Two springs of constant density A spring of constant density $\delta$ lies along the helix

$$\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 2 \pi$$

a. Find $I_{z}$

b. Suppose that you have another spring of constant density $\delta$ that is twice as long as the spring in part (a) and lies along the helix for $0 \leq t \leq 4 \pi .$ Do you expect $I_{z}$ for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calculating $I_{z}$ for the longer spring.

## Discussion

## Video Transcript

Okay, so we are working. Problem number 35 Section one of chapter 16 And we are given we are to find the mass a wire with characterized by occur in two different scenarios, but densities different. So we have a wire that lays on the car given by r t equal to the square root of two times t I plus the square root of two times t j plus the was four times for sorry for minus t square times. Okay, there we go. And then, um, this is for tea between zero one. So we have a wire that lays on this curve, okay? And they tell us, um, so we are too. Find the mass for two different scenarios. A When the density delta equals three t be when Delta is What? So one thing to take notice right away in scenario. Be here. All right. If the density is constant in a wire, then really, what we're doing is we're just calculated parkway, right? So that makes sense. So hopefully hopefully, you know, if you thought about this for a little bit, you can't look, these woke density constant density, right, doesn't it? If you have, like, a straight line, right? And it was constant density. Really. That would just mean that the mass was just however long, you know, the line was the line. Second waas. Right. Okay, so then the in general, calculating mass from the density varies is a little more difficult, right? So But on page nine overboard, we have this formula for the mass. Uh, a wire, some mass, uh, wire on her. See, with density. Delta is so the masses the integral over see of your density function times your differential are clay. Uh, so here's the formula. Right? So now the next thing you might be thinking is all right. Well, this section's about lining rules. So how does this match up with the formula for a line in a world? So let's go ahead and, um, go ahead and write down the formula for a line in bro of s. So we have, um ah, function defined on. So let's just say it's a smooth Kurt. See right? Bye, Cem formulas and Parametric equation. Let's say our of TIAs ex of t I plus why of t j plus Z up t. Okay. And that's for T between some A and B. Right? Okay, so then the line integral of Yeah, over C is integral around. See, uh, yeah. Uh, X Why Z d s. All right. Okay, So here's the line interval formula. Right. Here's a fearsome parametric equations that give us the curve. See? Right. So the question is, how do you actually compute with this? Right, So this is supposed to be a definition, but how do you actually compute with us? Right, So it turns out that our D s all right, So d s if you look in your textbook, you'll find a derivation of this, but the differential are cling. You can represent this way as the magnitude with length of the derivative vector, our prime of tea. DT. So when when these first derivatives of X, t y t and Z f t exist, you can take the derivative vector are. And if those air on continuous than this is all nice, good, even represent d s this way. Um so it turns out that that's d s right. And then what you're supposed to do is in this act of X Y z, right? You plug in your ex of T y T and Z f t. So that's how you're think of that is that's how you're restricting the values at just just to take the values about on the curve seat. So actually, computing this thing you integrate with respect the tea over the interval A to be, uh uh, ex t y t z f t. All right, so this is just up with except T y t z f t l put in here, but instead of d s. We do this with respect to the derivative effect, the length of the derivative vector like that. Right? Okay. So hopefully you're seeing how what we're given is, um what I mean is, hopefully you're seeing how functions we were given this dealt equals three t and dealt equals one. How that's gonna match up. We go back to this whiteboard, right? So we're given dealt the equals three t and Delta equals one, right? You might be thinking yourself well in the line. Integral section right. The functions are always usually or they're usually given has right functions of x, y and Z. And then I plugged with the Parametric equations for the curve. See in for X Y and Z, right? So But now I'm just given Delta equals three t, right? What does that mean? Well, so this Delta is already paramour tries for us, right? So don't let that be a stumbling block. But this is already paramount tries, so we don't need right. There's no function to plug the Parametric equations square rooty square square two times t squared two times t and then four minutes T square. There's no need to plug those in for some X y and Z, right? That's already served done for us in the problem. And then this d s. This is the thing we started. Need to calculate We need to figure out what's d s so d s right. Let's do that down here. It's a deep because we're gonna need d s for both problems. So d s is right. We said it was the length of the derivative vector, our prime a T. And in times DT, Right. So let's calculate our prime of tea. So your book calls are prime a t this be a t right? It does. D r over DT is vey empty. So our prime of tea due to do so. That's going to be the square root of two times I, plus the square root of two times J and then minus two tea. Okay, so now if we take the length of that, that's the We're gonna take a square root, right, Because we're doing Cleon distance. And then if we square the square to we get to So we have two plus two, right? We're squaring each of the components on our prime empty, and we're adding them all together. And then negative to T square is who were t square, right? So if you put negative Tootie and parentheses and then you square you get or t square and this right? So I'm gonna go ahead and rewrite it right away to save some space. This two plus two inside here, I'm gonna rewrite us or right way. So this is four plus 40 square, and now we can factor the whore out right? There's a common four said before times one plus t squared under the square root. But then I can pull the four out of the square root and write it is too right. So that would split in the square root of four times square root one plus t squared. But the square force too. So the length of the derivatives vectors are T reds of the derivative, the vector derivative function, Artie, or Star Prime A. T. Its length function is given by two things. The square root of one plus T square. Okay, so we will need d s for both of the integral we're going to set up. So let's open up a new white board. So now we have This is D s. And this is going to for this particular problem. This is D s right? This is going to help us quite a bit. Right? So now we're gonna take our mass right there. Sorry. Our density functions going to you times D s. We were going to integrate over occur given by our t up here. So what is that book? What? Right. So now for part A where Delta equals three tea, the mass should be given by the integral from zero toe one. Right. Those air the bounds for the curve are, uh, sit. Delta is three t. So we write three t and then d s waas two times the square root of one plus T square DT. But so you know, simplifying. We can take this two times. Three. That's a six. We can bring it to the front of the integral. So let's do that. Six tanks. The interval from 01 Uh, he times the square root of one plus t square tea too. OK, And from here, Right, So you might be thinking yourself. Okay, so now I've done this, right? I've made it this far That maybe was hard to track down all those connections. Right? To understand this is a line into girl. Set up the right into growth and then get to this point. So now here. Right? Okay. Think back. Integration techniques help, you know, maybe copped to. Maybe it's the same semester, depending on how you chunk up calculus. But think back to integration techniques. So here we have integral from 0 to 1, I'd ignore the six enter go from 0 to 1. We have integral of tea Times square with one dusty square. If we could somehow to get rid of this tea and turn this into, like, square root of variable and then integrate with respect to variable. We could do that right? If it was just the inner grow of square root of you, for instance, do you? Well, that's easy. That's just like a power rule. We know how to do it, right? So But then the question is, how do we How could we do that? Right. Well, if we set the inside to be you, who set one plus t square the inside of square to the U, then the derivative is to tea, and we have a multiple of the derivative. All right, so we have a multiple up to t right over here with this tea that we can then replace. So we're gonna set up a U substitution. So let's do Let's set you to be one plus t square. Let's calculate, do you That's too t d, too. And then let's go ahead and figure out what you could do this a couple ways. Let's go ahead and let's just do d'you over to t is DT. What? Solve her DT right away. That's sort of guarantees you that you can just plug in. Things will cancel appropriately. You can move on there. Okay, so now I'm gonna change the integration bounce right away. Well, so to do that, right? So to do that, we pluck in zero for tea. And when we do that right, So you zero is one. And when you plug in one for tea you want is to sow the new integration downs. Air 12 and the integral looks like this. So M is six times the integral from one two. Ah, so let's go ahead and let's let's leave the tea right. But then that's your place. One plus t squared with you and then d t Let's write, as do you over two tea. And now that good thing that he's cancelled, right? And then this d'you this 1/2 Do you or do you over to this 1/2 to you, this one a half come to the outside six. So this is my six over to. So it gives three interval from 1 to 2. Uh, and and so instead of writing at a square root EU, let's prepare ahead right, because we'll need to integrate. So that's you to the 1/2 power, do you? And then when you integrate that so now we have three and When you integrate, right, you add one of the power, and then you divide. So you divide by three halves. That's multiplying by 2/3. Three times 2/3 you to the three has, and we don't need to do a plus seeks. We're evaluating. We're doing a definite so did that from 1 to 2. And these threes cancel. And when you plug in, you end up getting to Times Square a minus one, and this ends up being mm. When Delta equals three tea. So this is when Delta equals three t. This is what we get right here, huh? So you get two times the square root of eight minus What? Okay, so we could distribute that. Or if you want to punch them, your calculator, right. To get an approximation, you go ahead. You do. Okay, so that's the first case when dealt equals three t, and it actually is gonna turn out from integration standpoint. From just a purely calculus standpoint, this was actually easier than what we could do in Delta's one. So it turns out when the density is uniform, when it's one right, So from a physical standpoint, that's easier and sense that that well, the density doesn't bury. It's just constant along this wire. Right? But it's gonna turn out that calculating just the art link, which will just be the integral of D s. And we want this three t out here and then this u substitution, which is probably what a lot of you feels sort of most comfortable with in terms of integration techniques we won't have available to us. And we'll have the whole some other integration tricks to get through. So let's go ahead and take a look at that case. So here, right, we have case be where Delta equals one right cats. And now again, we have. Mm. It is going to be the Inter girl from zero toe one. Right? And this time it's just gonna be one times Yes. Right. Okay, so now we plug in what? We know D s s a d s. Was that two times the square root of one plus t squared. So you this? Yeah, he s was to Times Square to one plus t square. D T. Right. Okay, So now looking at this as I said so from a purely calculus standpoint, this is interesting, because from a physical standpoint, this is maybe easier in a sense, the densities uniform. It doesn't change throughout the wire. However, from a calculus standpoint, this is actually, at least in my opinion, slightly harder in the sense that for most students, I thought, um, they usually feel a little bit more comfortable with something like you substitution. Then they do sort of tricks we're going to do to get to the end of this problem. So okay, what do we do with this? Well, first up, let's bring the two players. Let's bring that out front. So let's write. This is two times the integral from 01 the square root of one plus T square. Okay, so now when you're looking at this right, focus on the Inter Grant here, Square one plus T square, you should be thinking you're so like OK, that's a square root. And then I've got a number that, like I could take the square of so I can think of one is like one square right? So it's the square root of once. Where was T Square? You like him? That's almost like square root of a squared plus B squared. So, like fifth agree and Kira, right triangles, right triangles. And the answer is yes. So whenever you're doing integration and you're thinking like right triangles, one thing you can always try to do is you can set try to set up a right triangle. So with an angle beta or how everyone a label, that angle. But let's call it will set up a right triangle here. All right, so you should be thinking in burst Trig Substitution Tze, that's where. Go right Why don't I call inverse trick substitution? Sze will they work a little differently than you substitution in the sense that, um, they really are inverse substitution variables But that difference is not entirely port help three here. So let's let's keep going. So right now we have a rate train. So this thing over here that I labeled with fifth question mark with angry in question mark right, that's good. That should play. What role in the triangle. So that should be the specially that Cotton's right. Hi. Pop news is the square root of something square. Plus something else squared. So right that there, then the question is okay now tea and one should be the legs. Which one do I put? Where? And I would recommend putting t across from data as much. Whenever you do these setups. If you have the option, right, put tea or put your your single variable across the angle when you can, Um, as opposed to Jason, just because it will probably give you expressions. You're maybe a little more comfortable with it. Doesn't really matter. Either way, we'll work out. You just have to follow the substitution of the calculus appropriately. But, uh, I I think this is probably easier way for most of us, so Okay, so now the question is right. What? What do we substitute here? Well, let's focus on just tea. Right. So what is tea in this scenario? Well, t is Well, it's t over one. So this is t over one, which in this triangle is the tangent of that angle. So we're gonna set t to be tangent of data, and then we're going to differentiate. We're gonna run a d d t both sides of this equation. So that's why it's a little different than use up right? In the sense that you see differentiate both sides with respect to tea and then you do an implicit differentiation. So on the left hand side of the equation, if you differentiate TV expected here one over here, right, you differentiate the outside function standard, get C can square data. But then you multiply by the derivative of the inside deep. They, uh d t Okay. And so then we rearrange this equation and let me let me go ahead and write it, uh, vertically, Yeah, let's go ahead and read vertically here. So we rearrange this equation and it gives us that d t. And the substitution should be C can't squared data. Steve There. Okay, so when you do inverse trick substitution Tze there's actually quite a bit to think about in terms of what is my substitution ballot and does it actually work? And that's all the theory and it's gonna turn out in this problem that the conditions are fine and that you could use it. Um, but I do just want to bring that up. Is that s O for the most part. Just setting up your substitution slogging through it's gonna work. But if you ever set up a substitution and you find that you're, like, trying to calculate, you know, trick values that don't make sense. You should maybe double check your substitution. Okay, so now let's go down here. So the other thing I want to do is s o instead of now, we have the tools to start replacing this integral over here. Right. We can replace DT with c can't square state of deep later, and we can replace the square root of one plus t square with the sea can't of data, and that'll give us the inner go see can't Cube. They don't eat data, however, right there. Still these integration bounce. This is a definite girl. So you could do this a couple ways. You could leave the integration bounds off completely and just do indefinite intervals and then go back. And, you know, do do the backwards substitution back to all t values at the end and then plug in t values. Or you can change the integration pounds to go. Just what I'm gonna do here. So how do you do that? So when t equals zero. So that's what you want to think. Okay, So when t equals zero, all right, that means the tangent of data is zero. Which means that we can take Seita to be zero. Right? So there's a couple. There's a few different angles where attention data zero. But we'll take it to be that one. And when t equals one, that means the tangent of data equals one. So we can take the data to be high over four. And so these this range of angles life squarely in the appropriate range or one of the appropriate ranges where that you can go ahead and substitute in tangent. Eso this substitution is all good to go. So now we just have to set up the integral. So let's do that. Okay, so now we're gonna do so now, M The mass is going to be two times the integral from zero to pi over for And so we're gonna replace Square root one plus t square. We're gonna place that with C can't data. That's the sea. Can't. They'd have been d t. We're gonna replace with c can't squared data deflator, huh? And so this gives us two times the integral from zero to high over four c can cubed data defense. Okay, so we're gonna need to open up a new whiteboard. But before we do that, let's go ahead. And, uh, let's go ahead and take a look at this right here. So in order to get to the end of this integral, you could play around with some trig identities, and, uh, you could go ahead and try some other stuff, but the fastest way to get done this is going to be to do an integration by parts. And so if you're looking at this, you can't Cube, you're thinking, Well, how do I integrate C? Can't you buy parts, right? Shouldn't I be using that trick reduction formulas? Actually, Serpico back here. And you look, this this is what tells you how to do immigration my parts. So our first part you were gonna be we're gonna set you to be the c camp of data we're gonna set Devi to be c can't square. They debate. So we'll do that on the next slide for the next white board. So here we go. So we have let me reread. The integral M is equal to the integral zero to high over four. See, you can't cube. They, uh do you think, Huh? So they said we're gonna set you to be see campaign. That means d'you is C Can't they, uh, tan theta deep there. Now Devi is c can't squared data d theta and the will take to the tangent data. Here we go. Okay, So now we're gonna go ahead and plug in the parts to you and to integral zero to high over for seeking. Cute. They are, do you think? Okay, so now it should be equal to si Cand Seita times Tangent data evaluated from zero to pi over four. Right. And then minus the integral from zero to high over four. All right, so now we plug in D you and then we do times be so that'll give us c Can't. She can't Fada cans Word data deflator. Okay, so here. How do what are we gonna handle this? Right. So now this is nice, because this is this weaken. Just evaluate right now. That's that's good. All right, but this right was still what we're gonna do with this other integral now. Okay, because that doesn't look super easy. Um, So what we're going to do is we're going to go ahead and replace this with eso Tangent square. You can write as c can't squared. They, uh, minus one. So when you do that, you get the following, and so you get that m is equal to it. C can't. They, uh, tan data evaluated from 0 to 2. Pi over four minus. And now, right. We're just gonna split this up. And so one in agro is gonna be the in n roll from zero to pi over four. All right, we'll have a c can't time. C can't square will give us a c can't cube data and then will have Ah, si Cand times minus one. So that'll be minus the integral of C. Can't data groups forget me, defeat us. Nobody is ever too much of a comfort. Not right deep. It just makes it clear D theta minus the integral from 0 to 2 pi over four of c Can't there deep there. Okay. From here, I This is going to give us Let's distribute the minus sign. Then we might have to do one more white board, huh? Now we have minus right. So you may be wondering, right if you've never seen this trick maybe wondering, Right? How does this help? Now? We have We just have another c can't cubed data D theta right that we're trying to integrate. How does that help? Well, what's gonna turn out is so this is now minus minus lus plus internal from zero pi over or she can't date a date. Okay, so what's gonna turn now is now, right? What's on the other side of this equal sign? What's on the other side of his equal sign is antidote from zero pie. Of course, he can't cube data data. So we can actually send this one this, senator Oh, here to the other side of the equation, the other side of the equal sign. And we can add those two together, right? And then we can divide by the factor, too. Will get over there, and we'll actually be able to do so this in ago. We can do integral of C camp. They'd, uh this has a nice closed form. That's easy. All right, this is something we can do, so we'll be able to get to a place where we can evaluate this case. One more white board. So now what? I'm gonna do when I go to the next white board is I'm gonna I'm gonna move this minus zero. The minus this integral from 0 to 5. Of course, he can't Cube data data. I'm gonna add it to the other side. So I'm the new next white board. I'm gonna have two times the integral zero pi over for a C can keep data. The data equals stuff that's left on the right side. So we have let me right. Mm is equal. Where? Let's let's leave up again for now. Let that be too complicated. Let's. So the equation we're getting is the following. We're getting two times the integral from zero. Who's excuse me? Go from zero to high over four C Can Cube data D theta is equal to C. Can't. They'd, uh, and they'd, uh, evaluated from zero too high over four. You're plus Sorry. There's a plus there. Plus the inter go from zero to pi over four. C can data if they're okay. So now let's go ahead and let's do this. Let's leave the two there for a second. Let's go ahead and evaluate. So let's evaluate and integrate and evaluate. Right? Okay, So when you plug in pi over four to see can't data you get a square root of two when you blood into tangible once is gonna give us square root too, and then minus when you plug zero into C Can't they get one when you put your own attendant zero right and then plus So if with the integral of the anti derivative of C can't data ends up being this natural log of the absolute value, C can't fade a close 10 and then we're gonna evaluate a 0 to 2 pi over four. So if you don't remember that one, take a look in your textbook. The fastest way thio t o get to a place where you can ingrate seek and later find its anti derivative is set up the indefinite in a row and you take your sea can't data and you strategically multiplied by C can't data lust, Andrew Data and bye bye See data plus tangent data. And then you can perform a U substitution around there, and that will do the trick. It's either a u substitution inversion by parts. I shouldn't conjecture, but it's a substitution of some sort that gets you to the end of that. But you take a look at that. Okay, So now let's go ahead and let's go ahead and evaluate. So we have Let's get two times the integral from zero to high over four. C can't. Cubed day, huh? Is equal to the square root of two. Plus the natural log. When you plug in pi over Ford C can't data, it's so I'll keep these absolute value bars. So what? I'm gonna drop the absolute value bars because what's gonna happen is everything in here is gonna be it's gonna be greater than or equal to zero. So we don't need the absolute value cars. So when you plug in pi over four to see, can't they You get square root of two and then when you plug it into tangent, you get once it will be natural log expiring too plus one and then minus natural Log of when you plug in pi over something plugged in zero to see Can't they get one When you plug in a standard get zero natural log one is zero So this ends up being square root of two plus natural log square root of two plus one. Okay, we're almost there. So this means now if we divide both sides by two, right. This means that the inner grow from zero to high over four, Uh, of C can't Cube data, which remember? What was that? Well, that was our massive ago. This is the thing we actually want. So now it ends up being the square two. May thinks that there we go. Ends up being the square root of two. Divided by two. Who, us? The natural log of square root of two plus one over too. But me figure out may make some space. I wanna Okay, let's go. We're here. Let me look at Michael. Signed me. You following right? So if we divide both sides, they're like to. And if we divide both sides here by two, we get the following. So the inner grow from zero to put over four c. Can't cube. They'd, uh, deflator is the square root of two over too. Plus the natural log. The square root of two plus one over, too. Which this was equal to mm. Or case be where Delta equals one. All right, so this is our answer right here. Square to 2/2, plus natural law of square two plus 1/2. All right, so already that'll do it. So, what did we do? Well, we found the mass. A wire laying on a bird. All right, let's go back to the original problem. We found the massive this wire that lays on this curve given by these parametric equations in two different scenarios. One where the mass was variable and one where's uniforms. And it turned out it was kind of interesting, because from a physical standpoint, the uniform masses easier toe conceptualize write it turned out that calculus was maybe a little a little more involved, right? Whereas with the variable mass, the calculus integration was just a simple use up to. So I hope you enjoy this and take care.

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