Hello everyone. In our last lecture, we described "flat" objects: lines and planes. But the world, and the physics that describes it, is full of curves. Think about the path of a satellite orbiting the Earth, a roller coaster moving along a track, or even a baseball flying through the air. How can we describe these paths mathematically?
To do this, we need a new kind of function. Instead of a function that takes in a number and outputs another number (like $y=x^2$), we need a function that takes in a number (like time, $t$) and outputs a vector that points to the object's position in space. This is the core idea behind a vector-valued function.
A vector-valued function, or simply a vector function, is a function whose domain is a set of real numbers and whose range is a set of vectors. We typically use the variable $t$ to represent the input, which we often think of as time.
In three dimensions, we can write a vector function $\mathbf{r}(t)$ in terms of its component functions $f(t), g(t),$ and $h(t)$:
A vector function $\mathbf{r}(t)$ is given by:
The functions $f, g,$ and $h$ are real-valued functions of the parameter $t$. The domain of $\mathbf{r}(t)$ is the intersection of the domains of its component functions.
The concepts of limits and continuity extend very naturally to vector functions. To find the limit of a vector function, you simply take the limit of each of its component functions.
A vector function $\mathbf{r}(t)$ is continuous at a point $t=a$ if $\lim_{t \to a} \mathbf{r}(t) = \mathbf{r}(a)$. This simply means that the function is continuous if and only if all of its component functions are continuous.
Consider the vector function $\mathbf{r}(t) = \langle \cos(t), \frac{\sin(t)}{t}, t+1 \rangle$.
The function is continuous for all $t \neq 0$. To find the limit as $t \to 0$, we look at each component:
Therefore, the limit is: $ \lim_{t \to 0} \mathbf{r}(t) = \langle 1, 1, 1 \rangle $.
So, what do these vector functions actually look like? As the parameter $t$ varies, the terminal point of the vector $\mathbf{r}(t)$ traces out a path. We call this path a space curve. The component equations, $x=f(t), y=g(t), z=h(t)$, are the parametric equations for this curve. It's a key insight to realize that because we have only one independent variable ($t$), the curve itself is a one-dimensional object, even though it may twist and turn through three-dimensional space.
One of the more familiar space curves is the helix, described by the vector function:
The $x$ and $y$ components, $x=\cos(t)$ and $y=\sin(t)$, should look familiar. If you recall from 10.1 we parameterized a circle in $\mathbb{R}^2$ as $(r\cos(t), r\sin(t))$. In our example, if we just look at the xy-plane, these are the parametric equations for a circle. So, the projection of this curve onto the xy-plane is a circle.
Meanwhile, the $z$ component, $z=t$, is steadily increasing. So, as the curve goes around in a circle, it's also rising upwards, like a spring or the thread of a screw. 🌀
We've already been working with vector functions without even knowing it! Recall the vector equation of a line from our last lecture:
This is precisely a vector function. If $\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle$ and $\mathbf{v} = \langle a, b, c \rangle$, then the function is $\mathbf{r}(t) = \langle x_0+at, y_0+bt, z_0+ct \rangle$. This is the vector function that traces out a straight line.
Find the vector function for the line segment from point $P(1, 2, 3)$ to point $Q(4, 0, 5)$.
Step 1: Find the direction vector. The direction vector $\mathbf{v}$ is the displacement vector from P to Q.
$$ \mathbf{v} = \vec{PQ} = \langle 4-1, 0-2, 5-3 \rangle = \langle 3, -2, 2 \rangle $$
Step 2: Write the vector function. We use $P$ as our starting point $\mathbf{r}_0$.
$$ \mathbf{r}(t) = \langle 1, 2, 3 \rangle + t \langle 3, -2, 2 \rangle = \langle 1+3t, 2-2t, 3+2t \rangle $$
Step 3: Restrict the parameter. Since we want only the segment from P to Q, we need to find the range for $t$. At $t=0$, we are at $P(1,2,3)$. We get to point $Q$ when $t=1$. Therefore, the parameterization for the segment is valid for $0 \le t \le 1$.
Problem: Find the vector function that represents the line segment from $A(-1, 5, 0)$ to $B(4, 4, 4)$.
Often, a space curve is defined not by a function, but as the intersection of two surfaces (like a cylinder and a plane). Our job is to find the vector function $\mathbf{r}(t)$ that traces this curve.
The key is to find a way to describe all three coordinates ($x, y, z$) in terms of a single parameter, $t$. Here is a reliable strategy.
Find a vector function for the curve of intersection of the cylinder $x^2 + y^2 = 1$ and the plane $y + z = 2$.
Step 1: Choose a Parameter. Look at the two equations. The equation for the cylinder, $x^2 + y^2 = 1$, is very familiar. It describes a circle of radius 1 in the xy-plane. We already know how to parameterize a circle! This is the perfect place to start. We'll use the standard parameterization for a unit circle:
$$ x(t) = \cos(t), \quad y(t) = \sin(t) $$
Step 2: Express Two Variables in Terms of $t$. We've already done this in Step 1 by parameterizing the circle. We have found expressions for both $x$ and $y$ in terms of $t$.
Step 3: Solve for the Final Variable. Now, we use the second surface, the plane $y + z = 2$, to find $z$. We can easily solve this for $z$:
$$ z = 2 - y $$
Since we already know from Step 1 that $y(t) = \sin(t)$, we can just substitute that in:
$$ z(t) = 2 - \sin(t) $$
Result: We now have all three components in terms of $t$. The vector function for the curve of intersection is:
$$ \mathbf{r}(t) = \langle \cos(t), \sin(t), 2 - \sin(t) \rangle $$
This curve is an ellipse, slanted on the plane $y+z=2$, whose projection on the xy-plane is a perfect circle.
Problem: Find a vector function for the curve of intersection of the parabolic cylinder $z = x^2$ and the plane $y = 3x$.
Today we've made the leap from static lines and planes to dynamic curves in space. Vector functions are the language of motion and paths in three dimensions. The key takeaway is that a vector function $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ packages the three parametric equations for a space curve into a single, powerful expression. Let's test your understanding with a few final problems.
Problem: Find the domain of the vector function $\mathbf{r}(t) = \langle \sqrt{t} \cos(3t), \sqrt{t} \sin(3t), \sqrt{4-t} \rangle$.
Problem: Which of the following vector functions parameterizes a circle of radius 4, centered at the origin, in the plane $z=10$?
Problem: Describe or sketch the curve represented by the vector function $\mathbf{r}(t) = \langle 3, \cos(t), \sin(t) \rangle$. What is its shape and orientation?