Line Integral Example - mistake or am I missing something?

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SUMMARY

The discussion centers on the application of the Fundamental Theorem for Line Integrals, specifically in calculating the work done by a gravitational field represented by the vector function \(\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x}\). The participant questions the textbook's solution regarding the scalar potential function \(f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}\) and its gradient. After some deliberation, they conclude that the scalar function is indeed an antiderivative, which allows for the correct computation of the gradient vector that yields the appropriate force. This clarification resolves their initial confusion about the dimensionality of the results.

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kostoglotov
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This is an example at the beginning of the section on the Fundamental Theorem for Line Integrals.

1. Homework Statement


Find the work done by the gravitational field

\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x}

in moving a particle from the point (3,4,12) to (2,2,0) along a piece wise smooth curve

Now, I think that there's a mistake in the solution given...but this textbook is pretty good, and plenty of times in the past I've thought it had made a mistake and really I was mistaken.

So, I understand all the concepts (edit: I obviously didn't), all good there. It's here where I'm scratching my head.

f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}

So, we need to get the scalar function of f, call it the potential function, we know that in a conservative vector field \vec{F} = \nabla f, no worries. However, shouldn't it be

\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x} = -\frac{mMG|\vec{x}|}{|\vec{x}|^3}\vec{u} = -\frac{mMG}{|\vec{x}|^2}\vec{u}

And so converting the vector form of F into a scalar field from which we can compute the grad vector, doesn't

|\vec{x}|^2 = x^2 + y^2 + z^2

not

\sqrt{x^2+y^2+z^2}

so shouldn't it be

f(x,y,z) = \frac{mMG}{|\vec{x}|^2} = \frac{mMG}{x^2+y^2+z^2}

how would one end up with

f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}

??

edit: both the 6th and 7th editions have the same example...so I'm guessing that I'm missing something.
 
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Can you work out the gradient of the f(x,y,z) you suggest? Does it give the right force?
 
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DEvens said:
Can you work out the gradient of the f(x,y,z) you suggest? Does it give the right force?

No, it gives N not N m.

I'm still not sure what's going on here though. Is it the case that we're NOT trying to just find the magnitude of the gravity vectors on a given domain, but rather searching for some other related function, that will produce a grad vec that when dot producted with a path vector will result in the dimensions that we need in our answer?

I still don't know how they got to their answer.

edit: Oh wait, I think I see what's happening...f(x,y,z) is essentially an antiderivative ?? Thus, we need to be able to get to the grad vec from that f(x,y,z)...??

edit2: yep, that worked, and it made sense...thanks :)
 
Last edited:
kostoglotov said:
This is an example at the beginning of the section on the Fundamental Theorem for Line Integrals.

1. Homework Statement


Find the work done by the gravitational field

\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x}

in moving a particle from the point (3,4,12) to (2,2,0) along a piece wise smooth curve

Now, I think that there's a mistake in the solution given...but this textbook is pretty good, and plenty of times in the past I've thought it had made a mistake and really I was mistaken.

So, I understand all the concepts (edit: I obviously didn't), all good there. It's here where I'm scratching my head.

f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}Standard result: if ##\vec{r} = x\vec{e}_x + y\vec{e}_y + z \vec{e}_z## with magnitude ##r = \sqrt{x^2 + y^2 + z^2}##, then
\vec{ \nabla} \frac{1}{r} = -\frac{\vec{r}}{r^3}
Just look at ##\partial r^{-1} / \, \partial x ## for example.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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