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- Astrophysics

|dw:1436077425511:dw| leggo

- Astrophysics

I didn't have the chance to till now, so now I'll dooooo it! \[W = \int\limits \vec F \cdot dr\]

- Empty

Good luck I'll be watching haha.

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## More answers

- Astrophysics

Oh that dr should be a vector but w/e ok lol

- Astrophysics

\[\frac{ \vec dr }{ dt } = \vec r'(t) \implies \vec r'(t) dt\]
\[\vec r'(t) = <1,-2(t-2)>\]
\[W = \int\limits <0,-g> \cdot <1,-2(t-2)> dt\]

- Astrophysics

|dw:1436077905456:dw| just seeing the drawing again XD
\[W = \int\limits_{0}^{4} (0) \vec i + (2g(t-2)) \vec j dt \]

- Michele_Laino

is:
\[\Large {\mathbf{F}} = \left( {0, - mg} \right)\]?

- Astrophysics

Got 0? It's a made up question haha

- Astrophysics

But yeah it would be -mg otherwise

- Michele_Laino

since, I see:
\[\Large {\mathbf{F}} = \left\langle {0, - g} \right\rangle \]

- Empty

Yeah I just made this question up on the fly last night really late, so yeah throw an m in there haha.

- Astrophysics

|dw:1436078560687:dw|
hey empty quick question, I can't really remember, but would I have just a constant in the i direction as I'm integrating respect to 0

- Michele_Laino

I got:
8g

- Astrophysics

Mhm sec, maybe I made a mistake

- Astrophysics

Oh im dumb

- Astrophysics

I messed up at distributing XD

- Michele_Laino

please note this:
\[\Large W = \int_0^4 {dt\left( {2gt - 4g} \right)} \]

- Astrophysics

ok let me redo it |dw:1436078733805:dw|

- Astrophysics

Thanks for checking @Michele_Laino !!

- Michele_Laino

so:
\[\Large W = \int_0^4 {dt\left( {2gt - 4g} \right)} = 16g - 16g = 0\]

- Michele_Laino

- Empty

Ok so you have the calculation done correctly which is excellent, but what is the significance of what you've done? @Astrophysics That's what I want you to explain to us.:D

- Michele_Laino

I think that a possible answer can be this:
"the work done by the weight force, along that trajectory is 0"

- Astrophysics

The work done by the field is pulling us down via trajectory giving us 8g, but if we had a mass it would be W = mg.

- Astrophysics

Well Michele put it much nicer than me

- Michele_Laino

we gave the same answer! :)

- Astrophysics

Haha, yeah well work problems are pretty fun

- Michele_Laino

yes! they are related to vector calculus, and I like very much the vector calculus! as you know @Astrophysics eh eh :)

- Astrophysics

But if we had a circle lets say, that's the field and the circle is going counter clockwise, how would we define the work exactly? We'll just assume x = cost, y = sint, w/e |dw:1436079570813:dw|

- Astrophysics

|dw:1436079731168:dw|

- Michele_Laino

I think that we have to use this parametrization:
\[\Large \begin{gathered}
x = R\cos t \hfill \\
\hfill \\
y = R\sin t \hfill \\
\end{gathered} \]
|dw:1436079812068:dw|

- Astrophysics

Haha yeah, you're probably right, all of these questions in this post are just made up out of our heads and I guess that's not very useful...xD but to say the least it's fun, I just want to know what happens with the field and the path taken

- Astrophysics

If we give it an interval [0,2pi], we would get -2 pi, so let me just ask you a question to clarify, aside from the math what exactly does this mean, it's opposing the field, or just saying it's going in the opposite direction?

- Michele_Laino

what is the vector field?

- Astrophysics

Oh yi-xj

- Empty

This is because you have a nonconservative field in this case. In the previous case, we could have written that particular vector field as the gradient of a scalar field while this one has no corresponding scalar field.
So think, what is the corresonding scalar field to the Force field from the first question? Just let yourself be stumped on this for a while so you will be satisfied when learning the answer since it is important and should be remembered, as it has a fairly well known associated equation. :)

- Michele_Laino

I got this:
\[\Large W = \int_0^{2\pi } {\left( {R\sin t,R\cos t} \right) \cdot \left( { - R\sin t,R\cos t} \right)} dt\]

- Astrophysics

Ah, I see thanks empty, I did not think of it as a nonconservative field, and no worries Michele, the math is fine, but I was just trying to figure out the theoretical part of the problem!

- Michele_Laino

oops.. I have made an error of sigh, here is the right integral:
\[\Large W = \int_0^{2\pi } {\left( {R\sin t, - R\cos t} \right) \cdot \left( { - R\sin t,R\cos t} \right)} dt\]

- Michele_Laino

sign*

- Astrophysics

It's ok :)

- Astrophysics

Thanks @Michele_Laino and @Empty !! :)

- Michele_Laino

we can note that your field can be rewritten as follows:
\[\Large {\mathbf{F}} = \left( {y, - x,0} \right)\]
now according to the observation of @Empty , we have:
\[\Large \nabla \times {\mathbf{F}} = \left( {0,0, - 2} \right)\]
namely our vector field is an irrotational field

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