Work done by vector field on straight path

[sa1988]

Homework Statement

Homework Equations

W= ∫F.dr

The Attempt at a Solution

I'm fairly sure I've done the right thing, however my lecturer hasn't uploaded any solutions to any of these problems (which is ridiculous - how am I supposed to learn if I don't know when I'm right or wrong?!) So I'm putting it over to physics forums for some general opinion on my answer.

The path between (1,4,2) and (0,5,1) is a simple move of (-1,-1,-1),

This parameterises to a move of (-t,-t,-t) for 0<t<1

and means x= -t, y= -t, z= -t

so dr = (-t,-t,-t) = (-1,-1,-1)dt

and F = (x, 3xy, -(x+z)) = (-t, 3t², 2t)

Which gives ∫F.dr = (-t, 3t², 2t )(-1,-1,-1)dt

= ∫ (t -3t² - 2t) dt

= [t²/2 - t³ - t²] , 0<t<1

= -3/2 units of work

So am I right? Or mostly right? Or mostly wrong? Or totally wrong?

Thanks a lot!

 

[pasmith]

Homework Statement

Homework Equations

W= ∫F.dr

The Attempt at a Solution

I'm fairly sure I've done the right thing, however my lecturer hasn't uploaded any solutions to any of these problems (which is ridiculous - how am I supposed to learn if I don't know when I'm right or wrong?!) So I'm putting it over to physics forums for some general opinion on my answer.

The path between (1,4,2) and (0,5,1) is a simple move of (-1,-1,-1),

This parameterises to a move of (-t,-t,-t) for 0<t<1

and means x= -t, y= -t, z= -t

No. The path in the question is a straight line from (1,4,2) to (0,5,1). The path (-t,-t,-t) with $0 \leq t \leq 1$ is a straight line from (0,0,0) to (-1,-1,-1).

You can always parametrize the straight line between $\mathbf{a}$ and $\mathbf{b}$ as $\mathbf{r}(t) = \mathbf{a} + (\mathbf{b} - \mathbf{a})t$ for $0 \leq t \leq 1$.

 

[sa1988]

No. The path in the question is a straight line from (1,4,2) to (0,5,1). The path (-t,-t,-t) with $0 \leq t \leq 1$ is a straight line from (0,0,0) to (-1,-1,-1).

You can always parametrize the straight line between $\mathbf{a}$ and $\mathbf{b}$ as $\mathbf{r}(t) = \mathbf{a} + (\mathbf{b} - \mathbf{a})t$ for $0 \leq t \leq 1$.

Hmm I think my method of parameterisation worked?

I actually made a mistake in the straight line. It's not (-1,-1,-1), it's (-1, 1,-1).

However, even in the form r(t) = a+(b-a)t, I still get dr=(-1, 1,-1)dt, just as I had before (apart from the sign error for one of the terms).

So I presume the part I'm wrong with is the bit where I suggested x=-t, y=-t, z=-t ?

I need to be able to set the the vector F so it's represented by the parameter t, so F[x(t),y(t),z(t)].

So then I can do the dot product F.dr = F(t).r(t)dt to bring out an integral purely in t.

How would I do that? I think my main problem now is giving F in terms of t.

Thanks

 

[pasmith]

Hmm I think my method of parameterisation worked?

I actually made a mistake in the straight line. It's not (-1,-1,-1), it's (-1, 1,-1).

That is indeed $\dfrac{d\mathbf{r}}{dt}$. What's $\mathbf{r}(t)$?

How would I do that? I think my main problem now is giving F in terms of t.

Having found $\mathbf{r}(t)$, and my previous post tells you all you need to find it, you substitute it into the given expression for $\mathbf{F}(\mathbf{r})$ in order to calculate $\mathbf{F}(\mathbf{r}(t)) \cdot \dfrac{d\mathbf{r}}{dt}$, which is the function you need to be integrating with respect to $t$ from 0 to 1.

 

[sa1988]

Aaaaaahhhhhhhhhhh.

From r(t) I get:

x= 1-t
y= 4+t
z= 2-t

So then I have

F[r(t)].(dr/dt)dt

= (x, 3xy, -(x+z).(-1, 1,-1)dt

= (1-t, 3(1-t)(4+t), 2t-3).(-1,1,-1)dt

= -3t²-10t+13

Then do the integral between 0<t<1

Final answer of 8.

Fantastic. My stumbling point was parameterisation of F(x,y,z) and now I understand. Thank you very much!