>> The plot below is a Lissajous figure created with the parametric equations X
of equals A cosine BT, and Y of T equals C sine DT.
We want to find possible positive values for A, B, C,
and D. So first notice how the X value is controlled by the cosine function,
and the Y values are controlled by the sine function, and we started before
with the basic cosine function graphed here in red,
as well as the basic sine function graphed here in black.
Notice for both the cosine and sine functions, the maximum function values are positive 1,
and the minimum function values are negative 1.
So by looking for the maximum and minimum X value, we should be able to determine A,
and by determining the maximum and minimum Y value,
we should be able to determine C. Remember, the absolute value of A
and C would be the amplitudes of our cosine and sine functions.
So to help us determine A, we'll project the graph onto the X axis,
and notice how if we do this, the minimum X value would be negative 4,
and the maximum X value would be positive 4.
And therefore, A, if we're using a positive value, would have to be positive 4,
meaning our cosine function would have an amplitude of positive 4.
So we know that X of T would be equal to 4 times cosine of BT, and now if we project our figure
onto the Y axis, notice how the minimum Y value would be negative 5,
and the maximum Y value would be positive 5, and therefore, C would be 5, meaning the amplitude
of our sine function would have to be 5.
So we know that Y of T equals 5 times sine DT.
And looking at our equation to far, remember B and D affect the period
of our cosine and sine functions.
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 what we'll do is trace out the figure and determine how many cycles
of the cosine function, which control X, and how many cycles of the sine function,
which control Y, it takes to trace the entire figure.
Before we do this, though, let's determine the point on the figure when T equals 0.
Well notice when T is 0, X should be 4 times cosine 0, which is 4,
and Y would be equal to 5 times sine 0, which is 0.
So when T is 0, we'd be at this point here, 4, 0.
So if we start here, in order to trace this figure, we'll have to backtrack, and therefore,
we'll determine how many cycles of cosine and sine it takes to trace this twice.
Let's begin by focusing on the X values of our figure and looking at the basic cosine function.
Notice how at an input value of 0, we're at a maximum,
and then once the function value returns to the maximum,
it represents one cycle of our cosine function.
So in this case, our maximum is 4 here.
So if we trace the figure in this direction here,
and notice how the function values decrease, we reach a minimum value of negative 4,
and from here we have to back track, and as we begin to backtrack, notice how we go back
to a maximum function value which is 4.
So this represents one cycle of our cosine function, and then we'll continue tracing
in this direction where we reach a minimum X value of negative 4, and then we backtrack again
to a maximum X value of positive 4.
So notice how it took two complete cycles of our cosine function to trace our figure twice.
So we wanted to trace this twice using the interval for T, from 0 to 2 pi.
Notice how the period of X of T, notice how the period of our cosine function would have
to be 2 pi divided by 2, and therefore, B, the coefficient of T, would be 2.
So we could say that this is equal to 4 times cosine of 2T.
Again this is assuming that we're using T values to trace our figure twice from 0 to 2 pi,
because B and D are not actually unique, but their ratio is,
which we'll talk about in just a moment.
So now I'll trace our figure again, focusing on the Y values to determine how many cycles
of the sine function it takes to trace our figure, and again,
because we're backtracking, we'll trace it twice.
But before we do this, let's take a look at the sine function.
Notice how when the input is 0, we start at a function value of 0, and we have to return
to the function value of 0 not once, but twice to make one complete cycle.
So again, we're tracing, focusing on the Y values.
So notice as the input increases, the sine function values increase,
and they'll reach a maximum value of 5.
So we would go from here to here, and now we'll start to backtrack.
As we backtrack, notice how we reach a Y value of 0 once, but again,
we have to reach this Y value twice for one complete cycle.
So we keep tracing.
Here we're at a minimum Y value of negative 5, and we start backtracking and we're back
to 0 a second time, which represents one complete cycle of the sine function,
and therefore, the period would be 2 pi divided by 1 and therefore, in this case D would be 1.
So we have Y of T equals 5 sine of 1 times T or just T. Now as I mentioned earlier,
B and D are not unique but their ratio is, meaning as long as B is twice the value of D,
we would still trace this figure twice, but it would be a different interval for T. Another way
to say that would be that as long as the period of our cosine function here is half
of the period of our sine function here, it would still create this graph.
It's just the interval for T to trace the figure would change.
I hope you found this helpful. ------------------------------ed7dcd6c0430--