Work done by vector field on straight path

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Homework Statement



4sn1hs.png



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!
 

Answers and Replies

  • #2
pasmith
Homework Helper
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Homework Statement



4sn1hs.png


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 [itex]0 \leq t \leq 1[/itex] is a straight line from (0,0,0) to (-1,-1,-1).

You can always parametrize the straight line between [itex]\mathbf{a}[/itex] and [itex]\mathbf{b}[/itex] as [itex]\mathbf{r}(t) = \mathbf{a} + (\mathbf{b} - \mathbf{a})t[/itex] for [itex]0 \leq t \leq 1[/itex].
 
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  • #3
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No. The path in the question is a straight line from (1,4,2) to (0,5,1). The path (-t,-t,-t) with [itex]0 \leq t \leq 1[/itex] is a straight line from (0,0,0) to (-1,-1,-1).

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

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
 
  • #4
pasmith
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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 [itex]\dfrac{d\mathbf{r}}{dt}[/itex]. What's [itex]\mathbf{r}(t)[/itex]?

How would I do that? I think my main problem now is giving F in terms of t.
Having found [itex]\mathbf{r}(t)[/itex], and my previous post tells you all you need to find it, you substitute it into the given expression for [itex]\mathbf{F}(\mathbf{r})[/itex] in order to calculate [itex]\mathbf{F}(\mathbf{r}(t)) \cdot \dfrac{d\mathbf{r}}{dt}[/itex], which is the function you need to be integrating with respect to [itex]t[/itex] from 0 to 1.
 
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  • #5
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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!
 

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