Arc Length Problem | Find Solution

In summary: But again, I don't see how that helps in terms of getting the equation back into the integrand...If this is homework, I'd suggest going to a TA or your professor and getting help walking through the integral. This will be a rather long integral, and I can't think of any ways to simplify it off the top of my head. It'd be better to have help in person, since leading you through every step on the forums is going to get arduous.Yea... my professor is apparently sick, so I was hoping to get some help from my fellow students. Only one has even tried, which brought me here.
  • #1
ohms law
70
0
Hi folks. I'm working on an arc length problem, and my answer doesn't match the answer that our professor gave, so I was hoping that someone could point out where I'm going wrong here.

Homework Statement


Find the arc length determined by x^2=y^3 from (0,0) to (1,1).all of the work is in the attached Word doc. I realize that I'm severely limiting the help that I'll receive by using an attachment, but copying all of that formatted work to bbcode (I assume) formatting isn't exactly trivial, so... *shrug* I'd very much appreciate any help or discussion about this, though.

Edit: I had to .zip the file because .docx isn't a recognized format here. Sorry again for that.
 

Attachments

  • Review Questions - Problem 5.zip
    46.1 KB · Views: 157
Physics news on Phys.org
  • #2
ohms law said:
Hi folks. I'm working on an arc length problem, and my answer doesn't match the answer that our professor gave, so I was hoping that someone could point out where I'm going wrong here.

Homework Statement


Find the arc length determined by x^2=y^3 from (0,0) to (1,1).

all of the work is in the attached Word doc. I realize that I'm severely limiting the help that I'll receive by using an attachment, but copying all of that formatted work to bbcode (I assume) formatting isn't exactly trivial, so... *shrug* I'd very much appreciate any help or discussion about this, though.

Edit: I had to .zip the file because .docx isn't a recognized format here. Sorry again for that.
I assume that unzipping that file gives a .docx Word file back, which many of us will not open.

Can you at least output your document to a pdf file and post that ?
 
  • #3
sure. uh... here you go.
 

Attachments

  • Review Questions - Problem 5.pdf
    502 KB · Views: 195
  • #4
It appears that you've incorrectly done u substitution here. In this step --

[tex] s = \int_{0}^{1} u^{1/2} du (u = 1 + 4/9 x^{-2/3}) [/tex]

-- you've implicitly assumed that the derivative of u is present in the integrand before substitution, but it isn't. When [itex] u = 1 + 4/9 x^{-2/3} [/itex], [itex] du = -8/27 x^{-5/3}dx [/itex], but that term isn't present in the unsubstituted integral. What you'll have to do is solve for dx in terms of u and du and substitute that into the integrand. It might help if you simply define [itex] u = x^{-2/3} [/itex] instead of the entire argument of the square root. You'll also probably have to do another round of u substitution and maybe use some trigonometric functions as well. Don't forget to solve for the new boundaries of integration. This is a pretty nasty integral.
 
  • #5
yea, I was kindof suspisious about that u substitution...
Thanks. I'll start working at it again.
 
  • #6
humm... ok, I'm not sure that u substitution is going to help here at all.
If [itex] u = x^{-2/3} [/itex], then [itex] u' = {2/3}x^{-5/3} [/itex], so... that doesn't really help anything.
 
  • #7
Just so people don't have to go hunting through this whole thread:
The problem is how to integrate

[itex] s = \int_{0}^{1} √(1+{4/9}x^{-2/3}) dx [/itex]
 
  • #8
ohms law said:
humm... ok, I'm not sure that u substitution is going to help here at all.
If [itex] u = x^{-2/3} [/itex], then [itex] u' = {2/3}x^{-5/3} [/itex], so... that doesn't really help anything.

You have [itex] du = -2/3 x^{-5/3} dx [/itex] and [itex] u = x^{-2/3} [/itex] which means [itex] x = u^{-3/2} [/itex]. Substitute u in for x in the equation for du and then solve for dx in terms of u and du. Then you'll have an integral that you can start to work with.

If this is homework, I'd suggest going to a TA or your professor and getting help walking through the integral. This will be a rather long integral, and I can't think of any ways to simplify it off the top of my head. It'd be better to have help in person, since leading you through every step on the forums is going to get arduous.
 
  • #9
Yea... my professor is apparently sick, so I was hoping to get some help from my fellow students. Only one has even tried, which brought me here.

This helps though, so thanks. I need to go eat something, and I'll come back and look at it again.
 
  • #10
ok, let's see... looking at these fractional exponents is making my eyes swim, but I'm working at it.

I don't see how [itex] x = u^{-3/2} [/itex] helps. I staired at it long enough (a few seconds, you know?) to see what you're getting at there, but I'm looking at the integral and I don't see how it fits back in there. I went back and did a couple of substitution examples too, just to make sure that I have that concept down. I don't know, if there's something there substitution wise (with that u), I'm just not quite seeing it.

One thing that I'm thinking about right now is simplifying out the square root by converting it to an exponent and distributing it, so that the integral looks like:
[itex]s=\int_{0}^{1} 1+{2/3}x^{-1/3} dx [/itex] (I'm fairly certain, at least... that took a little work). However, that basically leaves me in the same place with a slightly different set of products... if [itex]f(x)=x^{-1/3}[/itex], then [itex]f'(x)={-1/3}x^{-4/3}[/itex], so [itex]u=x^{-1/3}[/itex] and [itex]du={-1/3}x^{-4/3}dx[/itex]... same place, basically.
 
  • #11
ohms law said:
I don't see how [itex] x = u^{-3/2} [/itex] helps. I staired at it long enough (a few seconds, you know?) to see what you're getting at there, but I'm looking at the integral and I don't see how it fits back in there. I went back and did a couple of substitution examples too, just to make sure that I have that concept down. I don't know, if there's something there substitution wise (with that u), I'm just not quite seeing it.

With [itex] x = u^{-3/2} [/itex] (and remembering [itex]du = -2/3 x^{-5/3} dx[/itex]), then we can substitute [itex] u^{-3/2} [/itex] for x into [itex]du = -2/3 x^{-5/3} dx[/itex], obtaining [itex] du = -2/3 u^{5/2} dx[/itex]. We can solve for an expression of dx in terms of u and du, and plug that expression in for dx into the integral. We then obtain an expression like [itex] \int_{0}^{1} \sqrt{1+4/9u}*(-3/2 u^{-5/2})du [/itex]. Do you see my point now?

In any case, I've found an easier method (so you can essentially ignore everything we've done up to this point): try taking the integral with respect to y. Solve for dx/dy -- the argument of the square root turns into a simple u substitution this way.
 
  • #12
In any case, I've found an easier method (so you can essentially ignore everything we've done up to this point): try taking the integral with respect to y. Solve for dx/dy -- the argument of the square root turns into a simple u substitution this way.

Of course! *facepalm*
OK, I'll work on that.

Anyway,
jmcelve said:
With [itex] x = u^{-3/2} [/itex] (and remembering [itex]du = -2/3 x^{-5/3} dx[/itex]), then we can substitute [itex] u^{-3/2} [/itex] for x into [itex]du = -2/3 x^{-5/3} dx[/itex], obtaining [itex] du = -2/3 u^{5/2} dx[/itex]. We can solve for an expression of dx in terms of u and du, and plug that expression in for dx into the integral. We then obtain an expression like [itex] \int_{0}^{1} \sqrt{1+4/9u}*(-3/2 u^{-5/2})du [/itex]. Do you see my point now?
...yea, sorta. I'm getting a little burnt out (been at this all day, and this isn't the only problem that I've been working), so I think that I'll set this aside for the night and approach it fresh in the morning. However, just from looking at it, I'm missing how: "obtaining [itex] du = -2/3 u^{5/2} dx[/itex]" is happening, for some reason. It's probably obvious, but right now it doesn't seem... correct (not that I'm doubting you, you understand).

I (very much) appreciate all of the help so far, though!
 

Question 1: What is an arc length problem?

An arc length problem is a mathematical problem that involves finding the length of a curved line, also known as an arc, within a given shape or curve. It is commonly encountered in geometry and trigonometry.

Question 2: How do I solve an arc length problem?

To solve an arc length problem, you need to know the radius of the curve, the angle of the arc, and the formula for calculating arc length, which is L = rθ, where L is the arc length, r is the radius, and θ is the angle in radians. Plug in the given values and solve for L.

Question 3: What is the difference between arc length and arc measure?

Arc length is the actual distance along the curve, while arc measure is the size of the angle that the arc subtends at the center of the circle. Arc length is measured in units of length, while arc measure is measured in units of degrees or radians.

Question 4: Can I use the same formula for any type of curve?

No, the formula for calculating arc length, L = rθ, only applies to circles and circular arcs. For other types of curves, such as parabolas or ellipses, different formulas will be needed to calculate arc length.

Question 5: How is arc length used in real-world applications?

Arc length is used in many fields, such as engineering, architecture, and physics. It can be used to calculate the distance traveled by a moving object along a curved path, the length of a wire needed to form a specific shape, or the size of a curved section of a bridge. It is also used in the design and construction of circular objects, such as wheels and gears.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
3
Views
1K
Back
Top