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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

Tigeree
2006-02-04 12:50:24

a book can't expect anything dizzy

mikau
2006-02-03 14:06:27

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.

irspow
2006-02-03 13:48:12

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.

mikau
2006-02-03 12:27:05

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.

irspow
2006-02-03 08:30:05

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.

mikau
2006-02-03 07:03:52

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?

ryos
2006-02-03 05:53:04

I'll look at it later today. I need to study for my Calc test anyway.

mikau
2006-02-02 18:36:37

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.

mikau
2006-02-02 16:15:36

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?

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