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# 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 with an outlaw right that X is a function of T was equal to 10 I believe if I did this a couple of hours ago so I think that's what I said 10 plus 5t + y is a function of T was equal to 50 minus 5t squared and the graph looks something like this let me redraw it never hurts that was the y-axis that's the x-axis and we saw that at t equals zero and we could try two equals zero 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 looks something like that and this was the path of of the of the actual car as it as it hurl hurtled to the ground but this parametric equation actually doesn't just describe this part of the 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 t equals minus 1 you get minus 5 so 10 minus 5 is 5 if you put a minus 1 here you get this becomes a plus 1 so minus 1 so you get 545 so you get that point right there and if you did -2 you're going to get a point that looks something like that and minus 3 you're going to get point something like that so the whole curve described by this parametric equation will look something like this do it in a different color it looks something like this and to end the direction as X increase as 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 say this is the path only for certain values of T in our example it was when we leave this point which is at T equals zero 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 zero so this in the last example before when we actually talked about a car careening off of a cliff it was between zero so T is greater than equal to zero and then we have to figure out what T what T value makes us hit the ground or when does y equals zero because when y equals zero Y is our altitude so when y equals zero we've hit the ground so if we say what T value makes y equals zero we just say zero equals fifty minus 5t squared so let's do that zero is equal to fifty minus 5t squared I just took this and set y equal to zero because that's when we're going to be hitting the floor at five T squared to both sides you get five T squared is equal to fifty divide both sides by five T squared is equal to ten T is equal to the square root of three and you can actually say plus or minus square root of three because this is we hit the ground right at t is equal to square root of oh sorry Square t is equal to square root of 10 t is equal to square root of ten my brain did three because I immediately said Oh square root of ten that's three point something something so this is Square t equals square root of 10 which is approximately equal to three point oh I don't know ones I don't know what it is I should probably take 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 ten out here there's a negative sign right there and if you could go backwards in time you would have hit 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 one 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 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 set and said that between time is equal to zero and time being less than the square root of ten 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 generalize 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 in that direction but can we express the set of parametric equation is just a normal 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 just 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 a 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 a 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 5 t 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 and let's simplify this so this is equal to switch colors 50 minus 5 times this is x squared over 25 minus SC 2/5 so it's 4/5 X plus 4 and then if you multiply so this is y out here so you get Y is equal to 50 and let's see minus 5 times y 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 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 we've progressed beyond that it describes the 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 card it 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 this was the path of the object and then I asked you where is the object after five seconds you don't know because this just tells you what the [ __ ] where X is relative to Y or where Y is relative to X it doesn't tell you where you are after five seconds so we've lost all of that information we've lost information that this is T equals zero this is T equals one this is T equals two this is T equals three this is T equals the square root of ten 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 you know just a Y and an X but also for you to appreciate that the parametric equation is useful and that you it does contain information that this just single equation won't contain anyway see in the next video