Looking for intuition on differentials and arc length

In summary, the conversation discusses the concept of the differential of an arc length and its relation to the Pythagorean theorem. It also explores the notation and notation changes in finding the length of a curve. The conversation concludes with a discussion on how to apply the concept to finding the length of a curve without a given formula.
  • #1
opus
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Please see the attached image. It's a part of a study guide for my final, but I didn't put it in the homework section because I already got the answer, I just don't know what it means.
The question has to do with the differential of an arc length. I made some drawings to see if I could make some sense out of it but I don't think I'm quite there.

So when we're talking about the differential of an arc length, what exactly does this mean?
Screen Shot 2018-12-09 at 7.06.17 PM.png
 

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  • #2
##d\vec s=dx~\hat i+dy~\hat j##
##ds =\sqrt{d\vec s \cdot d\vec s}=\sqrt{(dx)^2+(dy)^2}=dx\sqrt{1+(\frac{dy}{dx})^2}##
What's ##(\frac{dy}{dx})^2~?##
 
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  • #3
Apologies, but I don't understand the notation. This is for Calculus I if that gives an idea. It does look like you're doing something with Pythagorean's though!
 
  • #4
Sorry, I thought you understood vectors. Yes, it has to do with the Pythagorean theorem. A two dimensional displacement ##ds## has a horizontal part ##dx## and a vertical part ##dy##. The two form a right triangle and ##ds=\sqrt{(dx)^2+(dy)^2}## is the hypotenuse.
 
  • #5
Ok that's much more clear, thanks!
So then, since our ##dx## is extremely small, we can pretty much say that the curve of the graph is a straight line in that ##dx##.
This would give us a right triangle with the base as ##dx##, the hypotenuse as ##ds##, and the side length as what? It would have to be the change in y values corresponding the the change in x values (##dx##), but in terms of notation, what would this be?
 
  • #6
Look at post #2. If ##ds =dx\sqrt{1+(\frac{dy}{dx})^2}##, what is ##ds## when ##y=2\sqrt{5x+1}?##
 
  • #7
##ds=dx\sqrt{\frac{5x+26}{5x+1}}##?
 
  • #8
Right. Does this answer your original question or is there more to it?
 
  • #9
I guess I don't understand what the question is really asking me, so I suppose there's more to it. In words, how would you ask the given question to me? For example, "here is a function, what is it doing from here to here?"
 
  • #10
The question is asking you to find ##ds## if ##y=2\sqrt{5x+1}## and you have done that. How would you approach the following problem:
You drive your car from x = 10 units to x = 40 units along the curve given by ##y=2\sqrt{5x+1}##. How much distance has been added to your car's odometer?
 
  • #11
The distance added to the odometer would be equal to the length of the curve, which is Arc Length ##= \int_{10}^{40} \sqrt{\frac{5x+26}{5x+1}}dx## which is a specific integral to the more general ##s(x)## or ##ds## too. I guess this notation is throwing me for barrel rolls. It keeps changing and I think it's distracting me from the picture.
 
  • #12
opus said:
The distance added to the odometer would be equal to the length of the curve, which is Arc Length ##= \int_{10}^{40} \sqrt{\frac{5x+26}{5x+1}}dx## which is a specific integral to the more general ##s(x)## or ##ds## too. I guess this notation is throwing me for barrel rolls. It keeps changing and I think it's distracting me from the picture.
It's standard notation, it's part of "the picture" and eventually you will get used to it (I hope).
 
  • #13
So then ##ds## is saying the change in the function ##s(x)## which is a function that gives the arc length of the function ##y##. Why are we putting it in terms of ##ds##? Why are we concerned with change at all when we are looking for the length of a curve? We had a perfectly good way to find the length when we had the formula ##s(x) = \int_{0}^{x}\sqrt{\frac{5t+26}{5t+1}}dt## What's the difference?
 
  • #14
opus said:
So then ##ds## is saying the change in the function ##s(x)## which is a function that gives the arc length of the function ##y##. Why are we putting it in terms of ##ds##? Why are we concerned with change at all when we are looking for the length of a curve? We had a perfectly good way to find the length when we had the formula ##s(x) = \int_{0}^{x}\sqrt{\frac{5t+26}{5t+1}}dt## What's the difference?
When you get to differential geometry, several basic formulas are expressed in terms of ds.

Also, check out https://en.wikipedia.org/wiki/Arc_length.
 
  • #15
opus said:
So then ##ds## is saying the change in the function ##s(x)## which is a function that gives the arc length of the function ##y##. Why are we putting it in terms of ##ds##? Why are we concerned with change at all when we are looking for the length of a curve? We had a perfectly good way to find the length when we had the formula ##s(x) = \int_{0}^{x}\sqrt{\frac{5t+26}{5t+1}}dt## What's the difference?
Suppose someone asked you to find the length of the curve represented by the polynomial ##y=x^4+3x^3+6x## from ##x=0## to ##x=8##. What then? No formula is given to you and clearly you can't use ##s(x) = \int_{0}^{x}\sqrt{\frac{5t+26}{5t+1}}dt##. This example shows how to derive and apply the appropriate formula.
 
  • #16
opus said:
This is for Calculus I if that gives an idea. It does look like you're doing something with Pythagorean's though!
Exactly.
opus said:
So then ##ds## is saying the change in the function ##s(x)## which is a function that gives the arc length of the function ##y##. Why are we putting it in terms of ##ds##? Why are we concerned with change at all when we are looking for the length of a curve? We had a perfectly good way to find the length when we had the formula ##s(x) = \int_{0}^{x}\sqrt{\frac{5t+26}{5t+1}}dt## What's the difference?
ds represents an infinitesimal change in the length along the curve. This length depends on both dx and dy, not just dx.
The formula for s(x) that you show is not a general formula for arc length. This integral comes from the general formula for arc length, in which ds is defined as the hypotenuse on a right triangle whose legs are dx and dy.

Here's a picture.
arclength.png

So ##ds^2 = dx^s + dy^2##, or ##ds = \sqrt{dx^s + dy^2}##. This latter equation can take different forms, using algebraic manipulations. One form is ##ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}~dx##, if y is given as a function of x. Another form has ##1 + \left( \frac{dx}{dy}\right)^2## inside the radical, for when x is given as a function of y.
 

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  • #17
The differential is the approximation to the change of the function along the tangent line ( Or, in higher dimension the tangent plane, etc.).
 

Related to Looking for intuition on differentials and arc length

1. What is the purpose of studying differentials and arc length?

The study of differentials and arc length is important in mathematics and physics because it allows us to understand and quantify changes in quantities such as position, velocity, and acceleration. This is especially useful in fields such as calculus, mechanics, and geometry.

2. What is a differential?

A differential is the infinitesimally small change in a variable, typically denoted by "dx" or "dy". It represents the slope of a tangent line to a curve at a specific point. Differentials are used in calculus to calculate rates of change and to approximate solutions to problems.

3. How is arc length calculated?

Arc length is the length of a curved line, and it can be calculated using integrals in calculus. The formula for arc length is L = ∫(√(1+(dy/dx)^2))dx, where dy/dx represents the slope of the curve at a given point. This formula can be modified for polar coordinates or parametric equations as well.

4. What is the relationship between differentials and arc length?

Differentials and arc length are related because the differential of a function f(x) is equal to the arc length formula for that function. This means that infinitesimal changes in a variable can be used to calculate the length of a curve, making it a useful tool in solving problems involving arc length.

5. How are differentials and arc length used in real-world applications?

The concepts of differentials and arc length have many practical applications in fields such as engineering, physics, and computer graphics. They are used to optimize designs, calculate trajectories of objects, and create realistic computer-generated images. They are also used in financial mathematics to model and predict changes in stock prices and interest rates.

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