Expert: Paul Klarreich - 3/13/2007

Question
Hi Paul,

Can you show me how to solve for the arclength of y=2x^(3/2) with x between 0 and (5/3)?

Answer
Questioner:   Jeramie
Category:  Advanced Math
 
Subject:  calculus
Question:  Hi Paul,

Can you show me how to determine the arclength of y=2x^(3/2) with x between 0 and (5/3)?
..........................................
Hi, Jeramie,

The formula for arc length is:

{ b
|   sqrt(1 + (dy/dx)^2 ) dx
} a

If  y = 2 x^(3/2), then

dy/dx = 2(3/2) x^1/2 = 3 sqrt(x)
(dy/dx)^2 = 9x

Your integral is:

{ 5/3
|   sqrt(1 + 9x) dx
} 0

Let  u = 1 + 9x,  du = 9 dx

The integral is now:

{
|   u^1/2 du/9
}

  1  u^3/2
=  -  -----
  9   3/2

  2
=  -  u^3/2
  27

  2
=  -  (1 + 9x)^3/2,  from x = 0 to x = 5/3
  27

= 2/27[ (1 + 15)^3/2  - (1)^3/2 ]

= 2/27[ (16)^3/2  - (1)^3/2 ]

= 2/27[ 64  - 1]
 
= 2/27[ 63]

= 2/3(7) = 14/3,  about 4.67