Arc Length Question

  • Thread starter kevtimc
  • Start date
  • #1
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Homework Statement



y = (2/3) * (x^2 - 1) ^ (3/2) 1 <= x <= 3
Length = ?


Homework Equations



L = [tex]\int\sqrt{1 + (dy/dx)^2} dx[/tex]


The Attempt at a Solution



dy/dx y = (2/3) * (x^2 - 1) ^ (3/2) =
2x * sqrt(x - 1)

Any ideas for a proper substitution? The answer on wolfram seems ridiculous. :bugeye:
 
Last edited:

Answers and Replies

  • #2
299
20
Hmm... it seems wolfram is being ridiculous on this problem. But you don't need a substitution here. Observe that the derivative is 2x√(x² - 1), not 2x√(x - 1). So once you go back and do the arithmetic correctly, you will find that 1 + (dy/dx)² is the square of a polynomial (specifically, it is (2x² - 1)²). After that, this problem should be a piece of cake.
 
  • #3
17
0
Hmm... it seems wolfram is being ridiculous on this problem. But you don't need a substitution here. Observe that the derivative is 2x√(x² - 1), not 2x√(x - 1). So once you go back and do the arithmetic correctly, you will find that 1 + (dy/dx)² is the square of a polynomial (specifically, it is (2x² - 1)²). After that, this problem should be a piece of cake.

I actually knew it was x2, just a typing error. I actually got (2x2 - 1)2
the first time I calculated it, but I didn't get the right answer for the definite integral at 1 through 3 (15.333).

My solution:

(2x)^2 * (x^2 - 1) + 1 = 4x^4 - 4x^2 + 1 = (2x^2 - 1)^2

= (2/3) * x3 -x + C ] (1-3) = 15.333 :rolleyes: Nevermind . . .

Thanks for your help :]
 
Last edited:
  • #4
299
20
Glad you figured it out. It looks like you have the theory right, and were just getting caught up with typos.
 
  • #5
17
0
Glad you figured it out. It looks like you have the theory right, and were just getting caught up with typos.
Story of my mathematical life.
 

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