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I am having some difficulties doing some line integrals exercises, could someone please check my working and solution ? The question is as follows : 1. A particle moves from point A = (0,0,0) to point B (2$\pi$,0,2$\pi$), under the action of the force F=x i + y j -z k (please note that i,j,k is the direction vector). i) Calculate the work done by the force F on the particle if it moves along the conic-helical curve :

r(t)= (t cost) i +(t sint)j +tk with t between 0 and 2$\pi$

my approach was as follows :

Find the integral $\int_C$F((x(t),y(t),(z(t)))r'(t)dt

I calculate r'(t) to be (cost-sint)i +(tcost+sint)j +k

$\int_0 ^{2\pi}$(tcost i + tsint j -k)( r'(t))dt, this reduced to [t$^3$ + t]$_0$$^{2\pi}$ = 2$\pi$(4$\pi^2$ +1)

ii) Find the parametric vector equation for the straight line connecting A to B , and calculate the work done by the force F on the particle as it moves along the straight line.

I think in this question my methods is not correct, but it was as follows :

x=t y=t z=t

r(t)= ti+tj+tk

I determined the tangent vector to be $\frac{dr}{dt}$ = i +j -k then determined the force field of the tracjectory:

$\int_0 ^{2\pi}$(ti+tj-tk)(i+j-k)dt = $2\pi^2$,

part iii) asks to determine whether it is conservative , which I don't think I will have a problem with once these are checked

Many thanks is advance , Sophia

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  • $\begingroup$ In the second part, the particle ends up at a point with zero y coordinate. So you can just set y to be identically zero $\endgroup$ – Triatticus Mar 16 '16 at 22:11
  • $\begingroup$ I am not quite understanding how to do part ii, could you give me some hints? Also is part i right ? Many thanks $\endgroup$ – AlphaSophia Mar 16 '16 at 22:39
  • $\begingroup$ Part I is also not correct, unfortunately I am not near a computer so I can't type it efficiently, if no one answers by the time I get home I'll give an answer to your question $\endgroup$ – Triatticus Mar 16 '16 at 22:44
  • $\begingroup$ Many thanks Dan ! $\endgroup$ – AlphaSophia Mar 16 '16 at 23:01
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Ok Lets start with part 1:

We want to calculate the work done by a force field on the particle along a path $$ \int \vec{F}(\vec{r}) \cdot \mathrm{d}\vec{r} = \int \vec{F}(\vec{r}(t))\cdot \vec{r}'(t) \mathrm{d}t $$ We are given that the path is a conical helix given by $$\vec{r}(t) = t \cos{t} \;\hat{i} + t \sin{t} \;\hat{j} + t \hat{k} \quad t \in(0,2 \pi)$$ And $$\vec{F}(\vec{r}) = x \; \hat{i} + y\;\hat{j} + z\;\hat{k}$$ Using the product rule we obtain for $\vec{r}'$: $$\vec{r}'(t) = (\cos{t} - t\sin{t})\; \hat{i} + (\sin{t} + t\cos{t})\; \hat{j} + \hat{k}$$ And $$\vec{F}(\vec{r}(t)) = t\cos{t}\;\hat{i} + t\sin{t}\;\hat{j} + t\;\hat{k}$$ We take the dot product: \begin{eqnarray*} \vec{F}(\vec{r}(t)) \cdot\vec{r}'(t) &=& (\cos{t} - t\sin{t})t\cos{t} + (\sin{t} + t\cos{t}) t\sin{t} + t \\ &=& t\cos^2{t} - t^2 \sin{t}\cos{t} + t\sin^2{t} + t^2\cos{t}\sin{t} + t \\ &=& 2t \end{eqnarray*} Thus the resulting integral is: $$\int_0^{2 \pi} 2t \mathrm{d}t = \left. t^2 \right|_0^{2\pi} = 4 \pi^2$$

For part 2: We want to parameterize the straight line from $(0,0,0)$ to $(2\pi,0,2\pi)$ in the variable t, you were close with your parameterization however you didn't need y so the correct line is: $$\vec{r}(t)= t \;\hat{i}+ t\;\hat{k} \quad t\in(0,2\pi)$$ You can likely carry on from there. But if you need more I can continue on

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  • $\begingroup$ Many thanks Dan. I will try to finish it when arrive home ,cheers $\endgroup$ – AlphaSophia Mar 17 '16 at 7:57
  • $\begingroup$ No problem it should be clear from then $\endgroup$ – Triatticus Mar 17 '16 at 8:21
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i'm also struggling with a question like this, the question actually says that F=xi+yj-zk, your answer has taken F to be equal to xi+yj+zk, I worked out part 1 to overall equal 0? Edit: My rep couldnt comment on the above answer, please make sure he finds this!

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    $\begingroup$ This is comment or follow-up for your own question. Please do not post it as an ANSWER. I suggest you have an EDIT portion in your Question section, and put this question below as a follow-up. $\endgroup$ – Yujie Zha Mar 18 '17 at 15:05
  • $\begingroup$ Wasnt my question $\endgroup$ – khazoona Mar 19 '17 at 14:51
  • $\begingroup$ Ah, right, sorry. Then I recommend you to delete this answer, and ask a new question, referring to this one. So the way this site works is that it is not like a continuum discussion of posts, but one thread is one questions + several answers. We vote to make good answer standout. In such way, in one thread, people see one Question + a few good Answers, in stead of long discussions in time order. $\endgroup$ – Yujie Zha Mar 19 '17 at 14:57

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