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Find the length of the curve y = x^(2/3) from x = 0 to x = 8.
The formula for arc length is ∫ √( 1 + (f'(x))^2 ) dx (from a to b)
y = x^(2/3) so dy/dx =2/3 x^(-1/3) square this and we have:
4/9 x^(-2/3)
so we've found (f'(x))^2
So the length is:
∫ √( 1 + 4/9 x^(-2/3)) dx from x = 0 to x = 8.
I can't seem to integrate this. My book said in a later lesson they'll show how to integrate expressions such as √(1 + 4x^2) (the arc length of x^2) using trigonometric functions, but in the meantime use an approximation when necessary. But the answer to this problem is an exact answer and not an approximation. So there must be a simple way to integrate without trig functions.
Any idea's?
Last edited by mikau (2006-02-01 17:16:39)
A logarithm is just a misspelled algorithm.
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Well I just had a long chat on AIM with a more advanced calculus student and he couldn't solve it either, and is curious to know how its solved.
A logarithm is just a misspelled algorithm.
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I'll look at it later today. I need to study for my Calc test anyway.
El que pega primero pega dos veces.
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Calculus..yes.....it effects us all...
Our lives revolve around it.
The odd thing is the integral can be written as a simple function. Does that garentee it can be integrated manually?
A logarithm is just a misspelled algorithm.
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I tried this one and I think that it is possible. I did a u substitution and was left with something that seemed to be able to done by integration by parts. However I was completely stumped as to what the limits of integration should be or what to do with the undefined(?) uv part.
I will let you guys off the hook though, my calculus book specifically states that finding arc lengths with the above formula is manually possible only in very rare circumstances. The problems that you work through in the book are specifically chosen because you can integrate them without a computer system.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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This IS a problem in my book, and the answer is a simple expression. Usually these problems are rigged. So there must be a way to do it.
A logarithm is just a misspelled algorithm.
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Ooooh. I didn't think to try trigonomic substitution. Hmm, trigonomic substitution....foggy and dim. I don't think that I have ever used that after learning it in college years ago. If I don't see an answer here for a while, perhaps I will review that material. I am not that interested though, because like I stated above, it hasn't proven very useful in my life.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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Nope, you don't need it. I don't learn trig subs untill a coming lesson. So my book doesn't expect me to know it at this point.
A logarithm is just a misspelled algorithm.
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a book can't expect anything
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
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