Finding an arc length, and why isn't latex working for me?

  • Thread starter Stevecgz
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  • #1
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Finding an arc length

I am attempting to find the arc length of y = cuberoot[x] between (1,1) and (8,2).

I solved the integral from 1 to 2 of sqrt[1+(3y^2)^2]dy. I used a formula from a table of integrals in my text to solve this integral. The solution I get is 68.19. I can see that this is not a reasonable answer. Is my setup incorrect or am I solving the integral incorrectly?

Thanks for any help.

Steve
 
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Answers and Replies

  • #2
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Error vv
 
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  • #3
Gokul43201
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The set-up is okay. Looks like you're solving the integral incorrectly. Perhaps, if you show the working, someone will spot the error.
 
  • #4
Fermat
Homework Helper
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Stevecgz said:
Why isn't latex working for me

I thought it wasn't working for me either. But it seems to be not working properly only in the preview pane, when you are composing your thread/response.

When posted, the latex code works properly (at least for me it did :))

You can hit the edit button on your post and edit it there as need be.
 
  • #5
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You're right Fermat, I was only trying it in the preview page, but it's working now.

Thanks Gokul, this is how I got to my answer.

This is the formula I used from my text:

[tex]\int \sqrt {a^2 + u^2} du = \frac u 2 \sqrt {a^2 + u^2} + \frac {a^2} {2} \ln{(u + \sqrt {a^2 + u^2})} + C[/tex]

I used a = 1 and u = 3y^2, and that is how I came up with 68.19.

I'm trying to find a method to solve this integral without using that formula but I am having trouble. I can't see any logical u substitution that would work. I have tried a trigonometric substitution, with [tex]3y^2 = \tan{\theta}[/tex], but when I replace [tex]dy[/tex] with [tex]d\theta[/tex] I'm left with a harder integral than I started with. Thanks again for any help.

Steve
 
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  • #6
Fermat
Homework Helper
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have you tried,

u = a.sinht ?
 
  • #7
Fermat
Homework Helper
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I think the integal you derived

[tex]\int \sqrt {a^2 + u^2}\ du[/tex]

is wrong.

If the integral is,

[tex]\int \sqrt {1 + (3y^2)^2}\ dy[/tex]

then the substitution u = 3y² gives,

[tex]\int \sqrt {1 + u^2}\ dy[/tex]

but

du = 6y dy

or

dy = du/6y = √3.du/(6√u)

which givves the integral as,

[tex]\int \sqrt {1 + u^2}\ \sqrt{3}du/(6\sqrt{u})[/tex]

Also, I put your origianl integral,

[tex]\int \sqrt {1 + (3y^2)^2}\ dy[/tex]

into this http://integrals.wolfram.com/" [Broken] and got back exotic formulas (I think they're called)

I don't know how to evaluate those, sorry :frown:
 
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  • #8
68
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Also, I put your origianl integral,
[tex]\int \sqrt {1 + (3y^2)^2}\ dy[/tex]
into this http://integrals.wolfram.com/" [Broken] and got back exotic formulas (I think they're called)
I don't know how to evaluate those, sorry :frown:

I tried it at a similar site and got back some crazy stuff that didn't mean much to me. :frown:

I'm still not getting anywhere with this, I'm really curious about what I'm doing wrong. I'm sure it's just a simple mistake.

I guess I'll start working on another method and see if I get anywhere trying to find:

[tex]\int_{1}^{8} \sqrt{1 + \left[\frac {1}{3\sqrt[3]{x^2}}\right]^2} dx[/tex]

Steve
 
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  • #9
68
0
Is this integral not elementary? Do I need to use approximate integration?

Steve
 
  • #10
68
0
btt.......
 

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