Parameterization of Curves in Three-Dimensional Space Examples 1

# 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)
\begin{align} \quad x^2 + y + 1 - xy = 2 \\ \quad x^2 + y - xy = 1 \end{align}

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

(2)
\begin{align} \quad t^2 + y - ty = 1 \\ \quad y - ty = 1 - t^2 \\ \quad y(1 - t) = 1 - t^2 \\ \quad y = \frac{1 - t^2}{1 - t} \\ \quad y = \frac{(1 + t)(1 - t)}{1 - t} \\ \quad y = 1 + t \end{align}

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

(3)
\begin{align} \quad t(1 + t) + z = 1 \\ \quad t + t^2 + z = 1 \\ \quad z = 1 - t - t^2 \end{align}

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)
\begin{align} \quad x^2 + y^2 = 2x - 4y -1 \\ \quad x^2 - 2x + y^2 + 4y = -1 \\ \quad (x - 1)^2 + (y + 2)^2 = 4 \end{align}

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

(5)
\begin{align} \quad z = 2(2 \cos t + 1) - 4(2 \sin t - 2) - 1 \\ \quad z= 4\cos t + 2 - 8 \sin t + 8 - 1 \\ \quad z = 4 \cos t - 8 \sin t + 9 \end{align}

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)
\begin{align} \quad z = \sqrt{1 - t^2 - (1 - t)^2} = \sqrt{1 - t^2 - 1 + 2t - t^2} = \sqrt{2t - 2t^2} = \sqrt{2(t - t^2)} \end{align}

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 $0 ≤ t ≤ 1$.