00:01
Okay, so we are working problem number 35 of section 1 of chapter 16.
00:10
And we are given.
00:12
We are to find the mass of a wire with parametorized by a curve in two different scenarios with density difference.
00:20
So we have a wire that lays on the curve given by r of t.
00:34
To the square root of 2 times t i plus the square root of 2 times t j plus the plus 4 times t squared times k there we go and then this is for t between 0 and 1 so we have a wire that lays on this curve.
01:06
And they tell us, so we are to find the mass for two different scenarios, a, when the density delta equals 3t, and b when delta is one.
01:32
So one thing to take note of right away, in scenario b here, if the density is constant in a wire, then really what we're doing is we're just calculates.
01:41
The arc length, right? so that makes sense.
01:46
So hopefully, you know, if you thought about this for a little bit, you kind of look at the way, well, density, constant density, right? it doesn't, right, if you had 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 segment was, right? okay, so then in general calculating mass when the density varies is, a little more difficult, right? so but on page 904 we have this formula for the mass of a wire.
02:26
So mass of wire on curve c with density delta is.
02:44
So the mass is the integral over c of your density function times your differential arc length.
02:55
So here's the formula, right? so now the next thing you might be thinking is, all right, well, this section is about line integrals.
03:04
So how does this match up with the formula for a line integral? so let's go ahead and let's go ahead and write down the formula for a line integral of f.
03:17
So we have f, a function, defined on curve.
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So let's just say it's a smooth curve.
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C, right? by some formula of some parametric equation.
03:42
Let's say r of t is x of t, i, plus y of t, j, plus z of t, k.
03:58
And that's for t in between some a and b.
04:10
So then the line integral of f over c is the integral around c of f of x, d, s, d, s, right? okay, so here's the line integral formula, right? here's some parametric equations that give us the curve, c.
04:45
So the question is, how do you actually compute with this? so this is supposed to be a definition, but how do you actually compute with this? so it turns out that our ds, right, so ds, if you look in your textbook, you'll find the derivation of this, but the differential arc length, you can represent this way as a the magnitude or the length of the derivative vector r prime of t d t so when when these first derivatives of x of t y of t and z of t exist you can take the derivative vector r and if those are all continuous then this is all nice and good you can represent ds this way um so it turns out that that's d s right and then what you're supposed to do is in this f of x y z, right? you plug in your x of t, y of t, and z of t.
05:54
So that's how you're, think of that is, that's how you're restricting the values of f, just to take the values of f on the curve c.
06:02
So actually computing this thing, you integrate with respect to t over the interval a to b of f of x of t y of t, z of t.
06:23
Right.
06:24
So this is just f with x of t, y, t, z of t, i'll put in here.
06:28
And instead of ds, we do this with respect to the derivative vector, the length of the derivative vector.
06:40
Like that, right? okay.
06:43
So hopefully you're seeing how what we're given is, what i mean is hopefully you're seeing how the functions we were given, this delta equals.
06:55
3t and delta equals 1 how that's going to match up if we go back to this whiteboard right so we're given delta equals 3t and delta equals 1 right you might be thinking to yourself well in the line integral section right the functions are always usually or they're usually given as right functions of x y and z and then i plug the parametric equations for the curve c in for x y and z right so but now i'm just given delta equals 3 t right what does that mean well so this delta is already parametized for us right so don't let that be a stumbling block this is already parametized so we don't need right there's no function to plug the parametric equations square root t square root two times t square two times t and then four minus t squared there's no need to plug those in for some x y and z that's already sort of done for us in the problem.
08:12
And then this ds, this is the thing we sort of need to calculate.
08:15
We need to figure out what's ds.
08:17
So ds, right, let's do that down here.
08:21
Right, so because we're going to need ds for both problems.
08:24
So ds is, right, we said it was the length of the derivative vector r prime of t and then times dt.
08:35
Right, so let's calculate r prime of t.
08:42
So your book calls r prime of t this v of t, right? it does dr over dt as v of t.
08:49
So r prime of t, do, do, do.
08:52
So that's going to be the square root of two times i plus the square root of two times j, and then minus 2tk.
09:05
So now if we take the length of that, right, that's the, well, we're going to take a square root, right, because we're doing clitian distance.
09:20
And then if we square the square root of two, we get two.
09:23
So we have two plus two, right? we're squaring each of the components on r prime of t, and we're adding them all together.
09:31
And then negative 2t squared is 4t squared, right? so if you put negative 2t in parentheses, and then you square it, you get 4t squared.
09:44
Okay.
09:44
Then this, right, so i'm going to go ahead and rewrite it right away to save some space.
09:52
This 2 plus 2 inside here, i'm going to rewrite as 4 right away.
09:59
So this is 4 plus 4 t squared.
10:03
And now we can factor the 4 out, right? there's a common 4, so it be 4 times 1 plus t squared under the square root.
10:12
But then i can pull the 4 out of the square root and write it as 2, right? so that would split in the square root of four times square root 1 plus t squared, but the square root of 4 is 2.
10:23
So the length of the derivative vectors r of t, right? so the derivative, the vector derivative function r of t, or sorry, our prime of t, its length function is given by 2 times the square root of 1 plus t squared.
10:43
Okay.
10:44
So we will need ds for both of the integrals we're going to set up.
10:50
So let's open up a new whiteboard.
10:53
So now we have, right, this is ds.
10:56
And this is going to, for this particular problem, this is d .s, right? and this is going to help us quite a bit.
11:03
Right.
11:04
So now we're going to take our mass, right, or sorry, our density functions.
11:08
We're going to do times ds and we're going to integrate over the curve given by r of t up here.
11:14
So what is? that look like? right.
11:18
So now for part a where delta equals 3t, the mass should be given by the integral from 0 to 1, right? those are the bounds for the curve r.
11:37
Right? so delta is 3t, so we write 3t.
11:42
And then ds was 2 times the square root of 1 plus t squared d t right so now 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 times the integral from zero to one of t times the square root of one plus t squared d t okay 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 just to track down all those connections, right? to understand this is a line integral, set up the right integral, and then get to this point.
12:34
So now here, right, okay, think back integration techniques.
12:38
You know, maybe calp 2, maybe it's the same semester depending on how you chunk up calculus.
12:44
But think back to integration techniques.
12:47
So here we have integral from zero to one, right? ignore the six, integral from zero to one, right? we have integral of t times the square root to 1 plus t square.
12:56
If we could somehow get rid of this t 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 integral of square root of you, for instance, du, well, that's easy.
13:12
That's just like a power rule.
13:13
We know how to do that, right? so, but then the question is, how do we, how could we do that, right? well, if we set the inside to be u, if we set 1 plus t squared, the inside of the square root to be u, then the derivative is 2t, and we have a multiple of the derivative, right? so we have a multiple of 2t right over here with this t that we can then replace.
13:38
So we're going to set up a u substitution.
13:41
So let's do, let's set you to be 1 plus t square.
13:47
All right, then let's calculate du.
13:51
That's 2t, dt.
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And then let's go ahead and figure out what...
13:59
You can do this a couple ways.
14:01
Let's go ahead and let's just do du over 2t is dt.
14:07
Let's solve for dt right away.
14:09
That sort of guarantees you that you can just plug in and things will cancel appropriately.
14:16
You can move on from there.
14:17
Okay.
14:17
So now i'm going to change the integration.
14:19
Bounds right away as well.
14:21
So to do that, right? so to do that, we plug in zero for t, and when we do that, right, so u of zero is one, and when you plug in 1 for t, u of 1 is 2.
14:35
So the new integration bounds are 1 and 2, and the integral looks like this.
14:41
So m is 6 times the integral from 1 to 2 of...
14:49
So let's go ahead and let's leave the t, right? but then let's replace 1 plus t squared with u.
14:57
And then dt, let's write as du over 2t.
15:04
And now that's a good thing, the t's canceled, right? and then this du, this one -half, du, right? d -u over two, this one -half du.
15:12
This one -half can come to the outside with the six.
15:15
So this is, right, six over two.
15:18
Which will give us a three, the integral from one to two.
15:22
And so instead of writing it a square root of u, let's prepare ahead, right, because we'll need to integrate.
15:28
So that's u to the one -half power, du.
15:33
And then when you integrate that, so now we have three.
15:38
And when you integrate, right, you add one to the power, and then you divide.
15:42
So when you divide by three halves, that's multiplying by two -thirds.
15:46
So three times two thirds of u to the three halves, right? and we don't need to do a plus c because we're evaluating.
15:56
We're doing a definite integral.
15:57
So we'll do that from one to two.
16:00
And these threes cancel.
16:03
And when you plug in, you end up getting two times square root 8 minus one.
16:13
And this ends up being m when delta equals 3t.
16:24
So this is when delta equals 3t, this is what we get right here.
16:32
So you get 2 times the square root of 8 minus 1.
16:36
So we could distribute that, right? if you want to punch that into your calculator, right, to get an approximation, you go ahead and you do that.
16:42
Okay.
16:43
So that's the first case when delta equals 3t.
16:46
And it actually is going to turn out from an integration standpoint, from just a purely calculus standpoint, this was actually easier than what we're going to do.
16:54
And delta is one.
16:56
So it turns out when the density is uniform when it's one, right? so from a physical standpoint, that's easier in the sense that, well, the density doesn't vary.
17:04
It's just constant along this wire, right? but it's going to turn out that calculating just the arc length, which will just be the integral of ds.
17:13
So we want to have this 3t out here.
17:15
And then this use substitution, which is probably what a lot of you feel sort of most comfortable with in terms of techniques we won't have available to us and we'll have to pull 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 b where delta equals 1 right okay so now again we have m it is going to be the integral from 0 to 1 right and this time it's just going to be 1 times ds, right? okay, so now we plug in what we know ds is.
18:05
So ds was that 2 times the square root of 1 plus t squared.
18:13
So we have this, yeah, ds was 2 times square root 1 plus t squared d t, right? okay, so now looking at this, so as i said, so from a purely calculus standpoint, this is interesting because from the physical standpoint, this is maybe easier in a sense that the density is uniform.
18:37
It doesn't change throughout the wire.
18:39
However, from a calculus standpoint, this is actually, at least in my opinion, slightly harder in the sense that for most students i've taught, they usually feel a little bit more comfortable with something like use substitution than they do sort of the tricks we're going to do to get to the end of this problem.
18:59
So, okay, what do we do with this? well, first off, let's bring the two layer.
19:03
Let's bring that out from.
19:05
So let's write this is 2 times the integral of 0 of 1 of the square root of 1 plus t squared.
19:13
Okay, d t.
19:15
So now when you're looking at this, right, focus on the integrand here, square root 1 plus t squared.
19:21
You should be thinking to yourself like, okay, that's a square root.
19:25
And then i've got a number that, like, i can take the square of.
19:29
So i can think of 1 as like 1 squared, right? so it's the square root of one squared plus t squared.
19:35
You're like, hmm, that's almost like square root of a squared plus b squared.
19:39
So like pythagorean theorem, right? triangles? right triangles? and the answer is yes.
19:49
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.
19:59
So with an angle theta or however you want to label that angle.
20:03
But let's call theta.
20:04
We'll set up a right triangle here.
20:07
So you should be thinking inverse trig substitutions.
20:11
That's where we're going, right? and why do i call them inverse trig substitutions? well, they work a little differently than new substitution in the sense that they really are an inverse substitution of variables...