# The function \vec{r}(t) traces a circle. Determine the radius, center, and plane containing the...

## Question:

The function \vec{r}(t) traces a circle. Determine the radius, center, and plane containing the circle {eq}\vec{r}(t)=1\vec{i}+(8\cos(t))\vec{j}+(8\sin(t))\vec{k} {/eq}.

## Parametric Curves

A parametric curve in space is given as a vector function of one parameter, {eq}\displaystyle \mathbf{r}(t)=\langle x(t), y(t), z(t) \rangle, t\in\mathbf{R}. {/eq}

To find a surface that contains a given curve, we will try to eliminate the parameter *t*

and obtain a relationship between the three Cartesian coordinates.

If there are sine and cosine terms in the parametric equation, we will combine them and use the trigonometric identity

{eq}\displaystyle \sin^2 t +\cos^2 t=1. {/eq}

## Answer and Explanation:

The curve given parametrically as {eq}\displaystyle \mathbf{r}(t)=\langle 1, 8\cos t, 8\sin t\rangle {/eq} has the *y* and *z* coordinates given as sine and cosine functions, so

{eq}\displaystyle y^2+z^2=64\cos^2 t+64\sin^2 t=64 \iff \text{ the projection of the curve onto the } yz-\text{plane, is the circle centered at the origin and radius }8. {/eq}

The *x*-coordinate is fixed, {eq}\displaystyle x=1, {/eq} therefore the curve is

{eq}\displaystyle \boxed{ \text{ in the plane } x=1, \text{and the circle has a radius of } 8 \text{ and centered at } (8,0,0)}. {/eq}

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from Precalculus: High School

Chapter 24 / Lesson 3