Question about Finding a Force with line integrals

[Mohamed Abdul]

Homework Statement

[/B]
F =< 2x, e^y + z cos y,sin y >
(a) Find the work done by the force in moving a particle from P(1, 0, 1) to Q(1, 2, −3) along a straight path.

(b) Find the work done by the force in moving a particle from P(1, 0, 1) to Q(1, 2, −3) along the curved path given by C : r(t) =< 1 + sin πt, 2 sin(πt/2), 1 − 4t >, 0 ≤ t ≤ 1.

Hint: Think.

Homework Equations

Integral of Fdr = Integral of F(r(t))*r'(t)

The Attempt at a Solution

[/B]
(a) For the first one, I parametrized my points to get vector r as <1,2t,1-4t> with range 0<t<1. I then plugged this into F to get F = <2,e^2t+(1-4t)cos(2t),sin(2t)>. I mupltiplied this vector by the derivative of r, <0,2,-4> and took the integral, and after computing the whole thing I ended up with [e+sin(2)-4sin(2)-2cos(2)+4cos(1)-3].

Now this answer looks a bit odd to me, so I'm not sure I went through the right process to solve it.

(b) This part is where I'm really stumped. I tried plugging r into F but ended getting an extremely long and complex vector function to take the integral of. The hint said to think, so I assume I'm not supposed to brute force the integral. Is there any property of either F or r that will allow me to simplify my work?

 

[LCKurtz]

Homework Statement

[/B]
F =< 2x, e^y + z cos y,sin y >
(a) Find the work done by the force in moving a particle from P(1, 0, 1) to Q(1, 2, −3) along a straight path.

(b) Find the work done by the force in moving a particle from P(1, 0, 1) to Q(1, 2, −3) along the curved path given by C : r(t) =< 1 + sin πt, 2 sin(πt/2), 1 − 4t >, 0 ≤ t ≤ 1.

Hint: Think.

Homework Equations

Integral of Fdr = Integral of F(r(t))*r'(t)

The Attempt at a Solution

[/B]
(a) For the first one, I parametrized my points to get vector r as <1,2t,1-4t> with range 0<t<1. I then plugged this into F to get F = <2,e^2t+(1-4t)cos(2t),sin(2t)>. I mupltiplied this vector by the derivative of r, <0,2,-4> and took the integral, and after computing the whole thing I ended up with [e+sin(2)-4sin(2)-2cos(2)+4cos(1)-3].

Now this answer looks a bit odd to me, so I'm not sure I went through the right process to solve it.

(b) This part is where I'm really stumped. I tried plugging r into F but ended getting an extremely long and complex vector function to take the integral of. The hint said to think, so I assume I'm not supposed to brute force the integral. Is there any property of either F or r that will allow me to simplify my work?

Have you checked $\nabla \times \vec F$, and do you know why I ask that question?

 

[Mohamed Abdul]

Have you checked $\nabla \times \vec F$, and do you know why I ask that question?

I haven't checked that yet, but I know that is the curl. Would it be easier to evaluate the matrix for the curl as opposed to manually computing the integral?

 

[LCKurtz]

We have nothing to talk about until you check it and tell us what you think.

 

[Mohamed Abdul]

We have nothing to talk about until you check it and tell us what you think.

I got that the curl = 0. Would that mean that since there is no curl, I would be moving along a straight path and so I would only need to find the vector PQ, plug it into F, and multiply it by the derivative of PQ then find the integral of all of that?

 

[LCKurtz]

I got that the curl = 0.

You mean curl = $\vec 0$.

Would that mean that since there is no curl, I would be moving along a straight path and so I would only need to find the vector PQ, plug it into F, and multiply it by the derivative of PQ then find the integral of all of that?

There is a standard theorem about line integrals $\int_C~\vec F\cdot d\vec r$ when the curl of $\vec F$ is zero. What does it tell you?

 

[Mohamed Abdul]

You mean curl = $\vec 0$.
There is a standard theorem about line integrals $\int_C~\vec F\cdot d\vec r$ when the curl of $\vec F$ is zero. What does it tell you?

Isn't that just that F is irrotational?

 

[LCKurtz]

I don't mean words that are different ways to describe curl free. I mean what does it do mathematically for you. Look up the theorem.

 

[Mohamed Abdul]

I don't mean words that are different ways to describe curl free. I mean what does it do mathematically for you. Look up the theorem.

Is that Stoke's theorem, where you can rewrite the line integral as the double integral over the surface s for the curl of F? So then the double integral of a zero curve would yield zero, right?

 

[LCKurtz]

No. You don't have a surface in this problem. It has to do with independence of path and potential functions etc.

 

[Mohamed Abdul]

No. You don't have a surface in this problem. It has to do with independence of path and potential functions etc.

So if the curl is zero, does that mean that the vector field is independent of path and that my integral would be the line integral of the gradient multiplied by derivative of r. I didn't learn this in class just yet, so I'm going off this: http://tutorial.math.lamar.edu/Classes/CalcIII/FundThmLineIntegrals.aspx

 

[LCKurtz]

So if the curl is zero, does that mean that the vector field is independent of path and that my integral would be the line integral of the gradient multiplied by derivative of r. I didn't learn this in class just yet, so I'm going off this: http://tutorial.math.lamar.edu/Classes/CalcIII/FundThmLineIntegrals.aspx

I think you are getting a bit ahead of yourself here. Since the integral is independent of path, you might be able to find a path what works out easier. But the problem is really about potential functions and being able to avoid doing the line integration at all. The $f$ in that theorem is not the same as the $\vec F$ in your problem. You need to learn about when potential functions exist and how to find them. I suggest you wait until your class gets to the topic.

 

[vela]

Is that Stokes' theorem, where you can rewrite the line integral as the double integral over the surface s for the curl of F?

Hint: If you consider the path in part (a) and the path in part (b) together, they enclose a surface in the xy-plane. (Draw a picture.)

So then the double integral of a zero curve would yield zero, right?

This sentence doesn't really make sense because you're mixing up the two sides of Stokes' theorem. Try to choose your words more carefully.

 

[Mohamed Abdul]

Hint: If you consider the path in part (a) and the path in part (b) together, they enclose a surface in the xy-plane. (Draw a picture.)This sentence doesn't really make sense because you're mixing up the two sides of Stokes' theorem. Try to choose your words more carefully.

But why would I use the path in part a if it's not mentioned in part b?

 

[vela]

That has to do with the "think" hint. You have all the bits and pieces needed in this thread. You just need to put them all together.

 

[Mohamed Abdul]

I think you are getting a bit ahead of yourself here. Since the integral is independent of path, you might be able to find a path what works out easier. But the problem is really about potential functions and being able to avoid doing the line integration at all. The $f$ in that theorem is not the same as the $\vec F$ in your problem. You need to learn about when potential functions exist and how to find them. I suggest you wait until your class gets to the topic.

That has to do with the "think" hint. You have all the bits and pieces needed in this thread. You just need to put them all together.

So I did some studying and found that if the curl is zero, that means that F is conservative. So then in order to find my value, I could just do f(r(1))-f(r(0)) where f would be my original function F, right? But then how would I find the final value since the difference of those two would yield a vector?

 

[LCKurtz]

So I did some studying and found that if the curl is zero, that means that F is conservative. So then in order to find my value, I could just do f(r(1))-f(r(0)) where f would be my original function F, right?

Wrong. I mentioned that in post #12. And I hadn't noticed that parts a and b use the same two points. What does independence of path mean?

 

[Mohamed Abdul]

Wrong. I mentioned that in post #12. And I hadn't noticed that parts a and b use the same two points. What does independence of path mean?

Independence of path means that the only thing that the integral relies on are the bounds. How would we prove that this function is independent of path, though. Is it enough to say it is independent of path if curl = 0?

 

[LCKurtz]

Independence of path means that the only thing that the integral relies on are the bounds. How would we prove that this function is independent of path, though. Is it enough to say it is independent of path if curl = 0?

Isn't that what the theorem tells you? And if so, what does that tell you about your answers to a and b? And, by the way, your answer to a isn't correct.

 

[Mohamed Abdul]

Isn't that what the theorem tells you? And if so, what does that tell you about your answers to a and b? And, by the way, your answer to a isn't correct.

Was my answer to a incorrect only or was my process wrong? Also, wouldn't the path independence make the answers the same for both since curlF is 0?

 

[Mohamed Abdul]

Isn't that what the theorem tells you? And if so, what does that tell you about your answers to a and b? And, by the way, your answer to a isn't correct.

So I finally learned about potential functions and I believe I may have the right idea as to solve b. Here is my work:

Also I believe my problem with part one is that when I tried making the two points into a fector I should have added the initial point to the unit vector between the points.