Paul's Online Notes
Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Parametric Equations and Curves
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 9.1 : Parametric Equations and Curves

13. Write down a set of parametric equations for the following equation.

\[{x^2} + {y^2} = 36\]

The parametric curve resulting from the parametric equations should be at \(\left( {6,0} \right)\) when \(t = 0\) and the curve should have a counter clockwise rotation.

Show Solution

If we don’t worry about the “starting” point (i.e. where the curve is at when \(t = 0\)) and we don’t worry about the direction of motion we know from the notes that the following set of parametric equations will trace out a circle of radius 6 centered at the origin.

\[\begin{align*}x & = 6\cos \left( t \right)\\ y & = 6\sin \left( t \right)\end{align*}\]

All we need to do is verify if the extra requirements are met or not.

First, we can clearly see with a quick evaluation that when \(t = 0\) we are at the point \(\left( {6,0} \right)\) as we need to be.

Next, we can either use our knowledge from the examples worked in the notes for this section or an analysis similar to some of the earlier problems in this section to verify that circles in this form will always trace out in a counter clockwise rotation.

In other words, the set of parametric equations give above is a set of parametric equations which will trace out the given circle with the given restrictions. So, formally the answer for this problem is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}x & = 6\cos \left( t \right)\\ y & = 6\sin \left( t \right)\end{align*}}\]

We’ll leave this problem with a final note about the answer here. This is possibly the “simplest” answer we could give but it is completely possible that you may have come up with a different answer to this problem. There are almost always lots of different possible sets of parametric equations that will trace out a particular parametric curve according to some particular set of restrictions.