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## Algebra (all content)

# Removing the parameter in parametric equations

Sal starts with parametric equations that give x and y as functions of t, and he manipulates the equations to get y as a function of x. Created by Sal Khan.

## Want to join the conversation?

- I still don't understand how you know the direction of the object's path.8:40to8:50(12 votes)
- We know, that time, t, always runs forward (in the physical world), so that t in this video's example is always increasing.

And with the parametric equations, we get a position, (x,y), at some time, t. And given that time always runs forward, we can say, that the graph is to be read in the "forward (time) direction".

Now, if we transform our parametric equations, x(t) and y(t), to y(x), consider this:

The car is running to the right in the direction of an increasing x-value on the graph. And you'd implicitly assume, of course, as x increases, t (time) increases. But he might as well have drawn the car running over the side of a cliff leftwards in the direction of a decreasing x-value. So to get the path of the car, you'd have to start at some x-value and then decrease it. If you still assumed, that time increases with an increasing x-value, you'd get the path followed by the car wrong in this case.(49 votes)

- I do not understand how you got the second term (the 4/5) when solving (x/5-2)^2(18 votes)
- I think Sal uses the quadratic equation (a+b)^2 = a^2 + 2ab + b^2

-4/5(x) is the 2ab part

x^2/25 is the b^2 part, and so on....(19 votes)

- My questions: Is it possible to do the opposite - to parametrize function from its usual form (y in terms of x)?? How?? Is it possible to parametrize any function??(10 votes)
- Yes, it's possible to parameterize any function. It might not be very interesting though. For any function, ie:
`y(x) = x^10 + log(x) + ...`

You can always just change that to:`y(t) = t^10 + log(t) + ...`

and add on`x(t) = t`

.(5 votes)

- At around5:40, Sal talks about substituting. Is this similar to solving a system of equations by substitution? Or am I misunderstanding something?(8 votes)
- The objective here is to get rid of the parameter t while the objective in a linear system is to get a 1 point answer. Its just that substitution can be used to do both.(4 votes)

- where did the -4x/5 come from and how?(2 votes)
- (x/5-2)^2 is (x/5-2)*(x/5-2), and if you multiply it out, you get (x/5)^2-2*x/5-2*x/5+4; Sal just simplified it to skip some simpler steps.(7 votes)

- 1/4^-1/2 = 2.

Now what is x^1/2? It is the square root of x.

So does this mean that x^-1/2 = the negative square root of x?(3 votes)- An fractional exponent means to take the root of that power. So, an exponent of ½ is a square root, an exponent of ⅓ is a cube root, and so on.

A negative in the exponent means to take the reciprocal of the base, Thus here is how you would solve the example:`¼^(−½)`

← take the reciprocal of the base and change the sign of the exponent

= 4^(½)`= 2`

← now take the square root since we had an exponent of ½(2 votes)

- Does an equation like the simple one Sal began with have a limit or can it increase infinitely? In other words wouldn't the object simply start falling straight down, and then at that point would you say that the value of x is the limit of this parameter? I believe yet another way to ask this question would be the lim of x as y approaches (-)infinity.(2 votes)
- If we approach it as a physics problem, as in the example of the falling car, we have to be realistic about time moving forward and the existence of barriers like the ground (although we do assume that there is no atmosphere or friction to make the problem simpler). These assumptions make some parts of the curve impossible.

If we tackle it as a purely mathematical problem, (if there is no reason to impose restrictions as in the physics point of view) then the equations represent an infinitely long curve going to both positive and negative ∞ on the x-axis, and negative ∞ on the y-axis.

In any case, while you still have t (time) in the equation, you have the additional information of the direction to which **** (something) moves, and the starting point of the motion.(4 votes)

- Do you have a video where you show us how to get the parametric equation for a given conic section?(4 votes)
- Though most are easy to find based on trig/logical thinking.(0 votes)

- so... do parametric equations allow you to graph more than 2 variables in 2 dimensions? You simply have an equation for each dimension? I can see that time is not actually on an axis... yet I can obviously find what horizontal and vertical distance the object is at "x" time... (so perhaps my terminology is not correct...). Is it possible to use more than one parameter? Relate more than 3 variables?(2 votes)
- I could like to know how to find a pair of parametric equation given a conic equation(1 vote)

- its still a little confusing on how exactly the path of the falling object was detected by time and its speed. you would think you would need more data(1 vote)
- It's pretty cool that you don't, but then again, let us think of another example first.

If I'm in a car and I start a point x and I drive east for 30 minuets at 60 mph, where am I? I'm exactly 30 miles east of where I was.

Take the same simple example and throw on another axis for your change in y value.

Now I'm starting at a height, h, of 50 meters. After 2 seconds, where am I and where will I be in exactly 1 second? We'll assume gravity, g, is 10 m/s^2 (for easier calculations).

After 1 second I will be going 10 m/s and after 2, I will increase by 10 m/s so I'm now going 20m/s. BUT! That means I already went 10 meters the first second and 20 meters the 2nd second, so I'm now 30 meters below where I started and am thus 20 meters above the ground.

In 1 second I will be going 30 m/s and will have moved 10 + 20 + 30 = 60 meters.

However my starting height was only 50 meters so I'm 10 meters under ground, but that doesn't make sense, so because I was bounded by h greater than or equal to 0, I reach h=0 at some point between 2 and 3 seconds.

Congratulations, you have just hit the earth going really fast.

So, long story short, you do need more data. You need your domains and if you have a starting velocity and in which direction that velocity was in.

But that's pretty much it.(2 votes)

## 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.