>> The plot below is a Lissajous figure created with the parametric equations X
of T equals A cosine BT and Y of T equals C sine DT.
So notice how the X coordinates of our figure are controlled by the cosine function
and the Y coordinates of our figure are controlled by the sine function.
So we should already be familiar with the basic cosine function graphed here in red,
as well as the basic sine function graphed here in black.
Notice both functions have a maximum function value of positive 1,
and a minimum function value of negative 1, which means by determining the maximum
and minimum X value of our figure and the maximum and minimum Y value of our figure,
we should be able to determine the value of A and C, where the absolute
of A would be the amplitude of our cosine function, and the absolute value
of C would be the amplitude of the sine function.
So to determine the value of A, we'll project our figure on to the X axis,
so if we project this onto the X axis, notice how the minimum X value is negative 4,
and the maximum X value would be positive 4.
That means the amplitude of our cosine function must be 4,
and since we're using a positive value for A, A equals 4.
So we know that X of T must equal 4 times cosine of BT, and now to find C,
we'll project our figure onto the Y axis to determine the minimum and maximum Y value.
So if which project this onto the Y axis, notice how the minimum Y value,
which occurs twice would be negative 3, and the maximum Y value,
which occurs twice, could be positive 3.
Which means the amplitude of our cosine function must be 3, and again,
because we're using a positive value for C, C is 3.
So we know that Y of T equals 3 times sine of DT.
Now from here, we'll determine the value of B
and D. These values are not unique, but their ratio is.
So let's just assume that we want to trace this entire figure using the interval
for T from 0 to 2 pi radians.
So we'll use this interval for T to find the value of B and D that would work,
and then we'll talk about why these values are not unique.
To do this, we'll trace our figure and determine how many cycles of the cosine function
and how many cycles of the sine function it takes to trace our complete figure.
Before we do this, though, let's find the point on our figure when T equals 0.
When T is 0, X would be 4 times cosine 0, which would be 4, and Y would be equal
to 3 times sine 0, which would be 0.
So when T 0 would be at the point 4, 0 here.
Notice how if we start here, in order to complete our trace, and therefore,
we can think of this as tracing it twice, because it will have one overlap of our figure.
So we'll first focus on the X values, which are controlled by 4 cosine of BT.
Before we do this, though, let's take a look at the basic cosine function.
Notice when the input is 0, we have a maximum function value,
and it's not until the function value returns to the maximum function value
that we have one complete cycle of our cosine function.
So notice how we're starting at our next value of positive 4.
So when we trace this, when we return to positive 4,
this represents one complete cycle of our cosine function.
Let's go ahead and begin tracing.
The orientation would be in this direction here.
Notice here we're at a minimum X value of negative 4.
Remember, we're focusing only on the X values, but from here we have to backtrack.
So we backtrack in this direction,
and notice how we do return back to an X value of positive 4.
So this represents once cycle of our cosine.
Notice how we've traced this piece twice.
As we keep tracing in this direction, notice how we go out to an X value of negative 4.
The minimum X value, but to keep it consistent, we'll retrace this path
so that the figure has been traced twice.
So if we retraced this and return back to the X value of positive 4,
we've just completed our second cycle of the cosine function, and therefore,
if we're using this interval for T, the period of our cosine function would be 2 pi divided
by 2, and therefore, B, the coefficient of T, would be 2.
So we have X of T equals 4 times cosine of 2T.
Again because we're using this interval for T, and now we'll trace our figure again, again,
with overlap, and determine how many cycles of the sine function we need to trace our figure,
and now we'll be focusing on the Y values.
Before we do this, though, let's take a look back at the basic sine function.
Notice how when the input is 0, the function value is 0, and it has to return
to a function value of 0, not once but twice for one complete cycle.
So here we're starting at a Y value of 0, and the second time we return to a Y value of 0,
we have one complete cycle of our assigned function.
So we'll begin tracing.
[ Silence ]
We have a Y value of 0 here once, and we keep tracing.
Now we begin to backtrack and notice how here we've returned to the Y value
of 0 a second time, which represents one cycle of our sine function, and we keep tracing,
and we're back to a Y value of 0 here for the first time of our second cycle.
So we keep tracing, and now we have a Y value of 0 for a second time.
So we have two complete cycles of our sine function,
and we keep tracing, and now we backtrack.
We're back to a Y value of 0 once, and then finally,
we're back to a Y value again a second time to complete our third cycle of our sine function,
and therefore, with this interval T, that period would have to be 2 pi divided by 3,
and therefore, the coefficient of T is 3, and therefore, Y of T equals 3 sine 3T.
Now I mentioned the values of B and D are not unique, as long as the ratio remains the same.
Meaning, as long as the cosine function completes two cycles.
In the same interval that the sine function completes three cycles.
The parametric equation would still create the same figure.
The interval for T would change, but it would still be the same figure,
but I think it would still be easier to use interval for T, which is the period
of the basic sine and the cosine function to determine the most obvious values of B
and D. I hope you found this helpful. ------------------------------e9154f36f07b--