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## Algebra (all content)

### Course: Algebra (all content)>Unit 14

Lesson 18: Parametric equations

# 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. to •  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.
• I do not understand how you got the second term (the 4/5) when solving (x/5-2)^2 • 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?? • At around , Sal talks about substituting. Is this similar to solving a system of equations by substitution? Or am I misunderstanding something? • where did the -4x/5 come from and how? • 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? • 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)= 4^(½) ` ← take the reciprocal of the base and change the sign of the exponent
`= 2 `← now take the square root since we had an exponent of ½
• 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. • 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.
• Do you have a video where you show us how to get the parametric equation for a given conic section? • 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? • 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.