How is this mathematically rigorous?

[EddiePhys]

In the solution of this problem(121), dW = (kmgcosΦ + mgsinΦ) ds, where ds is the differential element along the curve. Now they have done kmg dscosΦ + mg ds(sinΦ) = kmgdx +mgdy. Makes sense intuitively, but I want to know how this is rigorous. What I'm thinking is, the curve is broken into N elements of the same length along the curve over which a Riemann sum gives a line integral. But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length? And if your explanation is that as Δx tends to zero the two become the same, then why can't we simply treat all differential elements as the same since they all tend to zero after all?

Sorry for the rambling at the end. But can someone just show me why such an operation is rigorous?

 

[jedishrfu]

Isn't $ds*cos(\phi) = dx$ and $ds*sin(\phi) = dy$ ?

You're breaking the $ds$ into components along x and y directions.

 

[EddiePhys]

Isn't $ds*cos(\phi) = dx$ and $ds*sin(\phi) = dy$ ?

You're breaking the $ds$ into components along x and y directions.

the curve is broken into N elements of the same length along the curve over which a Riemann sum gives a line integral. But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length?

 

[Mark44]

What I'm thinking is, the curve is broken into N elements of the same length along the curve over which a Riemann sum gives a line integral. But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length?

No, they wouldn't. In portions of the curve where the angle is less than 45°, $\Delta x$ will be larger than $\Delta y$. In other portions, where the angle is greater than 45°, $\Delta y$ will be larger. The curve is broken up into equal-length segments, with each segment being approximately the hypotenuse of a right triangle. Even though these segments of arc length are the same size, along the curve, that doesn't imply that the horizontal and vertical legs of the triangles will be equal in length.

 

[EddiePhys]

No, they wouldn't. In portions of the curve where the angle is less than 45°, $\Delta x$ will be larger than $\Delta y$. In other portions, where the angle is greater than 45°, $\Delta y$ will be larger. The curve is broken up into equal-length segments, with each segment being approximately the hypotenuse of a right triangle. Even though these segments of arc length are the same size, along the curve, that doesn't imply that the horizontal and vertical legs of the triangles will be equal in length.

I think you misread what I posted. I posted: "But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length?"

Now since the Δx's and Δy's everywhere would vary with Φ, how can we do a Riemann sum along x and y when the interval that we're using itself is varying?

 

[Mark44]

I think you misread what I posted. I posted: "But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length?"

Now since the Δx's and Δy's everywhere would vary with Φ, how can we do a Riemann sum along x and y when the interval that we're using itself is varying?

In a Riemann sum the intervals along the curve don't have to be the same size, but as the number of subintervals increases, the lengths of $\Delta s$, $\Delta x$, and $\Delta y$ grow smaller as well, but again, that doesn't mean they all are the same size.

 

[EddiePhys]

In a Riemann sum the intervals along the curve don't have to be the same size, but as the number of subintervals increases, the lengths of $\Delta s$, $\Delta x$, and $\Delta y$ grow smaller as well, but again, that doesn't mean they all are the same size.

But here if Δx is changing Φ with and we don't know the curve either, we can't do Δx = l/N. How would we evaluate this Riemann sum?

 

[Mark44]

But here if Δx is changing Φ with and we don't know the curve either, we can't do Δx = l/N. How would we evaluate this Riemann sum?

If you know $\Delta s$ and $\phi$, you can get $\Delta x$ and $\Delta y$ -- simple trig.

 

[EddiePhys]

If you know $\Delta s$ and $\phi$, you can get $\Delta x$ and $\Delta y$ -- simple trig.

But we don't know how Φ varies since we don't know the curve

 

[Mark44]

In the picture, there is a force $\vec{F}$ that is always tangent to the curve. That's enough to get you the angle $\phi$, since $\phi$ is the angle between $\vec{F}$ and the horizontal. Although not stated, I believe the assumption is that the direction $\vec{F}$ points would be known at each point along the curve.

 

[EddiePhys]

Although not stated, I believe the assumption is that the direction $\vec{F}$ points would be known at each point along the curve.

How would we know that if we don't know the curve?

 

[Mark44]

How would we know that if we don't know the curve?

Assume that the curve is given by y = f(x), which you can assume to be a differentiable function. At any point (x, f(x)) on the curve, you can find the tangent to the curve, right? I.e., at the point (x, f(x)), the slope of the tangent line is f'(x). From that you can write an expression for the angle $\phi$.

 

[Mark44]

Before going any further, what's your background in mathematics? IOW, what classes have you taken?

 

[EddiePhys]

Before going any further, what's your background in mathematics? IOW, what classes have you taken?

I know single variable calculus from math but I do know a bit about line integrals, and a little bit of vector calculus because of the physics I'm trying to learn.

 

[Mark44]

OK, so do you understand what I write in post #12?

 

[Demystifier]

OK, so do you understand what I write in post #12?

Let me try to anticipate what he will say next: "Yes, but how can I know that f(x) is differentiable if I don't know what f(x) is?"

 

[EddiePhys]

OK, so do you understand what I write in post #12?

We can write f'(x)= tanΦ but we don't know f(x).

 

[Demystifier]

We can write f'(x)= tanΦ but we don't know f(x).

I was close.

You need to think in an abstract way. That allows you to say a lot about things even if you don't know what they are. For instance one grbljh + one grbljh equals two grbljh even if you have no idea what grbljh is.

 

[EddiePhys]

I was close.

Haha :p

 

[Demystifier]

Haha :p

Seriously, the point is that if something is true for any f(x), then you don't need to know what is f(x) to make a rigorous argument about f(x).

 

[Stephen Tashi]

In the solution of this problem(121), dW = (kmgcosΦ + mgsinΦ) ds, where ds is the differential element along the curve.

Makes sense intuitively, but I want to know how this is rigorous.

It isn't mathematically rigorous. Reasoning with "infinitesimals" isn't mathematically rigorous. Reasoning with "differentials" can be made mathematically rigorous in advanced mathematical contexts, but that is not what was done in this problem.

To interpret the answer to the problem in terms of concepts that have a rigorous mathematical definition we need to find an interpretation that is free of infinitesimals. If it makes sense as physics, it should imply a differential equation. What is the differential equation that is implied? In that equation, some symbols represent functions. What variables are the arguments of these functions?

What I'm thinking is, the curve is broken into N elements of the same length along the curve over which a Riemann sum gives a line integral. But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length?

If the integral $\int_a^b F(s) ds$ had a defined value, then every symbol involved in expressing $F(s)$ must be a constant or a function of $s$. In this problem, $\Phi$ is a function of $s$.

Think of computing $\int_a^b F(s) ds$ as approximating the area under the graph of $F(s)$ by using a "regular" Riemann sum (where all rectangles having the same length base "$ds$". This is done on a graph of the function in the $F$-$s$ coordinate system (i.e. Force vs arc length). No $dx$ or $dy$ is involved in those coordinates. The shape of the function $F(s)$ is not the same as the shape of the function $y= f(x)$ that describes the profile of the hill in the $x$-$y$ coordinate system.

The goal of the "infinitesimal reasoning" is to express the graph of $F(s)$ (which is the derivative of $W(s)$ with respect to $s$ , i.e. "$\frac{dW}{ds}$") by using the graph of $y = f(x)$, which is the profile of the hill. To do that, we assume we are "at" $(F,s)$ on the graph of $F(s)$. This corresponds to being at some point $(x,y)$ on the graph of $y = f(x)$. If we "move" to $(F+dF,s+ds)$ on the graph of $F(s)$ then we move to $(x+dx,y+dy)$ on the graph of $y = f(x)$ but we don't assume that using equal increments of $ds$ will produce equal increments of $dx$ and $dy$. Both $dx$ and $dy$ will, in general, be functions of both $s$ and $ds$.

 

[Biker]

You can either think of it as infinitesimal or just think of it as this, If you have two points apart, in order to travel from point a to point b then you have to move a specified dx and dy. Now imagine any pathway that connects between these two points, What should be the total sum you get in the x direction? Just dx. If you move more or less then you "Overshoot". So that integral must be equal to dx and that applies to all functions.

And yes, when you choose ds to be really small and assume that the function is constant over that interval, the dx doesn't have to be equal in each step. You might get negative, postive and even zero but the sum of this all must be equal to the distance between these two points along the x axis.

 

[EddiePhys]

It isn't mathematically rigorous. Reasoning with "infinitesimals" isn't mathematically rigorous. Reasoning with "differentials" can be made mathematically rigorous in advanced mathematical contexts, but that is not what was done in this problem.

To interpret the answer to the problem in terms of concepts that have a rigorous mathematical definition we need to find an interpretation that is free of infinitesimals. If it makes sense as physics, it should imply a differential equation. What is the differential equation that is implied? In that equation, some symbols represent functions. What variables are the arguments of these functions?If the integral $\int_a^b F(s) ds$ had a defined value, then every symbol involved in expressing $F(s)$ must be a constant or a function of $s$. In this problem, $\Phi$ is a function of $s$.

Think of computing $\int_a^b F(s) ds$ as approximating the area under the graph of $F(s)$ by using a "regular" Riemann sum (where all rectangles having the same length base "$ds$". This is done on a graph of the function in the $F$-$s$ coordinate system (i.e. Force vs arc length). No $dx$ or $dy$ is involved in those coordinates. The shape of the function $F(s)$ is not the same as the shape of the function $y= f(x)$ that describes the profile of the hill in the $x$-$y$ coordinate system.

The goal of the "infinitesimal reasoning" is to express the graph of $F(s)$ (which is the derivative of $W(s)$ with respect to $s$ , i.e. "$\frac{dW}{ds}$") by using the graph of $y = f(x)$, which is the profile of the hill. To do that, we assume we are "at" $(F,s)$ on the graph of $F(s)$. This corresponds to being at some point $(x,y)$ on the graph of $y = f(x)$. If we "move" to $(F+dF,s+ds)$ on the graph of $F(s)$ then we move to $(x+dx,y+dy)$ on the graph of $y = f(x)$ but we don't assume that using equal increments of $ds$ will produce equal increments of $dx$ and $dy$. Both $dx$ and $dy$ will, in general, be functions of both $s$ and $ds$.

If $dx$ and $dy$ are functions of $s$ and $ds$, then how is it possible to evaluate the Riemann sum along $x$ or $y$ since we don't know how $\Delta x$ varies with $s$ since we don't know the curve(and hence we don't know the slope at any point and hence the Φ). Also in a Riemann sum we usually take $\Delta x = l/N$. Would this Riemann sum where $\Delta x$ varies with $s$ simply reduce to a normal integral along x $\int{f(x)dx}$ where f(x) is the horizontal component of force. If so, how can we show this mathematically?

Could you solve the above sum rigorously using Riemann sums? It would be really illustrative.

 

[Mark44]

If $dx$ and $dy$ are functions of $s$ and $ds$, then how is it possible to evaluate the Riemann sum along $x$ or $y$ since we don't know how $\Delta x$ varies with $s$ since we don't know the curve(and hence we don't know the slope at any point and hence the Φ).

Did you read what Demystifier wrote in post #20?

Seriously, the point is that if something is true for any f(x), then you don't need to know what is f(x) to make a rigorous argument about f(x).

Also in a Riemann sum we usually take $\Delta x = l/N$. Would this Riemann sum where $\Delta x$ varies with $s$ simply reduce to a normal integral along x $\int{f(x)dx}$ where f(x) is the horizontal component of force. If so, how can we show this mathematically?

Could you solve the above sum rigorously using Riemann sums? It would be really illustrative.

If you're asking us to do your work for you, we can't do that, per the rules of this forum. See https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/, under Homework Guidelines.

 

[EddiePhys]

Did you read what Demystifier wrote in post #20?

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If you're asking us to do your work for you, we can't do that, per the rules of this forum. See https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/, under Homework Guidelines.

This is clearly not a homework problem. I simply want to know why what I'm doing is rigorous and I don't know how to do that myself.

 

[Stephen Tashi]

If $dx$ and $dy$ are functions of $s$ and $ds$, then how is it possible to evaluate the Riemann sum along $x$ or $y$

What Riemann sum are you talking about? As I pointed out, the problem is focused on evaluating $\int_a^b F(s) ds$ in the $F$-$s$ coordinate system. There is no $dx$ or $dy$ involved in the Riemann sum for that integral.

since we don't know how $\Delta x$ varies with $s$ since we don't know the curve(and hence we don't know the slope at any point and hence the Φ).

You don't know the mass "m" either. The solution is based on being given the mass "m" and the function that describes the profile of the hill. Unless a particular curve for the profile of the hill is taken as given, the problem has no defined solution. If you assume the profile of the hill is given by a particular curve $H(x)$ then we can compute the slope at any point.

Also in a Riemann sum we usually take $\Delta x = l/N$.

Yes, that particular type of Riemann sum is a convenient way to think about approximating integrals. But don't forget that the general (modern) definition of "Riemann sum" doesn't require we use approximating rectangles whose bases have equal lengths.

Would this Riemann sum where $\Delta x$ varies with $s$

Be specific about which Riemann sum you are talking about. What function of what variable are you wanting to approximate by a Riemann sum?

simply reduce to a normal integral along x $\int{f(x)dx}$ where f(x) is the horizontal component of force. If so, how can we show this mathematically?

The total work done includes the work done by both the horizontal and vertical components of force. Let the cross section of the hill be given by the function $H(x)$. If you let $f(x)$ be the horizontal component of the force exerted by the person pulling the object at (x,H(x)) then, yes, the total work done by that horizontal force would be $\int_a^b f(x) dx$.

Could you solve the above sum rigorously using Riemann sums? It would be really illustrative.

Which sum is "the above sum"? As I said before, reasoning with infinitesimals is not rigorous mathematical reasoning.

Reasoning intuitively, we can say that the horizontal component of force needed to pull the object at $(x,H(x))$ on the hill can be computed as if the object were sitting on an incline plane whose slope matches the tangent line to $H(x)$ at $(x,H(x))$. The slope of the plane at $(x,H(x))$ is $H'(x)$. Can you find the expression that gives $f(x)$, the horizontal component of the force exerted by the person pulling the object? It doesn't require calculus. It is a typical inclined plane scenario. The only unusual thing is that a typical inclined plane problem would ask for the force the person exerts parallel to the plane to pull the object instead of asking only for the horizontal component of that force.