Work done along the curve in 3 dimensional space

[brkomir]

Homework Statement

A vector field $\vec{F}(x,y,z)=(e^y+z^2cosx,xe^y,2zsinx)$ is given and a curve $\Gamma$ with parametrization $\vec{r(t)}=(sin(3t),cost(5+sin(10t)),sint(5+sin(10t)))$. Calculate the work done along the curve $\Gamma$ for $t\in \left [ 0,2\pi \right ]$

Homework Equations

The Attempt at a Solution

I first checked if $\vec{F}$ is potential:

$rot\vec{F}=\begin{pmatrix} \frac{\partial }{\partial x} &\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ e^y+z^2cosx& xe^y & 2zsinx \end{pmatrix}=(0,2zcosx-2zcosx,e^y-e^y)=(0,0,0)$

So yes, $\vec{F}$ is potential!

Therefore the work done along the curve $\Gamma$ can be calculated $\int \vec{F}d\vec{r}=\int grad(u)dr=u(b)-u(a)$

So I somehow have to get the equation of scalar field u...

Since we have a potential field I can say that $u_x=e^y+z^2cosx$.

That gives me $u=xe^y+z^2sinx +C(y,z)$

To determine $C(y,z)$ I derive the last equation: $u_y=xe^y+C_y=xe^y$. The last equality comes from components of vector field $\vec{F}$.

We can now see, that $C(y,z)$ is actually a function of z only! Let's check:

$u_z=2zsinx+C_z=2zsinx$ now we can see that C is not a function of z either, there fore $C=const$ let's set C to 0.

Scala field is than given as $u(x,y,z)=xe^y+z^2sinx$ and the parametrization $\vec{r(t)}=(sin(3t),cost(5+sin(10t)),sint(5+sin(10t)))$ give us

$u(t)=sin(3t)e^{cost(5+sin(10t))}+sin^2(5+sin(10t))^2sin(sin(3t))$ but because sin is a periodic function and equal to 0 for $k2\pi$ where$k\in \mathbb{Z}$ the final result is than also zero!
Which can only be explained IF the curve $\Gamma$ is connected.

Is everything written above ok? Is there a way I can check if $\Gamma$ is really connected?

I appreciate all the help!

Cheers

 

[haruspex]

$\Gamma$ with parametrization $\vec{r(t)}=(sin(3t),cost(5+sin(10t)),sint(5+sin(10t)))$. Calculate the work done along the curve $\Gamma$ for $t\in \left [ 0,2\pi \right ]$
the final result is than also zero!

Do you mean
$\vec{r(t)}=(\sin(3t),\cos (t(5+sin(10t))),\sin (t(5+sin(10t))))$?
If so, is it not clear that $\vec{r(0)}=\vec{r(2\pi)}$?

Which can only be explained IF the curve $\Gamma$ is connected.

No. If $\Gamma$ is a closed loop then the work will be zero, but the work could happen to be zero even if $\Gamma$ is not a closed loop.

 

[Chestermiller]

Try this: dr = (dx, dy, dz). Take the dot product F dot dr and see what you get. (It's going to be an exact differential).

Chet

 

[brkomir]

Do you mean
$\vec{r(t)}=(\sin(3t),\cos (t(5+sin(10t))),\sin (t(5+sin(10t))))$?
If so, is it not clear that $\vec{r(0)}=\vec{r(2\pi)}$?No. If $\Gamma$ is a closed loop then the work will be zero, but the work could happen to be zero even if $\Gamma$ is not a closed loop.

No, the parametrization is $\vec{r(t)}=(\sin(3t),(5+sin(10t))\cos t,(5+\sin (10t))\sin t)$ .

Aha, ok, I forgot about that. So $\Gamma$ can be a closed loop or it can be anything else, but it just so happens that after $2\pi$ the function is at the same potential as in the beginning.

Try this: dr = (dx, dy, dz). Take the dot product F dot dr and see what you get. (It's going to be an exact differential).

Chet

$\int \vec{F}d\vec{r}=\int (e^y+z^2\cos x)dx+\int xe^ydy+\int 2z\sin xdz=2xe^y+2z^2\sin x$

Sorry to ask you (i feel so stupid at the moment), but, why is this important?However, we all agree that work done along that curve is 0, right?

 

[Chestermiller]

$\int \vec{F}d\vec{r}=\int (e^y+z^2\cos x)dx+\int xe^ydy+\int 2z\sin xdz=2xe^y+2z^2\sin x$

Sorry to ask you (i feel so stupid at the moment), but, why is this important?


However, we all agree that work done along that curve is 0, right?

I didn't say to write it as an integral. I said to write it as
F_xdx+F_ydy+F_zdz=(e^ydx+xe^ydy)+(z²cosxdx+2z sinx dz)=d(xe^y+z²sinx)

 

[brkomir]

And because this is an exact differential, than F is potential?

 

[Chestermiller]

And because this is an exact differential, than F is potential?

Sure. More precisely, F is a force described by a potential.

 

[vela]

Sure. More precisely, F is a force described by a potential.

That's what brkomir showed in the first post.

 

[haruspex]

No, the parametrization is $\vec{r(t)}=(\sin(3t),(5+sin(10t))\cos t,(5+\sin (10t))\sin t)$ .

OK, but the question still applies: is it not clear that r(t+2π) = r(t)?