Evaluate ∫C(x-y+z-2) d s where C is the straight-line segment x=t, y=(1-t), z=1, from (0,1,1) to

Evaluate ∫C(x-y+z-2) d s where C is the straight-line...

Key Concepts

Parametrization of Curves

Parametrization involves expressing the coordinates of points on a curve as functions of one or more parameters. This process allows one to represent complex curves in a more tractable form for analysis and integration by reducing the problem into finding the behavior of the curve with respect to a single independent variable.

Differential Arc Length (ds)

The differential arc length, denoted as ds, is a measure of an infinitesimally small segment along a curve. It is computed as the magnitude of the derivative of the position vector with respect to the parameter, and it serves as the weight when integrating functions over the length of the curve.

Line Integrals of Scalar Functions

A line integral of a scalar function over a curve computes the accumulation of the function’s values along the curve, weighted by the infinitesimal arc lengths. This technique is essential in various applications, including physics and engineering, where one needs to sum contributions along paths in space.

Transcript

00:01 Okay, folks, so in this video, we have this line integral that we need to evaluate.

00:06 We have this integral along c of x minus y plus z minus 2.

00:16 And we're doing the integral with respect to s, the variable s.

00:24 And so the way we're going to do this is we're going to, as usual, we're going to convert this d .s into something with respect to t.

00:31 If you remember, d .s is really just dx, dt squared plus dy, dt squared plus dz, dt squared, multiplied by dt outside of the square root.

00:53 And so now we're going to evaluate x and y and z, evaluate the derivative with respect to time.

01:00 So x as a function of t is just t.

01:03 So x prime of t is one and x prime i mean y prime of t is minus one because y is one minus t and z is a constant so when you take the derivative of the of a constant with respect to a variable you get zero so z prime of t is zero so we're going to cross this out because it's zero now we plug this in here we get ds equals square root of 1 squared, which is 1, and a negative 1 squared is also 1, multiplied by d t.

01:42 So we have root 2 d t.

01:44 So we're going to plug this back into the integral, which is right here.

01:49 And so our integral becomes x as a function of time is t minus y, y is 1 minus t, plus z.

02:00 Z is a constant, z is 1, and then minus 2.

02:03 So this is the integrand.

02:06 The ds becomes now root 2dt.

02:10 All right, that's good.

02:12 So we're going to evaluate what this whole thing is.

02:15 So it's t minus 1 plus t plus 1 minus 2.

02:20 Route 2 d t.

02:22 Okay, so now i'm going to do what's this? minus 1 plus 1.

02:26 Okay, so we have 2t minus 2.