Find the mass of a thin wire lying along the curve 𝐫(t)=√(2) t 𝐢+√(2) t 𝐣+(4-t^2) 𝐤 0 ≤t ≤1, if

Find the mass of a thin wire lying along the curve 𝐫(t)=√(2)...

Key Concepts

Line Integral for Mass

The mass of the wire is calculated by taking the line integral of the density function over the wire's arclength. This involves integrating the product of the density function and the arc length differential over the parameter interval, thereby accumulating the mass of small elements of the wire to yield the total mass.

Density Function

The density function describes how mass is distributed along the wire. It can vary with the parameter, as in the case of a function like 3t, or be constant. This function must be integrated against the arc length differential to account for varying mass contributions along the wire.

Curve Parameterization

The wire is described by a vector-valued function of a parameter, which defines the trajectory of the wire in space. This parameterization is essential for expressing other quantities, like density and arc length, in terms of the parameter, and forms the basis for setting up integrals along the curve.

Arc Length Differential

The differential element of arc length, denoted as ds, is computed as the magnitude of the derivative of the parameterized curve. It represents an infinitesimally small segment of the curve and is used to convert integrals over the curve into integrals over the parameter, capturing the true geometric length.

Transcript

00:01 Okay, folks, so in this video, we're going to take a look at problem number 35, where we're asked to find a mass of a thin wire lying along the curve given by the function, r of t.

00:15 Okay, and so there's really two parts of this problem, so we're going to look at part a first.

00:22 So for part a, we have a density function, we have delta.

00:27 My handwriting is a little bit off.

00:30 Delta of t is 3t.

00:32 We have this delta of t, the way you find the mass of the wire is, well, the way you find the mass of anything is just density multiplied by a volume element for a 3d object or like a line element, infinitesimal line segment for a wire.

00:52 So for this problem we're going to do delta ds.

00:55 The ds stands for a really small, you know, infinitesimal line segment.

01:02 And so that's the mass.

01:04 We're going to be integrating this over, i mean, along the wire.

01:08 All right.

01:08 So let's plug in delta as a function of t.

01:12 So we have 3t for part a.

01:15 And ds, well, d .s can be rewritten in terms of x, y, and z, but x, y, and z are all functions of t.

01:25 So let's go ahead and rewrite ds as x dot squared plus y dot squared plus z plus z.

01:33 X dot squared multiplied by d t and the integration limits is between zero and one for t all right so this now i should let you know that um that i'm using a notation here that you might not be very familiar with the dot on top of the letters means time derivative or derivative with respect to t okay so this is this is really just d x d t squared plus d y d t squared plus d z d t squared plus d z d t squared and we're given along the curve well the curve the the the usefulness of the curve r of t lies in the fact that that this is basically just three functions for for for x and y and z respectively r of t which is a vector basically gives gives us three functions okay so we have r of t for example the x component of r of t is going to be a function so we have root 2 t that's the x component and the y component is root 2t as well and the z component is this 4 minus t squared um and when you take the time derivative of three of these functions all three of them and you plug it in you're going to get 0 1 3 t 2 plus 2 plus minus 2 squared d t okay i skipped through the part where i'm supposed to take the derivative because i think that's too trivial, and you can handle that yourself.

03:14 All right, so we have 301t4 multiplied by t squared plus 1dt.

03:26 And this is trivial, but it requires, you know, a little bit of a special technique, not so special, because i'm not, because i'm sure you guys have all heard of it.

03:37 We're going to do a u substitution, where i define a new variable u, and i'm going to define it as t squared plus 1...

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