# Parameterize the curve of intersection of the surfaces x^{2} + y^{2} = 9, and z = x + 2y (The...

## Question:

Parameterize the curve of intersection of the surfaces {eq}x^{2} + y^{2} = 9{/eq}, and {eq}z = x + 2y{/eq} (The projection of the moving particle on the {eq}xy{/eq}-plane should move counterclockwise. The point {eq}(3, 0, 3){/eq} should correspond to the value {eq}0{/eq} of the time parameter {eq}t{/eq} to that point.)

{eq}x ={/eq}

{eq}y ={/eq}

{eq}z = {/eq}

{eq}0 \leq t \leq 2 \pi{/eq}

## Parametric Equations:

In three dimensions we typically use a vector parameterization to describe a curve. Since this curve exists in real physical space, we call them space curves. The intersection of two surfaces is generally a space curve, and to find an equation for it, it's usually best to start with something familiar.

## Answer and Explanation:

We immediately recognize {eq}x^2 + y^2 = 9 {/eq} as a circle with radius 3, and so we write

{eq}\begin{align*} x &= 3 \cos t \\ y &= 3 \sin t \end{align*} {/eq}

where we need {eq}t \in [0, 2\pi ] {/eq} to get all the way around. Note that this is oriented counterclockwise. Then

{eq}\begin{align*} z &= x+2y \\ &= 3 \cos t + 6 \sin t \end{align*} {/eq}

And so we are also at the point (3,0,3) when {eq}t = 0 {/eq}. So the parameterization of the intersection is

{eq}\begin{align*} \vec r (t) &= \left< 3\cos t, 3\sin t, 3\cos t + 6\sin t \right> \end{align*} {/eq}

#### Learn more about this topic:

Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3
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