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## Parametric equations

# Removing the parameter in parametric equations

## Video transcript

In the last video, we used a
set of parametric equations to describe the position of a car
as it fell off of a cliff, and the equations were x as a
function of t was-- and I will write that-- x as a function of
t was equal to 10, I believe. I did this a couple of
hours ago, so I think that's what I said. 10 plus 5t. And y as a function of t was
equal to 50 minus 5t squared. And the graph looks
something like this. Let me redraw it. It never hurts. That was the y-axis. That's the x-axis. And we saw that at t equals 0,
and we could try t equals 0 here, and we'll get the
point 10 comma 50. So the point 10 comma
50 was right there. That was at t equal to 0,
and then we plotted a few points in the last video. I think t equals 1 was there. t equals 2 was, like, there,
and t equal 3 looked something like that, and this was the
path of the actual car as it hurtled to the ground. But this parametric equation
actually doesn't just describe this part of the curve. It describes a curve that goes
in both directions forever. So it describes a curve that
does something like this. If you actually plot t equals
minus 1, what do you get here? You get minus 5. So 10 minus 5 is 5. If you put a minus 1 here,
this becomes a plus 1. So minus 1, so you get 5, 45. So you get that
point right there. And if you did minus 2, you're
going to get a point that looks something like that. And minus 3, you're
going to get a point something like that. So the whole curve described by
this parametric equation will look something like this. Let me do it in a
different color. It looks something like this. And the direction as x
increases-- sorry, as t increases, looks
like this, right? This is t is equal to minus 3,
minus 2, minus 1, 0, 1, 2, and so forth and so on. So, in the last example, our
path was actually just a subset of the path described by
this parametric equation. And I'm saying all of this
because sometimes it's useful to just bound your parametric
equation and say this is a path only for certain values of t. And in our example, it was when
we leave this point, which is at t equals 0, all the way
to when we hit the ground. So if we want to know those
boundaries, well, we know the first boundary is t equals 0,
so this in the last example, for when we actually talked
about a car careening off of a cliff, it was between 0, so t
is greater than or equal to zero. And then we have to figure
out what t-value makes us hit the ground? Or when does y equal 0? Because when y equals 0-- y is
our altitude, so when y equals 0, we've hit the ground. So if we say what t-value
makes y equal 0? We just say 0 equals
50 minus 5t squared. So let's do that. 0 is equal to 50
minus 5t squared. I just took this and set y
equal to 0 because that's when we're going to
be hitting the floor. Add 5t squared to both sides. You get 5t squared
is equal to 50. Divide both sides by 5. t
squared is equal to 10. t is equal to the
square root of 3. And you can actually say plus
or minus square root of 3 because this is-- we hit the
ground right at t is equal to square root of-- oh, sorry. t is equal to
square root of 10. t is equal to
square root of 10. My brain did 3 because I
immediately said, oh, square root of 10, that's 3 point
something, something. So this is t equals square root
of 10, which is approximately equal to 3 point--
I don't know. I don't know what it is. I should probably get my
calculator going, but that's not important for this. And if we actually took the
positive and negative square root, this would be t is
equal to the negative square root of 10. Out here there's a negative
sign right there, and if you could go backwards in time,
you would have kind of hit the ground there. But remember, in our
example, we weren't going backwards in time. If we went backwards in time at
t minus 1, we were actually on a plateau, and then there would
have been a different set of parametric equations that would
have described the car when it was just driving
on the plateau. So if we really want to
describe the path of the car, we should have bounded it and
said that between time is equal to 0 and time being less than
the square root of 10, this would have been our
parametric equations. Fair enough. Now, the next thing that's
interesting is, you know, we took these parametric
equations, and we generalized it, and we just said what
it would look like if we didn't bound it. If this wasn't the case, then
the parametric equation would just keep going like that, and
it would have this kind of direction, right? It would move in
that direction. But can we express the set of
parametric equations as just a normal equation where y is
expressed as a function of x, or x is expressed as
a function of y? And let's try. So the easiest thing to do is
let's just rewrite them, just so we clean up our board. So if you have x is equal to 10
plus 5t and y is equal to 50 minus 5t squared, if we want to
write this system as just an equation with just y and x,
what we can do is we could just solve for t in terms of either
x or y and then substitute back into the other equation. So the easier one to solve is
this top equation because it's a linear equation. This is a little
more complicated. We'll have, you know, t is
equal to the square root of y, and it becomes a lot more
complicated, so let's just use this. So if we want to solve
for t, let's subtract 10 from both sides. You get x minus 10
is equal to 5t. Divide both sides by 5. You get x over 5
minus 2, right? 10 divided by 5 is equal to t. And now, we can take
this and substitute it. This is t, right? So we can take this and
substitute it back here for t because we say
this is equal to t. So our equation becomes y is
equal to 50 minus 5 times x minus x over 5 minus 2 squared. Now, let's simplify this. This is equal to--
let's switch colors. 50 minus 5 times x squared
over 25 minus-- let's see, 2/5, so it's 4/5x plus 4. And then if you multiply--
and this is a y out here. So you get y is equal to 50. And let's see, minus 5 times--
minus 5 over 25 is minus 1/5, so minus x squared over 5. Minus 5 times minus
4/5, that's plus 4x. And then minus 5
times 4 is minus 20. We're almost there. We can add the 50 to the
negative 20, and we get y-- that's too dark-- y is equal
to minus x squared over 5 plus 4x plus 30. And we're done. We just, I guess we could
say, simplified this set of parametric equations to this
one equation, where y is a function of x. And so you might be saying,
Sal, why didn't we just always have it this way? This is so much more simple,
and we can just graph this, and you know how to deal with it,
and it's just one equation with just y and x. We don't have this
third variable t. And that might be OK if
you just want to know the shape of the path. So if you took this equation,
and if I told you to graph it, you would indeed get a
parabola, kind of a downward-opening parabola that
looks something like that. But when you go from these two
equations to this one equation, you lose information. If I told you that this
describes the path of an object, some object, it
doesn't have to be the car. We've progressed beyond that. It describes a path
of some object. Just with this equation alone,
you don't know if the object is-- is the object
going this way? Is it going like that? Or is the object going
in the other direction like our car did. Is the object going that way? We don't know because we
don't know how it moves according to time. Likewise, if I told you this
was the path of the object and then I asked you where is the
object after 5 seconds, you don't know. Because this just tells you
what the shape-- where x is relative to y or where
y is relative to x. It doesn't tell you where you
are after 5 seconds, so we've lost all of that information. We've lost information that
this is t equals 0, this is t equals 1, this is t equals 2,
this is t equals 3, this is t equals the square root of 10. And that's why parametric
equations are useful. But anyway, I wanted to do this
example mainly to show you that you can express the same shape
this way and also to show you how to express a parametric
equation in terms of a y and an x, but also for you to
appreciate that the parametric equation is useful and that it
does contain information that this just single
equation won't contain. Anyway, see you in
the next video.