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Homework Help: Find the arc length of the given function

  1. Aug 3, 2005 #1
    This is my last homework problem and I feel that I almost have it solved. The problem is as followed:

    [tex] f(x) = \sqrt{4-x^2} [/tex]

    Find the arc length of the given function from x=0 to x=2.

    I know that I am supposed to use this formula to solve for arclength:

    [tex] \int_{0}^{2} \sqrt{1 + (f\prime(x))^2} dx[/tex]

    Ok so I take the derivative of

    [tex] f(x) = \sqrt{4-x^2} [/tex]

    and I get:

    [tex] f\prime(x) =\frac{-x}{\sqrt{4-x^2}} [/tex]

    I then plug this into the equation get:

    [tex] \int_{0}^{2} \sqrt{1 + (\frac{-x}{\sqrt{4-x^2}}) ^2} [/tex]

    [tex] \int_{0}^{2} \sqrt{1 + \frac{x^2}{4-x^2}} [/tex]

    I get stuck here because I can't integrate this. I can't use any integral tables and I even try some algebraic manipulation inside of the square root and get:

    [tex] \int_{0}^{2} \sqrt{\frac{4}{4-x^2}} [/tex]

    Any help? Thank you.
     
  2. jcsd
  3. Aug 3, 2005 #2
    Well, if [tex]f(x)=\arcsin (x) \Rightarrow f'(x)= \frac {1}{\sqrt{1-x^2}}[/tex]

    Then .....
     
  4. Aug 3, 2005 #3

    robphy

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    Visit integrals.com and enter Sqrt[4/(4-x^2)]
    Don't just take the answer [In addition, remember it's just a computer.]
    ... learn from the answer to figure out how you could have gotten it yourself.

    The answer has an (x/2) in it. Hmmm. maybe a substitution or two is in order.
     
  5. Aug 3, 2005 #4
    you can break it down into partial fractions
     
  6. Aug 3, 2005 #5
    i second the inverse trig antiderivative!
     
  7. Aug 3, 2005 #6
    yeah...but then it's a problem of looking up integration tables again, and not necessarily finding anything, either.
     
  8. Aug 3, 2005 #7
    Alright I think I got the solution. Is this right?

    [tex] \int_{0}^{2} \sqrt{\frac{4}{4-x^2}} [/tex]

    [tex] \int_{0}^{2} \frac{\sqrt{4}}{\sqrt{4-x^2}} [/tex]

    [tex] 2\int_{0}^{2} \frac{1}{\sqrt{4-x^2}} [/tex]

    then using an integral table.....

    [tex] 2\arcsin{\frac{x}{2}} |_{0}^{2} [/tex]

    [tex] 2\arcsin{1} [/tex]

    Does that look right?
     
  9. Aug 4, 2005 #8
    well done iggy
    well done
     
  10. Aug 4, 2005 #9

    HallsofIvy

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    Tninking in terms of formulas or looking things up in a table, you miss most of what's interesting about this problem.

    Looking at something like [tex] 2\int_{0}^{2} \frac{dx}{\sqrt{4-x^2}} [/tex] (DON'T forget the "dx"), the first thing that should come to mind is a trig substitution (I hate integral tables!). Let x= 2sin(θ) so that dx= 2 cos(θ) and 4- x2= 4- 4sin2(θ)= 2(1- sin2(θ))= 2 cos(θ). When x= 0, 0= 2sin(θ) so θ= 0. When x= 2, 2= 2sin(θ) so θ= [tex]\frac{\pi}{2}[/tex]. The integral becomes [tex]2\int_0^{\frac{\pi}{2}}d\theta[/tex]. That's pretty easy, isn't it!

    Yet another way to do this is to put the formula in parametric equations. [tex]y= \sqrt{4- x^2} [/tex] is (almost) the same thing as y2= 4- x2 or x2+ y2= 4, a circle of radius 2.
    Parametric equations for that are x= 2cos(t), y= 2 sin(t) where t ranges from 0 to [tex]\frac{pi]{2}[/tex]. [tex]ds= \sqrt{(\frac{dx}{dt})^2+ (\frac{dy}{dt})^2} = 2dt[/tex]. x goes from 0 to 2 as t goes from 0 to [tex]\frac{\pi}{2}[/tex] so the arc length is [tex]2 \int_0^{\frac{\pi}{2}}dt[/tex] which is also pretty darn easy!

    Finally, it should be obvious that you are talking about 1/4 of a circle of radius 2. Find the circumference and divide by 4.
     
    Last edited by a moderator: Aug 4, 2005
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