>> The plot below is Lissajous figure created with parametric equations X of T equals cosine
of BT and Y of T equals C times sine DT.
We want find of possible positive values for A, B, C, and D. Before we do this, though,
we should be familiar with the basic cosine functioning graph here in red,
and the basic sine function in graph here in black.
Notice how the X values are controlled by the cosine function,
and the Y values are controlled by the sine function.
Again, referring back to our basic cosine and sine functions,
notice how both have a maximum function value of positive 1, and a minimum function value
of negative 1, which means, by analyzing the X values of our figure,
we should be able to determine the value of A, where absolute value of A would be amplitude
of our cosine function, and by analyzing the Y coordinates of our figure,
we should be able to determine the value of C where the absolute value
of C would be the absolute of the sine function.
So let's first focus on determining the value of A, where X of T equals A times cosine of BT.
So we'll project this figure onto the X axis to determine the maximum X value
and the minimum function value, which will help us determine A. So if we project this figure
onto the X axis, notice how the minimum X value is negative 3,
and the maximum X value is positive 3, and that means, A,
since we're using a positive value, would have to be 3.
Meaning the amplitude of our cosine function would be 3.
So X of T must equal 3 times cosine of B times T, and now we'll analyze the Y coordinates
of our figure to determine the value of C. if we project this figure onto the Y axis,
notice how the minimum Y value would be negative 4 here,
and the maximum Y value would be positive 4 here, and therefore, again,
if we're using a positive value for C, C would have to be positive 4.
Meaning the amplitude of our sine function,
which controls the Y coordinates, would have to be 4.
So we have Y of T equals 4 times sine of BT, and notice that B and T are going
to affect the period of our cosine and sine function.
Where if we have Y equals sine B theta, or Y equals cosine B theta, the period would be equal
to 2 pi divided by B, and the frequency would be the reciprocal of the period.
So now what we'll do is trace our figure and determine how many cycles of the cosine function
and how many cycles of the sine function would match the figure exactly once.
Before we do this though, let's determine the point on the figure when T equals 0.
Notice when T is 0 ,X would be equal to 3 times cosine 0, which would be 3, and Y would be equal
to 4 times sine 0, which would be 0, and therefore, when T equals 0,
the point would be 3, 0, which is here, and now as T increases,
the orientation of the graph would be in this direction here.
[ Silence ]
And to simplify thing slightly, let's say we want to trace this entire figure
from T equals 0 to T equals 2 pi radiance.
It doesn't have to be true, but it will help us determine the value of B and D,
because B and D are not unique, but their ratio is, which we'll talk about later.
So now we'll trace the figure, focusing on just the X values and how many cycles
of the cosine function it takes to trace our figure.
Before we do this, if we take a look at the basic cosine function,
notice when the input is 0, we have a maximum function value,
and then when the function value returns back to the maximum function value,
we have one complete cycle of our cosine function.
Since we have a maximum function value of positive 3,
when the X coordinate returns the positive 3,
this will represent once complete cycle of our cosine function.
Let's begin tracing.
Again, we're waiting until we return back to positive 3.
Notice here we're at negative 3, a minimum function value,
which would be half a complete cycle.
So we keep tracing.
[ Silence ]
And notice how now we're finally back at positive 3, and the figure is completely traced,
and therefore, it only takes one complete cycle of the cosine function over this interval for T
to trace our figure, and therefore, the period would be 2 pi divided by 1, and therefore, B,
the coefficient of T, would be 1.
So now we know that X of T equals 3 times cosine of 1 times T or just T.
And now we'll trace a figure again focusing on just the Y values, and determine how many cycles
of the sine function it takes to trace our figure.
But again, before we do this, let's take a look at our basic sine function.
Notice how when the input is 0, we're at the midline or a function value of 0.
We have to return to a function value of 0, not once but twice
for one complete cycle of our sine function.
So let's begin tracing.
Notice how we have a Y value of 0, and then we return back to a Y value of 0 here,
with a return to a function value of 0 twice for one complete cycle.
So we keep tracing.
Here we are back to a Y value of 0 twice.
So this represents one complete cycle of our sine function.
And we keep tracing.
We're back to 0 once, and finally back to 0 twice.
So we have two complete cycles of our sine function, and therefore, the period would have
to be 2 pi divided by 2, and therefore, our coefficient D must equal 2.
So we have 3 times sine of 2T.
As I mentioned earlier, this interval of T, these equations would trace our figure,
but as long of the ratio of B and D stayed the same, meaning the period
of the sine function is half the period
of the cosine function, we would still trace the figure.
It would just take a different interval for T to trace the figure once.
So then as long as D is twice the value of B, we would still create the same figure
from the parametric equations, but I think in most cases, it's easier to use the basic period
of the cosine and sine function for the interval of T to determine B
and D. I hope you found this helpful. ------------------------------6e75b2ac7225--