# Give a piecewise smooth parametrization for the curve which starts at the origin, follows a...

## Question:

Give a piecewise smooth parametrization for the curve which starts at the origin, follows a straight line segment to the point (1, 0, 1), and then follow another line segment to the point (-2, 3, 1).

## Parametric Equations:

To make a piecewise smooth parameterization for a curve, we need to break the curve up into pieces so that each piece is continuous and smooth. It is important to remember that the path is allowed to have kinks (since they are impossible to avoid: the path is the path), we just need each *piece* to be smooth and continuous, which we get for free when we parameterize a line.

## Answer and Explanation:

First, we need to get from (0,0,0) to (1,0,1). We can easily parameterize this on {eq}t \in [0,1] {/eq} as

{eq}\begin{align*} \vec r (t) &= \left< t,0,t \right> \end{align*} {/eq}

Next we need to get from (1,0,1) to (-2,3,1). We need to subtract 3 from {eq}x {/eq} to get it from 1 to -2, we need to add 3 to {eq}y {/eq} to get it from 0 to 3, and we need to leave {eq}z = 1 {/eq}. To parameterize on {eq}t \in [1,2] {/eq}, we need to make up for the fact that this starts at {eq}t = 1 {/eq} by writing {eq}t \to t-1 {/eq}, so this time we have

{eq}\begin{align*} \vec r (t) &= \left< 1 - 3(t-1), 3(t-1), 1 \right> \\ &= \left< 4 - 3t, 3t-3, 1 \right> \end{align*} {/eq}

Then our piecewise smooth parameterization for the curve is

{eq}\begin{align*} \vec r (t) &= \begin{cases} \left< t,0,t \right> & \text{ if } 0 \leq t \leq 1 \\ \left< 4 - 3t, 3t-3, 1 \right> & \text{ if } 1 \leq t \leq 2 \end{cases} \end{align*} {/eq}

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#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3