# Parameterization of Curves in Three-Dimensional Space Examples 1

We saw some examples of parameterizing described curves in $\mathbb{R}^3$ on the Parameterization of Curves in Three-Dimensional Space page. We will now look at some more examples of parameterizing curves like this.

## Example 1

**Parameterize the curve of intersection of the surfaces $x^2 + y + z = 2$ and $xy + z = 1$.**

The second equation can be rewritten as $z = 1 - xy$. Plugging this into the first equation gives us:

(1)Now let $x = t$. Then we have that:

(2)Plugging these values of $x$ and $y$ into either of the original equations - let's choose the second equation, gives us:

(3)Therefore a parameterization of the curve of intersection of these two surfaces is given by $\vec{r}(t) = (t, 1 + t, 1 - t - t^2)$.

## Example 2

**Parameterize the curve of intersection of the paraboloid $z = x^2 + y^2$ and the plane $2x - 4y - z = 1$.**

The second equation can be rewritten as $z = 2x - 4y - 1$. Substituting this into the first equation gives us:

(4)Let $x = 2\cos t + 1$ and $y = 2 \sin t - 2$. Then $z = 2x - 4y - 1$ implies that:

(5)Therefore the curve of intersection of the original two surfaces can be parameterized as $\vec{r}(t) = (2\cos t + 1, 2 \sin t - 2, 4 \cos t - 8 \sin t + 9)$.

## Example 3

**Parameterize the curve of intersection of the top unit hemisphere $z = \sqrt{1 - x^2 - y^2}$ and the plane $x + y = 1$.**

Let $x = t$. Then the second equation has that $y = 1 - t$. Plugging these values of $x$ and $y$ into the second equation gives us:

(6)Therefore the curve of intersection of the original two surfaces can be parameterized as $\vec{r}(t) = (t, 1 - t, \sqrt{2(t - t^2)} )$. Note that $t - t^2 ≥ 0$, that is $t ≥ t^2$. This happens for $0 ≤ t ≤ 1$.