Find the length of the curve

  • Thread starter Caldus
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  • #1
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Can't seem to finish this problem:

Find the length of the curve:

[tex]x = 3y^{4/3}-((3/32)y^{2/3})[/tex]

And -343 <= y <= 125.

I used the formula:

[tex]\int_{a}^{b}\ (1 + (x')^{2})^{1/2}dx[/tex]
[tex]\int_{-343}^{125}\ (1+(4y^{1/3}-(1/16)y^{-1/2})^{2})^{1/2}dx[/tex]

But how do you find this integral? Is this even the right integral to use? Thanks.
 

Answers and Replies

  • #2
Tide
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First, you want [itex]\left(y'\right)^2[/itex] under ther radical. Then use the relationship between x and y to find y in terms of x from which you can find y'(x) and you should be able to complete the integration.
 
  • #3
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But isn't it just easier to do it the way I did (since it is given that y is between two values and not x)?
 
  • #4
Zurtex
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Have you tried expanding it out?
 
  • #5
T@P
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can i suggest a calculator?
 
  • #6
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What do you mean by expanding it out?

And the calculator will not give me an answer since 0 is not in the domain of y^(-1/3)...
 
  • #7
Zurtex
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When I said expand out I mean like this:

[tex]\left(1+\left(4y^{\frac{1}{3}}-\frac{1}{16}y^{-\frac{1}{2}}\right)^{2}\right)^{\frac{1}{2}} = \left(1 + 16y^{\frac{2}{3}} - \frac{1}{2}y^{-\frac{1}{6}} + \frac{1}{128}y^{-1}\right)^{\frac{1}{2}}[/tex]

Anyway I messed about with this and found it wasn't intergratable in terms of elementary functions sorry. Which is a bit of a shame as both your limits are cubes so their were some nice substitutions I tried 1st. I tried rearranging it as [itex]y=f(x)[/itex] and found that [itex]1 + y'^2[/itex] wasn't intergratable in terms of elementary functions as well.

Any way it occurred to me while I was coming up with a numerical solution that the function is not continuous at y=0 therefore the integral doesn't exist at y=0
 
  • #8
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So then how do I go about coming up with a numerical solution to it? Am I going to have to use limits?
 
  • #9
Zurtex
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Caldus said:
So then how do I go about coming up with a numerical solution to it? Am I going to have to use limits?
There is no numerical solution, your function isn't continous for [itex]y \leq 0[/tex]

Have you tried getting to plot it out? You'll notice generally for y<0 your function isn't even real.
 
  • #10
HallsofIvy
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It has been suggested repeatedly that you "multiply it out". Did you even try that?

The whole point is this: [itex]\frac{dx}{dy}= 4y^{\frac{1}{3}}-\frac{1}{16}y^{\frac{-1}{3}}[/itex]. so [itex]\(\frac{dx}{dy}\)^2= 16y^{\frac{2}{3}}- 1/2+ \frac{1}{256}y^{\frac{-2}{3}}[/itex].

Now add 1 to that and the only change is that the "-1/2" in the middle becomes "+1/2". What's the square root of that?
 
  • #11
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I can't figure it out. What is the square root of it? Do you try to form a perfect square with it all or what?
 
  • #12
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Oh wait I think I know what you mean now. You just complete the square or whatever and then cancel out the square with the square root. So now I can do the integral. Problem is that I get a negative answer which doesn't make sense for the length of an arc. :\

Guess I'm still doing it wrong. Here's exactly what I did:

[tex]\int_{-343}^{125}\ (1+16y^{\frac{2}{3}}- 1/2+ \frac{1}{256}y^{\frac{-2}{3}})^{1/2}dx[/tex]

[tex]\int_{-343}^{125}\ (16y^{\frac{2}{3}}+ 1/2+ \frac{1}{256}y^{\frac{-2}{3}})^{1/2}dx[/tex]

[tex]\int_{-343}^{125}\ ((4y^{1/3}+(1/16)y^{-1/2})^{2})^{1/2}dx[/tex]

[tex]\int_{-343}^{125}\ 4y^{1/3}+(1/16)y^{-1/2}dx[/tex]

And then from there I got a negative answer and it's incorrect according to this program I'm using. :\
 
  • #13
HallsofIvy
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Well, there is one obvious mistake:

That "[itex]y^{-\frac{1}{2}}[/itex]" should be "[itex]y^{-\frac{1}{3}}[/itex]
 
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  • #14
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Yeah that was a typo. I actually used y^1/3 there. My final answer ended up being -5330.25 which makes no sense. :\
 
  • #15
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How can an arclength be negative? Is that even possible?
 

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