# Thread: Work done by a force field on a particles...

1. ## Work done by a force field on a particles...

So the question says:
Find the work done by the force field F = (2x+y-yz, x+y-xz, -xy-3z^2)
on a particle moved along the straight line from (0,2,1) to (1,4,0)

Soooo what I've done is:

the straight line is parametrized by:
x = 0.5 t
y = t + 2
z = -0.5 t + 1
for 0<t<2

Then F = (t+t+2 - (t+2)(-0.5t+1), 0.5t +t+2-0.5t(-0.5t+1), -0.5t(t+2)-3(-0.5t+1)^2)

So then F = (0.5t^2+2t, 0.25t^2+t+2, -1.25t^2+2t-3).

Therefore work = integral(0 to 2) of F dot ds

So
= integral(0 to 2) of (0.5t^2+2t, 0.25t^2+t+2, -1.25t^2+2t-3) dot (0.5t, t+2, -0.5t+1)

which works out to be 127/6

Is this correct? I'm not at all sure about the parametrization, is there an easier way of doing this???
Thanks in advance for any help!

2. Originally Posted by MeganKnox
So the question says:
Find the work done by the force field F = (2x+y-yz, x+y-xz, -xy-3z^2)
on a particle moved along the straight line from (0,2,1) to (1,4,0)

Soooo what I've done is:

the straight line is parametrized by:
x = 0.5 t
y = t + 2
z = -0.5 t + 1
for 0<t<2
Well, I would have used x= t, y= 2t+ 2, z= 1- t, 0< t< 1, to avoid the fraction, but that's a perfectly valid paramaterization.

Then F = (t+t+2 - (t+2)(-0.5t+1), 0.5t +t+2-0.5t(-0.5t+1), -0.5t(t+2)-3(-0.5t+1)^2)

So then F = (0.5t^2+2t, 0.25t^2+t+2, -1.25t^2+2t-3).

Therefore work = integral(0 to 2) of F dot ds

So
= integral(0 to 2) of (0.5t^2+2t, 0.25t^2+t+2, -1.25t^2+2t-3) dot (0.5t, t+2, -0.5t+1)

which works out to be 127/6

Is this correct? I'm not at all sure about the parametrization, is there an easier way of doing this???
Thanks in advance for any help!
I don't know that it is easier, but you could not that this is a conservative force field. That is, $\displaystyle (2x+y-yz)_y= 1= (x+y- xz)_x$, $\displaystyle (2x+y-yz)_z= -y= (-xy-3z^2)_x$, and $\displaystyle (x+y-xz)_z= -x= (-xy-3z^2)_y$. That means that the integral is independent of the path and there exists a function F(x,y,z) such that $\displaystyle dF= (2x+y-yz)dx+ (x+y-xz)dy+ (-xy-3z^2)dz$. You could find that function, F, evaluate it at (0, 2, 1) and (1, 4, 0) and subtract.

$\displaystyle \frac{\partial F}{\partial x}= 2x+ y- yz$ so $\displaystyle F= x^2+ xy- xyz+ g(y,z)$. The "constant of integration" may be a function of y and z since they are treated as constant when differentiating with respect to x.

Differentiating that with respect to y, we get $\displaystyle \frac{\partial F}{\partial y}= x- xz+ \frac{\partial g}{\partial y}= x+ y- xz$ so $\displaystyle \frac{\partial g}{\partial y}= y$. (The terms involving x cancel out- that had to happen and is a consequence of this being a conservative force field.)
We can integrate that with respect to y go get $\displaystyle g(y)= \frac{1}{2}y^2+ h(z)$ where, now, the "constant of integration" may be a function of z.

Now we have $\displaystyle F(x, y, z)= x^2+ xy- xyz+ g(y,z)= x^2+ xy- xyz+ \frac{1}{2}y^2+ h(z)$. Differentiating that with respect to z we have
$\displaystyle \frac{\partial F}{\partial z}= -xy+ \frac{dh}{dz}= -xy-3z^2$ so that $\displaystyle \frac{dh}{dz}= -3z^2$. From that $\displaystyle h(z)= -z^3+ C$ where C really is a constant.

That is,
$\displaystyle F(x,y,z)= x^2+ xy- xyz+ \frac{1}{2}y^2+ h(z)= x^2+ xy- xyz+ \frac{1}{2}y^2- z^3+ C$.

Evaluate that at (0, 2, 1) and (1, 4, 0) and subtract. Of course, the "C" will cancel.

By the way, this really has nothing to do with "Linear and Abstract Algebra" and should have been posted in either the "Applied Math" or "Calculus" areas.

3. If to apply linear algebra , something is sample

$\displaystyle W=\vec{F}\cdot \vec{s}=<\vec{F}, \vec{s}>=\int\nolimits_C <\vec{F}, d\vec{s}>$
$\displaystyle \vec{v}_1=\begin{bmatrix}0\\2\\1\end{bmatrix}~,~ \vec{v}_2=\begin{bmatrix}1\\4\\0\end{bmatrix}~,~ \vec{s} = \vec{v}_1(1-t)+\vec{v}_2t ~,~ 0\leq t \leq 1 ~,~ \vec{w} = \vec{v}_2-\vec{v}_1 ~,~d\vec{s}=\vec{w}\,dt$

$\displaystyle W=\int\nolimits_0^1 <\vec{F}(\vec{s}), \vec{w}\,dt> = \int\nolimits_0^1 <\vec{F}(\vec{s}), \vec{v}_2-\vec{v}_1> \,dt$

$\displaystyle =\int\nolimits_0^1 <\vec{F}(\begin{bmatrix}t\\2+2t\\1-t\end{bmatrix}), \begin{bmatrix}1\\2\\-1\end{bmatrix}> \,dt =12$