Find the work done by 𝐅= (x^2+y) 𝐢+(y^2+x) 𝐣+z e^x 𝐤 over the following paths from (1,0,0) to (1

step 1 Hello, this is from 29. It's asked us to find the work done by that view over different path, A,

Find the work done by 𝐅= (x^2+y) 𝐢+(y^2+x) 𝐣+z e^x 𝐤 over...

Find the work done by $\mathbf{F}=$ $\left(x^2+y\right) \mathbf{i}+\left(y^2+x\right) \mathbf{j}+z e^x \mathbf{k}$ over the following paths from $(1,0,0)$ to $(1,0,1)$. a. The line segment $x=1, y=0,0 \leq z \leq 1$ b. The helix $\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(t / 2 \pi) \mathbf{k}, 0 \leq t \leq 2 \pi$ c. The $x$-axis from $(1,0,0)$ to $(0,0,0)$ followed by the parabola $z=x^2, y=0$ from $(0,0,0)$ to $(1,0,1)$

Find the work done by $\mathbf{F}=$ $\left(x^2+y\right) \mathbf{i}+\left(y^2+x\right) \mathbf{j}+z e^x \mathbf{k}$ over the following paths from $(1,0,0)$ to $(1,0,1)$.
a. The line segment $x=1, y=0,0 \leq z \leq 1$
b. The helix $\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(t / 2 \pi) \mathbf{k}, 0 \leq t \leq 2 \pi$
c. The $x$-axis from $(1,0,0)$ to $(0,0,0)$ followed by the parabola $z=x^2, y=0$ from $(0,0,0)$ to $(1,0,1)$

Key Concepts

Work as a Line Integral

Work done by a force field along a path is computed as a line integral of the vector field. This involves integrating the dot product of the force vector and the differential displacement vector along the given curve. This concept is fundamental in relating forces to energy in physical systems.

Vector Field

A vector field assigns a vector to every point in space and is used to represent physical quantities such as force or velocity. Understanding the components of a vector field is essential when setting up integrals to compute work or flux through a region.

Parameterization of Paths

Parameterizing a path involves expressing the coordinates of points along the curve as functions of a single parameter. This technique converts the line integral into an ordinary integral over the parameter, making it possible to evaluate the work done by the field along different, possibly piecewise, paths.

Path Dependence

When a vector field is not conservative, the work done moving between two points depends on the chosen path. In such cases, different paths connecting the same endpoints will generally result in different amounts of work done by the force field. This concept is crucial for correctly interpreting the results of line integrals in non-conservative fields.

Transcript

00:01 Hello, this is from 29.

00:03 It's asked us to find the work done by that view over different path, a, b, and c.

00:11 Each one going to be from less complicated to bigger complicated.

00:17 Lye segment from a, the helix b and c, as in the picture, is going on the line and then going on a parabolic following the blue line.

00:27 So, do you really want to do this? that's just so.

00:32 First of all, i don't want to do that.

00:35 Let's try to find something that we can reduce this problem.

00:40 Number one, we notice that for a, b, c, people going from point zero.

00:46 Sorry, 1 -0.

00:49 That's the b, all going from 1 -0.

00:53 If we're just talking about point, to the final point is 101, for all a, b, and c.

01:00 And second, the domain is for our tree.

01:06 So it's kind of good domain though.

01:09 So now if we somewhat like similar earlier problem, if you know that f is exact, all right? and on our tree, therefore it is conservative, right? then the work, which we know that this is f, what is the work? f, dr over dt, work to have dt.

01:42 And if we simply find it, it's just going to be f .dr.

01:46 This is going to be independent of part, independent of path, if this f is exact.

01:58 And in that case, you only care about the beginning value and the ending value of the little potential function.

02:06 So let's see, is your big f, is it really going to be conservative? this is your m, this is your n, and this is your p.

02:17 Let's check it over here.

02:20 So i'm going to parcel of p with respect to what is we check partial p with respect to y.

02:29 That is zero, right? and it is the same parcel of n with respect to z.

02:36 And then we check partial of m with respect to the y.

02:41 That's also zero.

02:43 Right.

02:44 And is it the same with partial p? okay, missing partial with respect to x...

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