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University of North Texas



Problem 54 Easy Difficulty

Let's consider the problem of designing a railroad track to
make a smooth transition between sections of straight
track. Existing track along the negative $x$ -axis is to be
joined smoothly to a track along the line $y=1$ for $x \geqslant 1$
(a) Find a polynomial $P=P(x)$ of degree 5 such that the
function $F$ defined by
$$F(x)=\left\{\begin{array}{ll}{0} & {\text { if } x \leqslant 0} \\ {P(x)} & {\text { if } 0<x<1} \\ {1} & {\text { if } x \geqslant 1}\end{array}\right.$$
is continuous and has continuous slope and continuous curvature.
(b) Usc a graphing calculator or computer to draw the
graph of $F .$


$P(x)=10 x^{3}-15 x^{4}+6 x^{5}$

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Video Transcript

{'transcript': "So if we want to find, um whatever p f x is here, we're gonna have to use a couple of things. So first, we could use the fact that curvature farm business. And actually, they also tell us the function is continuous as continuous, slow and continuous curvature. So let me actually go ahead and write this up. Um, so, you know, affects continuous, um, are curvature. KFX is continuous, and we also will have where, um are derivative. F prime of X is continuous because they say continuous slope. So these are the three things we're going to need to use, and we can use this to help us, hopefully some kind of system of equations. So let's go ahead and do that. So let's start with this first from here, So f of X is continuously well f of X is continuous Means these endpoints for piecewise function should link up. So that means we're going to have p of zero is equal to zero, and p of one is equal to one. Okay, so we'll leave that like that for now. Um, for the derivative, um, well, we would actually Well, let's take the derivative of this really quickly, so we can kind of see what that might look like. So for the most part, F Prime of X is going to look like zero if X is less than or equal to zero, and then the derivative of one is also going to be zero when it's great to have a good one. And then we have P prime of X, um, for X between zero and one. Okay, that's what this is essentially telling us now is that at zero and one, just like before, Says it needs to be continuous. This also has to be, um, zero at the end points for the slope. So we're going to have p Prime of X is equal to ah zero. Oh, from the, uh, the prime of zero is 0 to 0, and p pride of one is equal to zero. Okay, so let's go ahead and see what we can figure out from this first, and then we'll come to the curvature being continuous. So let me just go ahead and bro this down here, and I'll kind of work off on the side with the information we have so far. So let me get these all in order. Okay, Now for this first one. So let's just start with some arbitrary, um, function. So what was it supposed to be? 1/5 degree polynomial. Yeah. So it's going to be p F X is equal to a X to the fifth plus b x to the fourth plus c execute plus d x square plus e x plus f. I think of what later came next. Well, if we plug in P of 00 that is going to give us the p of zero is Well, everything is going to be zero except for us. So this is just for the zeros, Beth. So we're good there and then for p of one being equal to one, this is just going to give us a plus B plus C plus D plus e. And then that last part here is just going to be zero. Okay, um, and then maybe we can use this later if we get four more equations. So if now we take the derivatives of this so p prime of X. Well, this is going to be five a x to the fourth plus four b execute plus three c x squared plus two d x plus e. And we now know if we plug in zero into this, this should be equal to zero. But that just means e zero, since everything else has a variable with it and then we have, um p prime of one is equal to zero. Well, that would mean five a plus four B plus three C plus two D is equal to zero. Actually, that means up here we can get rid of this and get rid of that. Yeah, Um, let me think if there's anything else we can get from this before going to the curvature, I don't think so. So, yeah, let's just go ahead and move on. So now let's get our curvature formulas. Let me move this down a little bit. So let's come up with a general, um, curvature for this. So since this is just in the XY plane, um, are all in terms of X, I should say we can go ahead and just use this formula that they give us, So we need to take the second derivative of this. Um, so let's find that really quickly. So the first derivative is this? Let me scoop this down that we could just take the second round of that so that would be f double Prime of X is equal to Well, it's going to be the same thing, but just 00 And then here we have p double prime effects the next lesson, or equal to 00 X to one and then x greater than one. So we have that. So if we think about this, so I'm just going to replace. So we know we have essentially, um, two different cases because we have different derivatives here. So in the case of it, being explicit or equal to zero notes that the 1st and 2nd derivatives are both just going to be zero um, so this would just be 0/1 0 squared, raised to the three halves, which is just zero. That would be for X is less than zero over here, the first derivative secondary for greater than one same thing. So that would just give us zero also be greater than equal to one. Actually, let me just erase this. Uh, I'll just move that over zero accessory, put zero. And now if we go ahead and plug in P prime. So that's going to be the magnitude of P double prime of X, all over one plus p prime of X squared, um, and then raised to the three house power. And this is from 1 to 0. So this is our curvature for this. Well, if this is continuous, we first need to make sure it's defined everywhere. And since it's the first derivative squared and then added to one that will always be greater than equal to zero are actually always strictly greater than zero. So we don't have to worry about divided by zero. So that means we just need are in points to match up like we did before. So essentially, this means, um that our second derivative here should be equal to zero at its in point, because again, we just want this to hook up. Um, so the release this So we need the second derivative at zero to be zero and then the second derivative at 12 b zero. Also right. So let's scroll back up here and pull this down. Okay, so let's take the second derivative. So that should give 20 a X cubed plus 12 B x squared plus six c x plus two. Oh, not too, too deep. So if we plug in zero into here, we should get out of zero. Which that means all of these is going to be zero and just give us two d, which would imply d zero. So now we can come up here and get rid of that term. Um, then we also need to plug in one. We have p double prime of one is equal to zero, which would give 20 a plus 12 B plus six c. Mm. Okay. So now we have all these equations. Um, so I'm just collect these really quickly, and then I'll be right back. Okay, so that just took a second for you. But it took me a little bit to collect all of this, so I just went ahead and wrote out all the equations without me kind of crossing anything out. So first, um, here we're told that PM zero equals zero. Battled us. That apple is just zero. So let's cross this out. So I'll check this off to say we've already used this here. We have p prime of 000 That gave us E zero. So that means we can come up here across that out and then down here, um, the second derivative need to be zero that gave us d was equal to zero. And go ahead and get rid of that. Okay, so now if we look to see what we have left, we still have. So I need to cross this out. That zero, um and then D is also zero. And India Pier is zero. Okay. And then was there anything else? I don't think so. So the three equations we still need to use we're going to be this one, this one and this one. So I'm just gonna write these off on the side. So this is gonna be a plus. B plus C is he could have won. Here. This is going to be five a plus, or B plus three c. Is he going to zero? And then this last one is 20. A plus 12 B or 60 is equal to zero. Um, now you can go ahead and solve this However you want. I'm just going to set up a matrix, uh, to do this and then just plug it into my calculator to kind of get the road reduction. So this is gonna be 111 one 5430 room 2012 60 Okay, so, um yeah, let me plug the syndrome fast, and I'll be right. Yeah. Okay, So after plugging this into my calculator, it said this would be row equivalent to 1006 010 negative. 15 001 10. And remember what this is saying. Um, there's a way we had this set up was this is a B and C. So this first equation is saying A is equal to six. B is equal to negative 15, and then C is equal to 10. And if we come back up here and look to see what p of X was originally so we don't need those last three terms, we just need this p of X. Well, I'm going to replace a here with six, be with negative 15, and then see with 10. And that would be our polynomial that has all of those properties. And so, um, I mean, if you haven't learned how to use matrices, you could just do you. Ah, months ago. I'm blanking on what the word is, but you essentially would just, like, isolate for all of one variable. Or you can add some of these, cancel like a B or C, get a two system and then kind of solved from there. Um, yeah. Just so the video isn't any longer than it really is at this point. Um, I'll just go ahead and use this fact here. Yeah, So that would be our polynomial with all of those special properties."}

University of North Texas
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