# Give parametric equations for the following curves: (a) A circle of radius 3 on the plane y = 1...

## Question:

Give parametric equations for the following curves:

(a) A circle of radius 3 on the plane y = 1 centered at (2,1,0) oriented clockwise when viewed from the origin.

(b) A line perpendicular to z = 2x-3y + 7 and through the point (1,-2,3).

(c) The curve y = (x+2)^3 oriented from (2,64) to (0,8).

(d) The intersection of the surfaces z^2 = x^2 + y^2 and z = 6-x^2 -y^2.

(e) The ellipse with major diameter 2a along the x-axis and minor diameter 2b along the y axis, centered at the origin.

## Parametric Curves

To find a parametrization of a curve given in Cartesian coordinates as {eq}\displaystyle f(x,y,z)=0, g(x,y,z)=0 {/eq}

we look for three functions of a single parameter {eq}\displaystyle t, {/eq} such that

{eq}\displaystyle x=x(t), y=y(t), z=z(t), a\leq t\leq b, {/eq} to verify both Cartesian expressions for all *t*.

If the curve is a conic, like a circle centered at {eq}\displaystyle (a,b) {/eq} and radius {eq}\displaystyle r: (x-a)^2+(y-b)^2=r^2, {/eq} we try to use sine and cosine functions and take

either {eq}\displaystyle x-a =r\cos t, y-b=r\sin t, \text{ for counterclockwise direction} {/eq} or {eq}\displaystyle x-a =r\sin t, y-b=r\cos t, \text{ for clockwise direction}. {/eq}

## Answer and Explanation:

(a) A parametrization for a circle of radius 3 on the plane {eq}\displaystyle y=1, {/eq} centered at {eq}\displaystyle (2,1,0), \text{ which in Cartesian coordinates is } (x-2)^2+z^2=9 \text{ and } y=1, {/eq} oriented clockwise,

we will choose sine for *x* and cosine function for *z*, like below

{eq}\displaystyle x-2=3\sin t, z=3\cos t, y=1\iff \boxed{\mathbf{r}(t)=\langle 2+3\sin t, 1, 3\cos t \rangle, 0\leq t\leq 2\pi} {/eq}

(b) A line perpendicular on the plane {eq}\displaystyle z=2x-3y+7, {/eq} through the point {eq}\displaystyle (1,-2,3) {/eq}

is a line whose direction vector is the normal vector to the plane, {eq}\displaystyle \langle 2,-3,-1\rangle {/eq}

given by the coefficients of *x*, *y*, *z* in one side.

So, the line is given parametrically as {eq}\displaystyle \boxed{\mathbf{r}(t)=\langle 1+2t, -2-3t, 3-t\rangle, 0\leq t\leq 2\pi}. {/eq}

(c) A parametrization for the curve {eq}\displaystyle y=(x+2)^3 {/eq} oriented from (2,64) to (0,8)

is given by choosing {eq}\displaystyle x=2-t: (t=0\implies x=2 \text{ and when } t=2 \implies x=0) {/eq}

and {eq}\displaystyle y=(4-t)^3, {/eq}

so a parametrization for the given curve is {eq}\displaystyle \boxed{\mathbf{r}(t)=\langle 2-t, (4-t)^3\rangle, 0\leq t\leq 2}. {/eq}

(d) For the curve of intersection of the surfaces {eq}\displaystyle z^2=x^2+y^2, z=6-x^2-y^2 {/eq}

we will first solve the system

{eq}\displaystyle z^2=x^2+y^2 \text{ and } z=6-x^2-y^2 \implies z^2+z-6=0 \text{ and } x^2+y^2=z^2 \implies z=2 \text{ and } x^2+y^2 =4 {/eq}

which is given by the following parametric equation

{eq}\displaystyle x=2 \cos t, y=2 \sin t, z=2 \iff \boxed{\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, 2\rangle, 0\leq t\leq 2\pi}. {/eq}

(e) The ellipse with major diameter 2*a* along the *x*-axis and minor diameter 2*b* along the *y* axis, centered at the origin,

is given in Cartesian coordinates as {eq}\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, {/eq}

is given parametrically in terms of multiples of sine and cosine function,

{eq}\displaystyle \boxed{\mathbf{r}(t)=\langle a\cos t, b\sin t \rangle, 0\leq t\leq 2\pi}. {/eq}

#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3