Understanding Line Integrals with Respect to Delta X and Delta Y

[Cyrus]

I took multivariable calculus a while back. It was all fine until we got to line integrals and surface integrals divergance etc. Then it seemed like I got the rug swept from beneath me. After taking physics 2 a lot of it makes so much more sense now though. I am reading through that stuff for a second time now, but am not sure about one thing for line integrals. I understand a line integral with respect to deltaS, or arc length. That can be visualized in terms of the area that the space curve "fence" so to speak makes on one side. But what in the heck is a line integral with respect to deltax, or deltay. They don't give you the same anwser as deltaS. I looked through 3 different books calc books, and not one of them goes into any detail about WHY we want to use these or what the hell they mean. So far I am only left with my intution telling me that it may somehow be related to a projection onto the xz plane where y has no effect for the delta x, or similarly the yz plane where x has no effect for delta y, I don't, you all are smarter than me please help.

The book does give some good examples of line integral applications in phyiscs, like center of mass of a wire, or the weight of a wire with varyin density, but all are with respect to deltaS, nothing to do with deltaX or deltaY!

 

[Hurkyl]

The easiest way to understand it, coming from a Riemann sum mindset, is that the d? part is part of the "function" -- when you're doing the line integrals with ds, the summand has a factor that is the difference in arclength (s) for that partition. When you're doing a line integral with dx, then you use the (signed!) difference in x instead.

 

[Tide]

If you had to integrate a function f(s) along a path

\int f(s) ds

that would be a simple enough matter. However, f is not usually given or known as a function along a path. Oftentimes, you will have a function f(x, y) to integrate. Unfortunately, mapping f(x, y) -> f(s) is usually not easy to do accomplish.

In those cases you have to resort to working with x and y and they are related by

ds = \sqrt {dx^2 + dy^2}

and, as a matter of convenience, you may work with it in this form:

ds = \sqrt {1 + \left(\frac {dy}{dx}\right)^2} dx

in which case your integral is

\int f ds = \int f(x, y) \sqrt {1 + \left(\frac {dy}{dx}\right)^2} dx

or something similar with the integral over y. You simply choose whichever fits your problem the best.

 

[dextercioby]

Sure,Tide,u're right.Everything depends on the way the path is given.There are some curves where either parametric or (plane) polar coordinates are given.In that case,you should should consider some line elements of this kind::
ds=\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}} dt
for the case of a non plane curve
x=x(t);y=y(t);z=z(t)
(think about a helicoid,it's given usually in parametric coordinates),and for the case of a plane curve (e.g.the cycloid,hypocycloid,astroid,...),simply give up "z" in the formula above.
As for a plane curve given in polar coordinates (think about a spiral (logarithmic,Archimedes,Cornu))
\rho=\rho(\phi)
,the line element is:
ds=\sqrt{1+(\frac{d\rho}{d\theta})^{2}} d\theta

I hope these examples (sorry,theory items ) can make u see a little more clear curvilliniar integrals of first kind.

Daniel.

EDIT:And thanks to Galileo,u're given a viewpoint about curviliniar integrals of the second kind (defined for vector fields) and their tremendous applications in phyisics.Maybe if he had mentioned something about the work and heat in thermodynamics,or Clausius principle,or 1 forms (exact or closed),or maybe that would have been too technical...

 

[Galileo]

But what in the heck is a line integral with respect to deltax, or deltay. They don't give you the same anwser as deltaS. I looked through 3 different books calc books, and not one of them goes into any detail about WHY we want to use these or what the hell they mean. So far I am only left with my intution telling me that it may somehow be related to a projection onto the xz plane where y has no effect for the delta x, or similarly the yz plane where x has no effect for delta y, I don't, you all are smarter than me please help.

The book does give some good examples of line integral applications in phyiscs, like center of mass of a wire, or the weight of a wire with varyin density, but all are with respect to deltaS, nothing to do with deltaX or deltaY!

They are very important in physics. I wondered the same thing at first as well, but it became clear later. I think you'll get to this part later in the course.

Per definition, the work done on a particle by a forcefield F is:

W=\int_C \vec F(x,y,z) \cdot \vec T(z,y,z) ds=\int_C \vec F(\vec r) \cdot \vec T(\vec r) ds
where C is the path the particle takes and T(x,y,z) is the unit tangent vector at (x,y,z) on C.

If the curve is given by the vector equation: \vec r(t)=x(t)\vec i + y(t)\vec j +z(t)\vec k (with a \leq t \leq b, then \vec T(t)=\vec r'(t)/|\vec r'(t)|.
Notice that ds=|\vec r'(t)|dt

Using we write the work done as:

W=\int_a^b \vec F(\vec r(t)) \cdot \vec r'(t)dt
this is often abbreviated and written as:

W=\int_C \vec F(\vec r(t)) \cdot d\vec r

Such an integral is very common in i.e. electromagnetism. Its called the line integral of F along C.

Suppose know that F is given by component functions:
F=P \vec i+Q \vec j+R \vec k

Now plug this in \int_C \vec F(\vec r(t)) \cdot d\vec r.
You'll see that:

W=\int_C \vec F(\vec r(t)) \cdot d\vec r=\int_C Pdx +\int_C Qdy+\int_C Rdz=\int_C Pdx +Qdy+Rdz

Which uses line integrals w.r.t. x, y and z.

 

[Cyrus]

Seems to make more sense. Tide, Let me just check to see if I am understanding you correctly. You are saying that if I am given a function in terms of a parameter t: x=x(t); y=y(t);z=z(t) and a function F(x,y,z), then I should use the formula with repect to arc length ds. But if I am given a function written as x=x; y=y(x); z=z(x); and a function F(x,y,z), then I should make everything in terms of x and dx. For example, a function in terms of x,y,z. If I were to choose to use dx to solve this problem, would this mean that y and z would HAVE to be functions of x, or could y and z not be functions of x. Also, let's say I could write a function in terms of dx, or dy, or dz. But I could also easily write it in terms of t and ds. If I were to do the line integral with respect to dx, dy, or dz, and ds, should I get the same anwser in all three cases? When it comes to visualizing a line integral with respect to dx,dy or dz, does it still look the same as ds? (ds being the area of one side of a "fence"). Would it look like the same exact space curve and the same exact fence?

 

[Tide]

Cyrus,

You had better get the same result regardless of which method you use!

Also, I'm not sure what you mean by "fence" and the line or path integral does not refer to area.

 

[Cyrus]

In my calc book it says that you can think of the curve as the base of a fence. The function f(x,y) is the height of the fence. And the line integral is the area of one side that makes up this fence.

 

[Tide]

Cyrus,

Oh, I see - that's kind of a neat way of looking at it.

 

[Cyrus]

Glad I caught you online now Tide, I got a slight problem I am not getting the same anwser I will post it in one sec.

 

[Cyrus]

In one calc book it says to evaluate:

\int_c f(x,y)dx

and

\int_c f(x,y)dy

if:
f(x,y)=xy^2 and C is part of the parabola y=x^2 from A(0,0) to B(2,4).

When they do with respect to dx, they get 32/3,
and with dy, 256/7

But these are not the same anwsers!?

 

[dextercioby]

And what's the problem??U have a function which is integrated wrt to different variables ("x" and then "y") and with limits which vary:when "x":0->2,when "y":0->4.And the fact that u're integrationg along a path simply requires substituting one variable wrt the other according to which variable of integrations is chosen.That is
\int_{C} xy^{2} dx=\int_{0}^{2} x^{5} dx =...=\frac{32}{3}
\int_{C} xy^{2} dy=\int_{0}^{4} y^{\frac{5}{2}} dy=...=\frac{256}{7}

You're actually integrating different functions along the same path.Remember that the variables "x" and "y" which appear in the function "f" are not independent,actually the fact that u integrate along a curve y=y(x) makes them depend one on another.Else,hadn't it been a curvilinear integral,u would have had:
\int_{0}^{2} xy^{2} dx=2y^{2}
\int_{0}^{4} xy^{2} dy=\frac{64x}{3}

Daniel.

 

[Cyrus]

Let me try to restate my problem a little more clearly.

First, we have a function, (lets assume for the moment), that depends on x and y. so:

z=f(x,y)

Now let's also assume we have a curve C.

Now we want to integrate this function, f=(x,y) over the space curve that C makes.

In order to do this, we must use the formula:

\int_C f(x,y)ds

but the problem now lies in the ds. We cannot integrate over ds, so we must use something we can integrate over. One possible way is to define the curve C with parametric equations. Then this reduces the line integral in terms of ds into:

\int^b_a f(x(t),y(t)) \sqrt( (\frac{dx}{dt})^2+ (\frac{dy}{dt})^2) dt

This is now trivial to solve, because everything is in terms of the parameter t.
So far so good, *I Think*, I understand this.

Now comes the integral in terms of dx and dy.
Lets assume that the curve C is in parametric form. But perhaps we can break the curve into piecewise-smooth segements. And for one segment it might be easier to write the function in terms of the variable x, and for the other segment it might be easier to write the function in terms of the variable y. In such a case the standard equation:
\int_C f(x,y)ds

reduces into the integral:

\int^b_a f(x,y(x))dx

Or we could do this:

\int^b_a f(x(y), y) dy

So this is what it means to write the equation in terms of the variable x, or the variable y, or the paramater t.

And if it were possible for me to choose any of the three integration methods, I should get the same anwser.

Isint this where a line integral in terms of dx or dy comes from?

Furthermore,

if you look at what I posted up above, one of my calculus books has an example:

"Evaluate \int_c f(x,y) dx and \int_c f(x,y)dy
if: f(x,y) = xy^2 and C is part of the parabola y=x^2 from A(0,0) to B(2,4).
But the book does it and gets two DISTINCT anwsers. For dx, it gets, \frac{32}{3} and for dy, it gets \frac{265}{7}.

I know I said it before, but I have to keep asking until I undersand this. WHY the difference? If I were to follow the logic that I stated before I got to the books example, I would EXPECT to get the same anwser. Why is this not working out?

 

[Cyrus]

Here is their work, but I don't agree with it, at least not the part they did for dy.

\int_c xy^2dx= \int^2_0 x(x^4)dx = \frac{32}{3}

\int_c xy^2dy = \int^2_0 x(x^4)2xdx = \frac{256}{7}

But its dy!?

Shouldent the second equation be this!?:

First y=x^2 so: dy=2xdx or: dx=\frac{dy}{2x}. If we plug this back into the origional equation we get:

\int_c \frac{xy^2dy}{2x}= \int^4_0 \frac{y^{2}dy}{2} = 32/3!.

This is EXACTLY he same anwser as if I integrated with respect to x!

What the hecks going on here?

Another example of this change between dx and dy:
If i integrate y=x^2 from 0,0 to 2,4 in terms of x or y, I get:

\int^2_0 x^2 dx = \frac{8}{3}

or:

we say y=x^2, then that means dy= 2xdx or: dx=\frac{dy}{2x}. If we plug this into the origional integral, we obtain:

x^2dx = \frac{x^2}{2x} dy = \frac {x}{2} dy

if we integrate this, we get:

\int^4_0 .5y^{1/2}dy = \frac{8}{3}

See, these are BOTH EQUAL! If I use dy or dx! Why isint the same true for the line integral!?

 

[Hurkyl]

This is EXACTLY he same anwser as if I integrated with respect to x!

That's because you did the first integral, not the second! Recall that the second equation says \int_C xy^2 \, dy, not \int_C xy^2 \, dx.

 

[Cyrus]

Oh, so how can they just interchange them like that with such ease? Is it even allowable to do that? It seems that doing this would voilate how you arrive at a solution in terms of dx or dy!

Maybe I am wrong, but to me, it seems that you can't just put a dx or a dy anywhere you would like! It seems that you have to first see if your function is defined in terms of x, y or t. Once you know how your function is defined, you can choose the variable that is simpliest to integrate, x,y or t, and find a solution. But even if you decide to go the long and hard route, choosing a variable that complicates things, you should STILL arrive at the same anwser.

 

[Hurkyl]

They're two different integrals. \int_C xy^2 \, dx and \int_C xy^2 \, dy have different integrands.

When you tried to do integral 2, you observed that dy = 2x \, dx, and started with

<br /> \int_c \frac{xy^2dy}{2x}<br />

Which is not integral 2 at all -- it's integral 1, after replacing dx with dy / 2x.

 

[Cyrus]

Ok, then can you explain the meaning of the two different integrals? This is what I am not understanding. If you look at the example I did, I started off with a line integral in terms of a paramater t. Then I reduced the equation to terms of x and dx, and then to terms of y and dy. They all lead to the same anwser. This means they all represent the same space curve, and that they all represent the same area of the "fence" so to speak that this function makes.

What does it mean to be in terms of x and y then? It seems that this is totally unrelated to the origional curve and function in terms of ds.

 

[dextercioby]

Ok, then can you explain the meaning of the two different integrals? This is what I am not understanding. If you look at the example I did, I started off with a line integral in terms of a paramater t. Then I reduced the equation to terms of x and dx, and then to terms of y and dy. They all lead to the same anwser. This means they all represent the same space curve, and that they all represent the same area of the "fence" so to speak that this function makes.
What does it mean to be in terms of x and y then? It seems that this is totally unrelated to the origional curve and function in terms of ds.

I believe that both I and Hurkyl told you that under the integration sing there were two totally different functions.That's the only explanation for the fact that the 2 integrals do not coincide.But the curve (the parabola y=x^{2}) is still the same.It's fixed.It cannot change.The arch of parabola along u integrate it's a geometrical locus.However,it's parametrization is not unique.You have proven that there can be found minimum 3 for the same simple curve:a parabola:
y(x)=x^{2};x(y)=\sqrt{y};x(t)=t;y(t)=t^{2}.

This integral u're evaluating doesn't have the significations of "area" under the graph of a curve.If someone told you that all first kind curvilinear integrals can be pictured as "areas" under certain curves,then he lied to you,as it is not true.
If,even now,you're still wondering why the two integrated functions are differrent,i'll repeat myself in telling u that the deciding 2 factors in this judgment are:
1)"y" and "x" are not idependent variables.This is due to the fact that the integral is evluated on a curve described by a function y=y(x).
2)The differentials of the 2 integration variables are different.Since the function to integrate is the same (xy^{2}) and the path is the same,it follows that the results are different.Now,it's handy to use the same variable of integration:let that be "x".Then the integral wrt to "y" must be converted into an integral wrt to "x" by merly an ordinary change of variables,which this time is not random/out of the blue,but is dictated by the path.From changing the variables,u can see for sure that the 2 functions which need to be integrated become different (cf.prior case),but this time the integration is made not only on the same curve,but also wrt to the same variable (let's call it "x").

Daniel.

PS.Think of the next example:
Evaluate \int x^{2}y^{2}z dz along the path:
R:x=2t^{2}+1;y=3t;z=3t+4;t=[3,5]

Solve it.

 

[Cyrus]

Can you provide an application where I would want to replace ds with dx?
Here is what my book says, maybe It can help:

"Two other line integrals are obtained by replacing \Delta S_i by either \Delta x_i = x_i - x_{1-2} or \Delta y_i = y_i - y_{1-2} in definition 2. They are called the line integrals of f along C with respect to x and y:

eq. 5 \int_c f(x,y)dx = lim_{\substack{n\rightarrow \infty}} \sum^n_{i=1} f(x^*_i,y^*_i)\Delta x_i

eq. 6 \int_c f(x,y)dx = lim_{\substack{n\rightarrow \infty}} \sum^n_{i=1} f(x^*_i,y^*_i)\Delta y_i

When we want to distinguish the origional line integral \int_c f(x,y)ds from those in Equations 5 and 6, we call it the line integral with respect to arc length.
The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t:x=x(t), y=y(t),dx=x'(t)dt, dy=y'(t)dt.

\int_c f(x,y)dx= \int^b_a f(x(t),y(t))x&#039;(t)dt

\int_c f(x,y)dy= \int^b_a f(x(t),y(t))y&#039;(t)dt

 

[Cyrus]

I take it that this line integral has NOTHING to do with the line integral in terms of ds, is that correct? I think what is confusing me is that the line integral with respect to ds, can be transformed into a form that has dx or dy in it. But this is not the same line integral with dx and dy that they are talking about?

 

[dextercioby]

Can you provide an application where I would want to replace ds with dx?

1.Compute the work done by a constant force in moving a body of mass "m" along the 4 sides of a square of side "a" at a constant velocity,knowing that the kinetic friction between the body and the path is "µ".(Piece of cake).
2.Compute the wotk done by the Lorentz magnetic force (F_{L,m}=q\vec{v}\times \vec{B}) on a electrically charged particle of mass "m" and charge "q",in the case the magnetic field is homogenous and constant and the angle between the direction of the particle's initial velocity vector and the direction of the (homogenous) magnetic field is \alpha.The initial velocity vector is \vec{v}_{0}(Typically a problem found in Jackson's book).
3.Compute the gravity forces's work made on a particle of mass "m" that falls in a nonuniform/variable gravitational field from the height "h".Assume all constants known.(In between the first 2 as level of difficulty).

Daniel.

PS.As u can see,i did gave only "theory" type problems.However,u ca insert numerical values in the final results.

 

[Hurkyl]

They spring up in a lot of places, both implicitly and explicitly.

Here's a cute one that, IMHO, makes for a great example: if R is a region of the plane, and C is the boundary of R, then the area of R can be given by:

<br /> \int_C x \, dy = -\int_C y \, dx = \frac{1}{2} \int_C x \, dy - y \, dx<br />

 

[dextercioby]

I take it that this line integral has NOTHING to do with the line integral in terms of ds, is that correct? I think what is confusing me is that the line integral with respect to ds, can be transformed into a form that has dx or dy in it. But this is not the same line integral with dx and dy that they are talking about?

The name for "curvilinear integral" can be seen both as "line integral".These two syntagmas depict the same mathematical notion.

What u did up there (prior post) was nothing but the standard procedure of computing such integrals.That is transforming the line integral in a Riemann integral (as u can see,u have used Riemann sums).It's the only way these types of integrals can be evaluated.

Daniel.

EDITurkyl's example can be computed (and hence proven) by passing to a Riemann integral,just by the recipe you mentioned.

 

[Cyrus]

Daniel,

Lets say I am given a line integral in terms of dx or dy, not in terms of ds. (if It were in terms of ds, I would have to convert it into a form of dx, dy, or dt in order to evaluate it, and I would get the same anwser in all cases.) It is possible for me to visualize the area that is spanned when I think of the line integral in terms of ds. The space curve makes a base for an imaginary fence. And at each point there is a value of z=f(x,y) that determines the height above the curve. So this will generate an area of one side of this fence (for a planar space curve). This is one way to visualize a line integral in terms of ds. How does one visualize a line integral in terms of dx or dy?

Ok, here's another way of asking it. Does every line integral in terms of dx, or dy come from an origional line integral in terms of ds?

 

[dextercioby]

Daniel,

Lets say I am given a line integral in terms of dx or dy, not in terms of ds. (if It were in terms of ds, I would have to convert it into a form of dx, dy, or dt in order to evaluate it, and I would get the same anwser in all cases.) It is possible for me to visualize the area that is spanned when I think of thline integral in terms of ds.

Nope.The only way to "visualize" the geometrical interpretation of the line integral is to transform it into a Riemann one,of whose geometrical interpretation you already aware.

The space curve makes a base for an imaginary fence. And at each point there is a value of z=f(x,y) that determines the height above the curve. So this will generate an area of one side of this fence (for a planar space curve). This is one way to visualize a line integral in terms of ds.

Nope,what y've discribed is nothing but the Riemann's integral geometrical interpretation.
\int_{C} f(x,y) ds
stands for the line integral,but only after passing it into a Riemann one u can interpret it an area of the domain beneath the function y=y(x)/x=x(y),obtained from f(x,y) through explicitation of one variable wrt to another.Assuming "smoothness" of the 2 variable function "f".

How does one visualize a line integral in terms of dx or dy?

The answer is simple,as you're asking about the geometric interpretation of a Riemann integral.That's simple and i think you're aware of it.

Ok, here's another way f asking it. Does every line integral in terms of dx, or dy come from an origional line integral in terms of ds?

No.Usually it' the other way arond.Under certain conditions (smoothness of the path),u can express any line integral through a Riemann one.

Daniel.

 

[Cyrus]

I talked with a math professor and I think it made things clearer. If I am wrong, please let me know!

It seems that line integrals with respect to arc length and line integrals with respect to x or y are very different.

A line integral with respect to arc length, ds, can be changed into a form of dt, dx, or dy. Even though a line integral with respect to arc length can be reparametrized with respect to dx, or dy, that does not make it a line integral with respect to x or y. When you solve the line integral with respect to dx, or dy, or dt you SHOULD get the same anwser in all cases.

The other case is a line integral with respect to x or y. These line integrals will be in terms of dx or dy, but depending on which one is used you will get a different anwser. I think this might be the key point to my problem. A line integral with respect to dx or dy does not result from reparametrizing a line integral with respect to ds. It is a result of a mathematical formula, (like doing the dot product mentioned by galileo). The dx and dy in that equation came about after doing the dot product. It seems that a line integral in terms of dx or dy always result from some mathematical operation, (like the dot product).

I think that you can almost always "Set up" a line integral with respect to arc length, but you can never "set up" a line integral with respect to dx, or dy. You will get a line integral with respect to dx or dy after performing some mathematical steps. This is why it is very difficult to think of any geometric interperation of a line integral with respect to dx or dy.

 

[Hurkyl]

All right, here's a thinker for ya, then -- "set up" an integral for the surface area of a curtain hung from a curve in 3-space.


Incidentally, in 2-space, line integrals wrt dx can generally be reparametrized in terms of dy as well, just like your integrals wrt ds.

 

[Cyrus]

I would have to resort to a 4d curtain hurkyl, one can only wish to see such a thing. A hyper curtain, lovely.

 

[Hurkyl]

No, I was thinking 3-D -- the curtain hangs from the curve in the -z direction.

 

[Cyrus]

Oh, I see. Then the function has to be one such that it is z= f(x,y). and the curve has to be defined as x=x(t); y=y(t); z=z(t). Now the curve does not lie on a plane, but none the less, you can still associate a point z=f(x,y) above the curve. (As long as the curve does not twist over itself, then things to go poopoo :-) .)

 

[Hurkyl]

So anyways, when you get ths integral set up right, you'll find that you're not integrating with respect to the arclength of the curve in 3-d space -- you'll integrate with respect to arclength in the projection on the xy plane. The key lesson I was trying to impart is that not all of the dimensions of the problem are relevant to the quantities you're trying to compute! That's why doing it WRT arclength isn't the "inherently right" way to do things.

 

[Cyrus]

I think I finally found the anwser I was looking for. I asked about line integrals with respect to x and y and why it was different from line integrals with respect to arc length. This was the anwser I was looking for: When you have a line integral with respect to a VECTOR field along a curve, then it can be reduced to three line integrals with with respect to x, y and z that are for a SCALAR function. This is where line integrals with respect o x,y and z come from. I kept asking if they are the of result doing a line integral with respect to arc length, and it turns out they are. Another mistake I was making was the fact that I kept asking about changing the line integral with respect to ds into a line integral with respect to dx or dy in order to solve it. But that was a big no no, because when I changed it to dx or dy, it was no longer a line integral. The integral was not along a curve, but now from end points a to b. I think that was the big thing I did not notice. When you reduce a line integral along a curve C into an integral with respect to x or y, its no longer a line integral. It becomes a regular definite integral. I was confusing this definite integral with that of the line integral with respect to x or y or z. They are two totally different things. It seems that line integrals with respect to x,y and z are the result of line integrals with respect to arclength of vector functions ONLY, Thats what they are and where the come from. But thank for trying to help. I think Galileo nailed it dead on, but I just dident process it until now. Thanks for all your help though dex and hurk.

 

[Hurkyl]

When you reduce a line integral along a curve C into an integral with respect to x or y, its no longer a line integral. It becomes a regular definite integral.

Only for special curves -- remember the example I gave before, where the area of a region R is computed by:

\int_C x \, dy

where C is the boundary of R. This isn't a regular definite integral at all!


Or, for example, the integral \int_C \, ds for the arclength of the unit circle becomes (after some very careful algebra):

\int_C -1/y \, dx

Which, again, is not just a regular definite integral.

 

[Cyrus]

Yes, hurkyl, but \int_C x \, dy does NOT come from the reduction of a line integral with respect to arclength for a scalar valued function, that's my point. It comes from the reduction of a VECTOR valued function when integrating with respect to arclength. Sorry If I did not type it clearly.

Anyways, another question:

In my book they write the following:

\int_c \nabla f * dr = \int^b_a \del f(r(t)) * r&#039;(t) dt
= \int^b_a (\frac{\delta f dx}{\delta x dt} + \frac{\delta f dy}{\delta y dt} + \frac{\delta f dz}{\delta z dt}) dt

I have a little trouble with their notation. If r(t) = x(t)i + y(t)j +z(t)k,

then that means:

\frac{dr}{dt} = \frac{dx}{dt}i + \frac{dy}{dt}j +\frac{dz}{dt}k

or:

dr = ( \frac{dx}{dt}i + \frac{dy}{dt}j +\frac{dz}{dt}k )*dt.

As for the gradient, shouldent It be written as:

\nabla f(r(t)) = \nabla f(x(t),y(t),z(t)) = ( \frac{\delta f}{\delta x} \frac{dx}{dt} + \frac{\delta f}{\delta y} \frac{dy}{dt} +\frac{\delta f}{\delta z} \frac{dz}{dt} )

Wont dotting (*) by the r'(t)dt term just add a redudant dx/dt dy/dt dz/dt to the gradient, no longer making both sides of the expression equal?!?

It looks as if you do the gradient of \nabla f(r(t)) alone is enough to equal \int_C f * dr

 

[Hurkyl]

As for the gradient, shouldent It be written as:

Nope; the gradient is simply the vector whose entries are the partial derivatives with respect to the corresponding argument. What you've written is the (total) derivative of f with respect to t.

 

[Cyrus]

AH, I see. I thought that when you did the gradient you had to do it with respcet to x,y,z and also in terms of t. So would this be correct?

x=t^2; y=t; z= 4t (for example)

and F(x,y,z) = x^2yz^4.

Then when I do the gradient of F, it would be:

2xyz^4i + x^2z^4j + x^2y4z^3k.
And then I would plug in the values of x,y and z in terms of t.
But I would not do the differentiation with respect to x=x(t).
For the (i) direction term I would use: 2xyz^4i = 2(t^2)(t)(4t)^4, instead of doing the derivative with respect to x=x(t) as well:
2(2t)(t)(4t)^4.

 

[Hurkyl]

AH, I see. I thought that when you did the gradient you had to do it with respcet to x,y,z and also in terms of t. So would this be correct?

Yep. The notation \grad F(x(t), y(t), z(t)) means that you take the gradient of F, evaluated at the point (x(t), y(t), z(t)). F itself doesn't care about t. It doesn't care about 'x', 'y', or 'z' either -- recall that

F(x,y,z) = x^2yz^4

means exactly the same thing as

F(p,q,r) = p^2qr^4
F(\alpha,\beta,\gamma) = \alpha^2\beta\gamma^4
and even
F(\spadesuit,\diamondsuit,\clubsuit) = \spadesuit^2\diamondsuit\clubsuit^4

The equation merely defines the function pointwise. In fact, you should strive to use different variables like this when you're confused. (IMHO, you should always do it, or at least until you're very comfortable with this abuse of notation)

 

[Cyrus]

Ive never seen the gradient of a royal flush before...interesting
Thanks for the help, Merry Christmas Hurkyl.

Cheers,

Cyrus Abdollahi

 

[Hurkyl]

They're my favorite TeX symbols. I knew a professor who liked to use smiley-faces for variables in his lectures, though!

 

[Cyrus]

Another question:

They say that:

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr

Now normally, It is written as:

f(x,y) =\int_C F*dr

Is the reason that they used the endpoints as the upper and lower limit because for a conservative vector field it does not matter which curve you use? So they decided to replace the C by simply the end points?

 

[Cyrus]

I think my math book has an error but ill let you guys decide. It trys to prove the conservation of energy based on the gradient of a line integral. I will just quote it below:

"Now let's further assume that F is a conservative force field;that is, we can write F = \nabla f. In physics, the potential energy of an object at the point (x,y,z) is defined as P(x,y,z) = -f(x,y,z), so we have F = - \nabla P. Then by theorem 2 we have:

W= \int_c F*dr = - \int_c \nabla P *dr

= - [P(r(b))-P(r(a))]
= P(A)-P(B)

Comparing this equation with Equation 16, we see that

P(A)+K(A) = P(B) + K(B) "

First point:

Isint this incorrect by definition, isint F =- \nabla f, by definition, where f is the potential function!?

It seems like they considered this when they said that P(x,y,z)= -f(x,y,z),
but when they added ANOTHER negative sign by saying that F =- \nabla P, they just canceled out their correction by using the first negative sign?

Its ok for me to factor out a negative sign if I do the gradient, so let's say I rewrite F= - \nabla P, in terms of the potential energy function f. Then I know that if P(x,y,z)= -f(x,y,z), then its also true that
\nabla P(x,y,z) = - \nabla f(x,y,z). Well then if I sub this back into the line integral, in the end I now get:
F(B)-F(A)

Oh, if I write that in terms of the potential P(x,y,z)= -f(x,y,z), I still get the same anwser. I see now. All this typing for nothing. But does anyone see my point. Isint it STUPID and CONFUSING to add in another damn negative when the definition is F= - \nabla f to start with!? This just makes things overly complicated when they need not be.

 

[Cyrus]

Another question:

They say that:

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr

Now normally, It is written as:

f(x,y) =\int_C F*dr

Is the reason that they used the endpoints as the upper and lower limit because for a conservative vector field it does not matter which curve you use? So they decided to replace the C by simply the end points?

Also, how is this true?

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr

When you evaluate this at the limits won't you get an anwser of:

f(x,y)-f(a,b)?

shouldent it be written as?:

f(x,y) - f(a,b) = \int^{(x,y)}_{(a,b)} F*dr

 

[Hurkyl]

They say that:

What exactly are they saying? This equation could be anything from a constraint on x & y to the definition of f.

Is the reason that they used the endpoints as the upper and lower limit because for a conservative vector field it does not matter which curve you use? So they decided to replace the C by simply the end points?

I have seen that notation used that way before.

 

[Cyrus]

They said, right before that equation, "Let A(a,b) be a fixed point in D. We construct the desired potential function f by defining" ... and then the equation.

But none the less, what ever happened to the evaluation of the lower limit f(a,b)?

 

[Hurkyl]

Why would there be an "evaluation at the lower limit"? This integral is a function of x and y, and their statement is defining f to be that function.


(Incidentally, what is f(a, b)?)

 

[Cyrus]

Let me restate the question Hurkyl. I thought about it and I wonder if this is the reason why. f is a potential function. But the potential of a function is always relative to some fixed point. f(a,b), the lower limit of the integral is that fixed point. This means that f(a,b) is where we define the potential energy to be zero, because when you evaluate the integral f(a,b) to f(a,b) is going to integrate into zero. Similarly, i think when they write f(x,y), they really mean that f(x,y)-f(a,b), but they don't write the f(a,b), because it is implied that this is the potential energery relative to the point at A(a,b). Yes, no?

 

[Hurkyl]

Are you trying to apply the fundamental theorem of calculus? If not, why should there be an f(a,b) on the LHS?

And yes, f(a, b) ... which is the integral of F*dr from (a,b) to (a,b) ... is indeed 0.

 

[Cyrus]

Yes, I was trying to apply the fundamental theorem of calculus for line integrals.

It says that:

\int_c \nabla f*dr = f(r(b))- f(r(a)).

But \nabla f = F

so this is equivalent to saying:

\int_c F*dr = f(r(b)) - f(r(a)).

but in our case we have r(b) = (x,y) so f(r(b)=f(x,y)
and r(a) = (a,b) so f(r(a))= f(a,b)

and then we have

f(x,y) - f(a,b)

but when they write:

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr what happened to -f(a,b) on the left hand side?

 

[krab]

Hi Cyrus: You seem to have gotten confused about the meanings of the symbols you are using. What could you possibly mean by something like r(b)=(x,y)? You seem to be going from a description of C being from (a,b) to (x,y), to one of going from a to b. That's completely different; don't get mixed up!

Go back to your first formula
\int_c \nabla f*dr = f(r(b))- f(r(a))
First, the gradient operator generates a vector, not a scalar. So your asterisk here is inappropriate. It should be a dot product between two vectors. IOW, dr is a vector as well.

Second, What is C? It is some curve in space, parameterized by some variable s. IOW, x=x(s) and y=y(s) so that x and y vary simultaneously as s in order to keep you on the curve C. (Aside: you can do this in some cases by just finding a function y=y(x), but that is not general and in particular does not work in 3D.)

Third, you can choose the start of the curve as s=0 (you are completely free to do so). You seem to use (x,y)=(a,b) as your starting point. That's fine; it just means x(0)=a and y(0)=b. But now be careful as to your end point. If you simply use (x,y), you can get confused because x and y are in a sense your dummy integration variables, and you'll be using them in two completely different senses. So don't do this unless you really know what you are doing.

Let us decide that curve C starts at (a,b) and ends at (u,v). Then we have
\int_C \nabla f\cdot d\vec{r}=f(u,v)-f(a,b)
Remembering that d\vec{r}=(dx,dy), we can show this directly:
\int_C \nabla f\cdot d\vec{r}= \int_C{\partial f\over\partial x}dx+{\partial f\over\partial y}dy=\int_C df=f(u,v)-f(a,b)

Or,
\int_{(a,b)}^{(u,v)} \nabla f\cdot d\vec{r}= f(u,v)-f(a,b)

In your text, they've switched from (u,v) back to (x,y). Technically this is fine, but as I said if they are also using with variables x and y as integration variables, it is confusing.

So we end up with
\int_{(a,b)}^{(x,y)} \nabla f\cdot d\vec{r}= f(x,y)-f(a,b)
So to answer your question, they've dropped f(a,b) because they've set the potentail at (a,b) to be zero. In physics, one is able to do this because none of the real observables that the potential function is used for depend on the value of the potential; they only depend on the difference in the potential at 2 different places, or on the derivatives of the potential. So they've pulled a fast one on you, but it doesn't have any physical effect.

Choosing where f=0 is like choosing where s=0; you are free to make this choice as long as you stick to it once it is made. Come to think of it, they are zero at the same location: f(a,b)=f(x(0),y(0)) with our chosen parameterization variable s.