HamiltonHardison’s Exploration of

Assignment 10:  Lissajous Curves

From Jim Wilson’sAssignment Page:  Consider the parametric equations:
              For 0 < t < 50.

Graph these for various values of a and b.  Describe fully.  Compare with

For an even integer b and a = 1, a pattern quickly emerges.  The number of loops in the graph is equal to b.  It also appears that the graph is bounded horizontally between -4 and 4 as a result of .  The graph is vertically bounded between -3 and 3, since .

If a is an even integer and b = 1, (the reverse of the previous cases), then the “rotation” of the above cases occurs with the some dilation due to the different amplitudes.  We might notice that, when a is even, the graph has a + 1 y-intercepts.  When b is 1, we have 1 x-intercept. 

For odd unit fractions, we have


We have a slightly different pattern occurring with the number of intercepts.  It appears that when a is one we have one y-intercept as expected; yet when b is odd we have b intercepts rather than b+1 intercepts as was the case above.

By examining more interesting combinations:

It seems that a determines the number of y intercepts.  If a is odd, then there are a-y-intercepts.  If a is even, there are a + 1 y-intercepts.  If b is odd, then there are b x-intercepts.  If b is even then there are b+1 x-intercepts.  It also appears that an even value for a or b results in a closed loop graph, whereas two odd values result in a graph that has “ends”.

It appears that the graphs of  are equivalent to the graphs of .

Overlays are shown for (a,b) = (3,4); (5,2); and (7,9)

For some reason, the quality of graph seems to be reduced in the second set of graphs due to the manner in which the calculator handles the expression.  I’m guessing that the t-step is too large to gain the needed precision.

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