Paul's Online Notes
Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Vector Functions
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 12.6 : Vector Functions

6. Identify the graph of the vector function without sketching the graph.

\[\vec r\left( t \right) = \left\langle {3\cos \left( {6t} \right), - 4,\sin \left( {6t} \right)} \right\rangle \]

Show All Steps Hide All Steps

Start Solution

To identify the graph of this vector function (without graphing) let’s first write down the set of parametric equations we get from this vector function. They are,

\[\begin{align*}x & = 3\cos \left( {6t} \right)\\ y & = - 4\\ z & = \sin \left( {6t} \right)\end{align*}\] Show Step 2

Now, from the \(x\) and \(z\) equations we can see that we have an ellipse in the \(xz\)-plane that is given by the following equation.

\[\frac{{{x^2}}}{9} + {z^2} = 1\]

From the \(y\) equation we know that this ellipse will not actually be in the \(xz\)-plane but parallel to the \(xz\)-plane at \(y = - 4\).