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Exact length of the curve analytically

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    a curve is given by y=(9-x^(2/3))^(3/2) for 1 ≤ x ≤ 8. Find the exact length of the curve analytically by antidifferentiation

    2. Relevant equations

    3. The attempt at a solution
    [tex]\int_a^b \sqrt{1+\frac{dy}{dx}} dx [/tex]
    I use this formula right?

    i took the derivative of the equation and i got \sqrt{9-x^(2/3)} times X^(-1/3)

    How do I integrate it with the negative exponent... I know its probably simple but i'm just not getting it
  2. jcsd
  3. Feb 21, 2010 #2


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    Your equation above is wrong. It should be

    [tex] \int_a^b \sqrt{1 + (\frac{dy}{dx}})^2 dx [/tex]

    When you use this form of the equation for the arc length, you will find a couple of things simplify.
  4. Feb 21, 2010 #3
    ok so now i have it down to integral of sqrt of 1-(9-x^(2/3)/(x^(2/3))

    but i still have a exponent in the denominator
  5. Feb 21, 2010 #4


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    Let me first fix your expression by putting in the "tex" commands:

    That doesn't look right.


    [tex] y = (9 - x^{\frac{2}{3}})^{\frac{3}{2}} [/tex]

    has a derivative that looks like

    [tex] \frac{dy}{dx} = -(9 - x^{\frac{2}{3}})^{\frac{1}{2}} x^{-\frac{1}{3}} [/tex]

    Now when you square that mess so you can put it into your expression for the arc length there will be simplifications that will allow you to do the integration.
  6. Feb 21, 2010 #5
    I got it down to the integral of sqrt of 1+ (27-x)/x

    but i still have an x in the denominator
  7. Feb 21, 2010 #6


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    Science Advisor
    Homework Helper

    It's 1+(9-x^(2/3)/(x^(2/3)), isn't it? Simplify the algebraically and then integrate.
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