House of Math

Parametrization of Curves with GeoGebra

You can use GeoGebra to draw curves using their parametrization:

If you need to find the parametrization of a line, see Item 4. When you find the parametrization, the line itself will also be drawn automatically.
If you want to draw curves in the plane, see Item 3.
If you want to draw curves in space, see Item 2.

GeoGebra Instruction 1

Find the Parametrization of a Line

1.

Open 3D Graphics and Algebra View under

View in

Menu.

2.

If you have a direction vector, enter it in Algebra View using the command

Vector(<Point>)

for the command in the next steps to work. Note the name of the vector in Algebra View, as you will need it later.

3.

Depending on the information you’re given, do the following in Algebra View:

If you have a point and a direction vector, enter

Line(<Point>, <Direction Vector>)

and replace <Point> with your point and <Direction Vector> with the name of the vector from Step 2.
If you are given two points, enter

Line(<Point>, <Point>)

and replace both <Point> fields with each of your points.

The line now appears in 3D Graphics. You can also see the parametrization in Algebra View.

Note! GeoGebra uses the Greek letter $λ$ for the parameter, instead of the usual $t$ .

4.

To move around in 3D Graphics, click Rotate 3D Graphics view

.

Example 1

Find the parametrization of the line through the point
$(x_{0}, y_{0}, z_{0}) = (5, - 6, 7)$

with direction vector
$\vec{r} = [1, 2, 1]$

Following Item 4 above, type

Line(<Point>, <Direction Vector>)

in its own line in Algebra View. Then type

Line(<Point>, <Direction Vector>)

and replace <Point> with (5, -6, 7) and <Direction Vector> with the name of the vector. You will then get the parametrization

(5, - 6, 7) + t (1, 2, 1) = (5 + t, - 6 + 2 t, 7 + t) .

Example 2

Find the parametrization of the line through the points

(- 3, 5, - 4)

and
$(- 1, - 1, - 1)$

Following Item 4 above, type

Line(<Point>, <Point>)

and replace the first <Point> field with (-3, 5, -4) and the second <Point> field with (-1, -1, -1). You will then get the parametrization

(- 3, 5, - 4) + t (2, - 6, 3) = (- 3 + 2 t, 5 - 6 t, 5 + 3 t) .

Screenshot of GeoGebra showing the parametrization of a line

GeoGebra Instruction 2

Drawing a Curve in the Plane Using a Parametrization

1.: Open Algebra View and Graphics View under View in Menu.
2.: Use the command

Curve(<Expression>, <Expression>, <Parameter Variable>, <Start Value>, <End Value>)

in Algebra View. Make sure you use the version where there are two <Expression> fields (there’s another one with three).
3.: Enter the expressions of $x (t)$ and $y (t)$ —in that order—in the <Expression> fields. Enter $t$ as <Parameter Variable>, and set the start value and end value in the <Start Value> and <End Value> fields. Press Enter.
Your curve will appear in Graphics View. Also, the parametrization will be printed in Algebra View.

Example 3

Draw the curve with the parametrization

x (t) = 3 t

and
$y (t) = t + 4 .$

with 0 as start value and 50 as end value of parameter $t$ .

Following Item 3 above, use the command

Curve(<Expression>, <Expression>, <Parameter Variable>, <Start Value>, <End Value>)

by replacing

The first <Expression> field with 3t
The second <Expression> field with t+4
<Parameter Value> with t
<Start Value> with 0
<End Value> with 50

The planar curve will appear in Graphics View and parametrization in Algebra View.

GeoGebra Instruction 3

Drawing a Curve in Space using a Parametrization

1.

Open Algebra View and 3D Graphics under

View in

Menu.

2.

Use the command

Curve(<Expression>, <Expression>, <Expression>, <Parameter-variable>, <Start>, <End>)

in Algebra View. Make sure you use the one where there are three <Expression> fields (there’s another one with three). Enter the expressions for $x (t)$ , $y (t)$ and $z (t)$ , in that order, for <Expression>. Enter $t$ as <Parameter Variable>, and set the start and end values to $t$ for <Start> and <End>. Press Enter.