# Homework Help: Work done along the curve in 3 dimensional space

1. Dec 26, 2013

### brkomir

1. The problem statement, all variables and given/known data
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 $t\in \left [ 0,2\pi \right ]$

2. Relevant equations

3. 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

2. Dec 26, 2013

### haruspex

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.

3. Dec 26, 2013

### Staff: Mentor

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

4. Dec 27, 2013

### brkomir

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.

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

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?

5. Dec 27, 2013

### Staff: Mentor

I didn't say to write it as an integral. I said to write it as
Fxdx+Fydy+Fzdz=(eydx+xeydy)+(z2cosxdx+2z sinx dz)=d(xey+z2sinx)

6. Dec 27, 2013

### brkomir

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

7. Dec 27, 2013

### Staff: Mentor

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

8. Dec 27, 2013

### vela

Staff Emeritus
That's what brkomir showed in the first post.

9. Dec 27, 2013

### haruspex

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

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