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# Removing the parameter in parametric equations (example 2)

Video transcript

Let's see if we can remove the
parameter t from a slightly more interesting example. So let's say that x is equal
to 3 times the cosine of t. And y is equal to 2
times the sine of t. We can try to remove the
parameter the same way we did in the previous video, where we
can solve for t in terms of either x or y and then
substitute back in. And I'll do that. But I want to do that first,
just to show you that it kind of leads to a hairy or
an unintuitive answer. So if we solve for--
let's solve for t here. We could do it either one,
they're equally complex. So if we solve for t here,
we would say divide both sides by 2. You'd get y over 2 is
equal to sine of t. And then you would take the
arcsine of both sides, or the inverse sine of both sides, and
you would get-- I like writing arcsine, because inverse sine,
people often confuse it with an exponent, taking it to
the negative 1 power. Arcsine of y over
2 is equal to t. Actually, let me do that
little aside there. I should probably do it at the
trigonometry playlist, but it's a good thing to hit home. Because I think
people get confused. So arcsine of anything,
let's say, y. The other way of writing
that is sine minus 1 of y. These two things are
equivalent, when they're normally used. But I don't like using this
notation most of the time, because it can be ambiguous. This could mean sine of y to
the negative 1 power, which equals 1 over sine of y. And arcsine and this are
definitely not the same thing. So you want to be very careful
there to make sure that you don't get confused when someone
writes an inverse sine like this. But they're not actually
taking sine of y to the negative 1 power. On the other hand, if someone
were to write sine squared of y, this is unambiguously the
same thing as sine of y squared. In fact, I wish this was the
more conventional notation because it wouldn't make people
think, oh, 2 and minus 1 there, and of course, that's
just sine of y squared. So it can be very ambiguous. And of course, if this was a
negative, this would be a minus 2, and then this really would
be 1 over sine of y squared. That's why, just a long-winded
way of explaining why I wrote arcsine, instead of
inverse sine right there. Needless to say, let's
get back to the problem. So we've solved for
t in terms of y. Now we can substitute
back here. And we've got an expression
for x in terms of y. So we get x is equal to 3
times the cosine of t. But we just solved for t. t
is this thing right here. So it's the cosine of
arcsine of y over 2. And we have eliminated the
parameter, but this is a very non-intuitive equation. We could have done
the other way. We could have solved for y in
terms of x and we would have gotten the sine of
the arccosine. It would have been equally
hairy or non-intuitive. But either way, we did remove
the parameters so I guess we could mildly pat
ourselves on the back. But that's not the
purpose of this video. The purpose of this video is to
see if there's any way we can remove the parameter that leads
to a more intuitive equation involving x and y. And what we're going to do is,
I guess you can call it a bit of a trick, but it's something
that shows up a lot. Especially when you deal
with polar coordinates. And you might want to watch
my polar coordinate videos, because this essentially
touches on that. But if I said-- let me rewrite
them. x is equal to 3 cosine of t and y is equal
to 2 sine of t. So what we can do is
just think, well, how can we write this? And you know, cosine
of t and [? kind ?] of t, how can we relate them? And the first thing that comes
to my mind is just the unit circle, or to some degree, the
most basic of all of the trigonometric identities. And that is that the cosine
squared of t plus the sine squared of t is equal to 1. This comes from
the unit circle. I explained it in the unit
circle video, and that's because the equation for the
unit circle is x squared plus y squared is equal to 1. The cosine of the angle is the
x coordinate, the sine of the angle is the y coordinate,
and so on and so forth. But this is our trig identity. You don't have to think about
it too much right now. Just, I guess, know that it's
true and watch some of the other videos if you want
it proven that it's true. But if we can somehow replace
this cosine squared with some expression in x, and replace
the sine or the sine squared with some expression of
y, we'd be done, right? And then we would
have it equaling 1. And that shouldn't be too hard. We can rewrite this. We can set cosine of t equal to
something in x, and we can set sine of t equal in
something in y. So let's do that. We divide both sides
of the equation by 3. You get x over 3 is
equal to cosine of t. And if you divide both sides of
this equation by 2, you get y over 2 is equal to sine of t. And then we can use this
trigonometric identity. Instead of the cosine of t,
we can substitute x over 3. Instead of the sine of t, we
can substitute y over 2. And you get x over 3 squared--
that's that, right there, that's just cosine of t
squared-- plus y over 2 squared-- that's just sine of t
squared-- is equal to 1. And now this is starting to
look a lot better than this. This, I have no
idea what this is. But hopefully if you've watched
the conic section videos, you can already recognize that this
is starting to look like an ellipse. We can simplify
it a little bit. We could say this is equal to x
squared over 9 plus y squared over 4 is equal to 1. And if we were to graph this
ellipse-- we will actually graph it-- we get--
let me draw my axis. I'm using this blue color
a little bit too much, it's getting monotonous. OK, let me use the purple. So that's our x-axis. That's our y-axis. The major axis is in the
x direction because the denominator here is
larger than that one. And it's the semi-major
radius-- this is going to be the square root
of this, it's 3. 1, 2, 3 in that direction. 1, 2, 3. I know I'm centered in
0, because neither of these are shifted. You should watch the conic
section videos if this sounds unfamiliar to you. And the semi-minor radius
is the square root of 4, so that's 2. So they get 1, 2. And 1, 2. Let me see if I can
draw this ellipse. So it looks something
like that. There you go. So just like that, by
eliminating the parameter t, we got this equation in a form
that we immediately were able to recognize as ellipse. When I just look at that,
unless you deal with parametric equations, or maybe polar
coordinates a lot, it's not obvious that this is the
parametric equation for an ellipse. But this, once you learn
about conic sections, is pretty clear. It's an ellipse. And it's easy to
draw that ellipse. But in removing the t and from
going from these equations up here, and from going from that
to that, like in the last video, we lost information. We lost, one, what is the
direction that we move in as t increases? And we also don't know what
point on this ellipse we are at any given time, t. So to do that, let's
make our little table. So let's take some values of t. So we'll make a little
table. t, x, and y. It's good to pick values of t. Remember-- let me rewrite the
equations again, so we didn't lose it-- x was equal to 3
cosine of t, and y is equal to 2 sine of t. It's good to take values of t
where it's easy to figure out what the cosine and sine are,
and without using a calculator. We're assuming the t is in
radiance, just for simplicity. So let's pick t is equal to 0. t is equal to pi over 2. That's 90 degrees in degrees. And t is equal to pi. And so what is x when
t is equal to 0? Well, cosine of 0 is
1 times 3, that's 3. What's x, when t is
just pi over 2? Cosine of pi over 2 is 0. 0 times 3 is 0. And what's x equal when
t is equal to pi? Cosine of pi is minus 1. Minus 1 times 3 is minus 3. Fair enough. Now let's do the y's. When t is 0 what is y? Sine is 0, 0. So 2 times 0 is 0. When t is pi over 2,
sine of pi over 2 is 1. 1 times 2 is 2. And when t is pi, sine of
pi-- that's sine of 180 degrees-- that's 0. 2 times 0 is 0. So let's plot these points. When time is 0, we're
at the point 3, 0. So 3, 0-- 3, 0 is right there. This is t equals 0. When t increases by pi over 2,
or if this was seconds, pi over 2 seconds is like 1.7
something seconds. So at t equals pi over 2,
we're at the point 0, 2. We're right over here. So this is at t is
equal to pi over 2. And then when t increases a
little bit more-- when we're at t is equal to pi-- we're
at the point minus 3, 0. We're here. So this is t is equal to
pi or, you know, we could write 3.14159 seconds. 3.14 seconds. And actually, you know, I want
to make the point, t does not have to be time, and we don't
have to be dealing with seconds. But I like to think
about it that way. I like to think about, maybe
this is describing some object in orbit around, I don't
know, something else. So now we know the direction. As t increased from 0 to pi
over 2 to pi, we went this way. We went counterclockwise. So the direction of t's
parametric equations is in that direction. And you might be saying,
Sal, you know, why'd we have to do 3 points? We could have just done
2, and made a line. If we just had that point and
that point, you might have immediately said, oh, we
went from there to there. But that really wouldn't
have been enough. Because maybe we got from
here to there by going the other way around. So giving that third point lets
us know that the direction is definitely counterclockwise. And so what happens if we just
take t from 0 to infinity? What happens if we bound t?
t is greater than 0 and less than infinity. Well, we're just going
to keep going around this ellipse forever. Multiple times. Keep writing over and
over, infinite times. If we went from minus infinity
to infinity, then we would have always been doing it, I
guess is the way to put it. Or if we just wanted to trace
this out once, we could go from t is less than or equal to-- or
t is greater than or equal to 0. All the way to t is less
than or equal to 2 pi. And in this situation,
t really is the angle that we're tracing out. If we were to think of this
in polar coordinates, this is t at any given time. And I just thought I would
throw that out there. It isn't always, but in
this case it really is. When you go from 0 to 2 pi
radius, you've made 1 circle. But this is about parametric
equations and not trigonometry. So I don't want to focus
too much on that. But anyway, that was neat. When we started with this,
if I just showed you those parametric equations, you'd
have no idea what that looks like. But by recognizing the trig
identity, we were able to simplify it to an ellipse,
draw the ellipse. And then by plotting a couple
of points, we were able to figure out the direction at
which, if this was describing a particle in motion, the
direction in which that particle was actually moving. Anyway, hope you enjoyed that.