Parameterization of Curves in Three-Dimensional Space

# Parameterization of Curves in Three-Dimensional Space

Sometimes we can describe a curve as an equation or as the intersections of surfaces in $\mathbb{R}^3$, however, we might rather prefer that the curve is parameterized so that we can easily describe the curve as a vector equation. We will now look at some examples of parameterizing curves in $\mathbb{R}^3$.

## Example 1

Parameterize the line that passes through the point $P(5, 15, 25)$ and $Q(10, 25, 30)$. Find a vector equation equation that represents this line. Find a vector equation that only represents the line segment $\overline{PQ}$.

Recall from the Equations of Lines in Three-Dimensional Space that all the additional information we need to find a set of parametric equations for this line is a vector $\vec{v}$ that is parallel to the line. This is simple to obtain though! Take the vector $\vec{PQ} = (5, 10, 5)$. Therefore we have that:

(1)
\begin{align} \vec{r} = \vec{r_0} + t \vec{v} \\ (x, y, z) = (5, 15, 25) + t (5, 10, 5) \end{align}

Therefore, the set of parametric equations that represent the line that passes through $P$ and $Q$ are $\left\{\begin{matrix} x = 5 + 5t \\ y = 15 + 10t \\ z = 25 + 5t \end{matrix}\right.$.

As a vector equation, we can write this line as $\vec{r}(t) = (5 + 5t, 15 + 10t, 25 + 5t)$.

If we wanted to write a vector equation of ONLY the line segment $\overline{PQ}$, then we will have to restrict $t$. Notice that the point $P$ corresponds to $t = 0$, and the point $Q$ corresponds to $t = 1$. Therefore $\vec{r}(t) = (5 + 5t, 15 + 10t, 25 + 5t)$ for $0 ≤ t ≤ 1$ represents the line segment $\overline{PQ}$.

## Example 2

Parameterize the line of intersection of the planes $x = 3y + 2$ and $y = 4z + 2$ by letting $x = t$. Write a vector equation that represents this line.

Let $x = t$. Then since $x = 3y + 2$, we have that $t = 3y + 2$ and so $y = \frac{t}{3} - \frac{2}{3}$.

Since $y = 4z + 2$, then $\frac{t}{3} - \frac{2}{3} = 4z + 2$, and so $z = \frac{t}{12} - \frac{2}{3}$.

Therefore the line of intersection can be obtained with the parametric equations $\left\{\begin{matrix} x = t\\ y = \frac{t}{3} - \frac{2}{3}\\ z = \frac{t}{12} - \frac{2}{3} \end{matrix}\right.$, or as the vector equation $\vec{r}(t) = \left ( t, \frac{t}{3} - \frac{2}{3}, \frac{t}{12} - \frac{2}{3} \right )$.

## Example 3

Parameterize the curve of intersection of $z = x^2$ and $z = 4y^2$ by using $y = t$ given that the curve passes through the point $(2, -1, 4)$.

Let $y = t$. Then since $z = 4y^2$ we have that $z = 4t^2$., and since $z = x^2$, then $4t^2 = x^2$ which implies that $x = \pm 2t$. Therefore there are two curves of intersection between the surfaces $z = x^2$ and $z = 4y^2$. We need to determine which curve contains the point $(2, -1, 4)$.

If $x = 2t$, then $t = 1$ corresponds to the point $(2, 1, 4)$.
If $x = -2t$, then $t = -1$ corresponds to the point $(2, -1, 4)$. Therefore we have $x = -2t$ as the given point lies on the curve.

Thus the parametric equations $\left\{\begin{matrix} x = -2t\\ y = t\\ z = 4t^2 \end{matrix}\right.$ represent the curve of intersection, or the vector equation $\vec{r}(t) = (-2t, t, 4t^2)$.

## Example 4

Parameterize the curve of intersection of $x^2 + y^2 = 9$ and $z = x + y$ in two different ways.

We first note that $x^2 + y^2 = 9$ represents a cylinder parallel to the $z$-axis and with radius $3$, while $z = x + y$ represents a plane that passes through the point $(0, 0, 0)$, and so the intersection will be an ellipse.

The easiest way to parameterize this curve of intersection is by letting $x = 3 \cos t$ and $y = 3 \sin t$. Therefore $z = 3 \cos t + 3 \sin t$. Therefore the parameterization $\left\{\begin{matrix} x = 3 \cos t\\ y = 3 \sin t\\ z = 3 \cos t + 3 \sin t \end{matrix}\right.$ represents our curve of intersection.

Now let's look at finding another parameterization of this curve of intersection.

Let $x = t$. Then $y =\pm \sqrt{9 - t^2}$, and so $z = t + \pm \sqrt{9 - t^2}$. Therefore the parametric equations $\left\{\begin{matrix} x = t \\ y = \sqrt{9 - t^2}\\ z = t + \sqrt{9 - t^2} \end{matrix}\right.$ and $\left\{\begin{matrix} x = t \\ y = -\sqrt{9 - t^2}\\ z = t - \sqrt{9 - t^2} \end{matrix}\right.$ for $-3 ≤ t ≤ 3$ represents the curve of intersection, each set of parametric equations represent half of the ellipse.

We should note that the first set of parametric equations is much nicer to work with.