.

Examples

Example 1: Basic Differentiation
Find the derivative of the vector-valued function

\mathbf{r}(t) = \langle t^2,\, 3t - 1,\, \sin(t) \rangle
.
\begin{align*} \mathbf{r}'(t) &= \left\langle \frac{d}{dt}(t^2),\, \frac{d}{dt}(3t - 1),\, \frac{d}{dt}(\sin(t)) \right\rangle \\ &= \langle 2t,\, 3,\, \cos(t) \rangle \end{align*}
This derivative vector gives us the instantaneous velocity at any time t. For example, at t = 0, the velocity is
\langle 0,\, 3,\, 1 \rangle
.


Example 2: Circular Motion
A particle moves in a circle of radius 2 in the xy-plane according to
\mathbf{r}(t) = \langle 2\cos(t),\, 2\sin(t),\, 0 \rangle
. Find its velocity and speed at any time t.
\begin{align*} \mathbf{r}'(t) &= \langle -2\sin(t),\, 2\cos(t),\, 0 \rangle \quad \text{(velocity)} \\ \text{speed} &= \|\mathbf{r}'(t)\| = \sqrt{(-2\sin(t))^2 + (2\cos(t))^2 + 0^2} \\ &= \sqrt{4\sin^2 t + 4\cos^2 t} \\ &= \sqrt{4(\sin^2 t + \cos^2 t)} \\ &= \sqrt{4} = 2 \end{align*}
The constant speed of 2 units shows the particle moves uniformly around the circle.


Example 3: Physical Application
A projectile is launched with initial velocity
\langle 2,\, 4,\, 3 \rangle
from the origin. Find its position vector accounting for gravity (g = -9.8 m/s²):
\begin{align*} \mathbf{r}(t) &= \left\langle 2t,\, 4t - \frac{9.8}{2}t^2,\, 3t \right\rangle \\ \mathbf{r}'(t) &= \langle 2,\, 4 - 9.8t,\, 3 \rangle \quad \text{(velocity)} \\ \end{align*}
The projectile reaches its maximum height when the y-component of velocity is zero:
\begin{align*} 4 - 9.8t &= 0 \\ t &= \frac{4}{9.8} \approx 0.41 \text{ seconds} \end{align*}
At this time, the position is approximately
\langle 0.82,\, 1.23,\, 0.82 \rangle
meters.