1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Length of Curve

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the length of the curve between x=0 and x=1. Note: can this be done without a calculator?


    2. Relevant equations
    y = sqrt(4-x^2)



    3. The attempt at a solution

    x=2sin∅
    dx = 2cos∅ d∅


    sqrt(4-4(sin∅)^2) ---> 2cos∅


    integral (0 to 1) sqrt(1+(-2sin∅)^2)

    integral (0 to 1) sqrt(1+4(sin∅)^2) (How do i integrate this without using a calculator?)
     
    Last edited: Mar 8, 2013
  2. jcsd
  3. Mar 8, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    The length of a non-parametrized curve such as yours is given by :

    ##L = \int_{a}^{b} \sqrt{1 + (f')^2} dx ##

    Where f' is the derivative of your function with respect to whatever variable it may be.
     
  4. Mar 8, 2013 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @whatlifeforme: It's too early to think about a substitution before you have the integral set up correctly, as Zondrina points out. Also, where did you get your limits of 0 to 1?
     
  5. Mar 8, 2013 #4
    sorry. it should say Find the length of the curve between x=0 and x=1. .
     
  6. Mar 8, 2013 #5
    is this correct integral for arc length?

    integral (0 to 1) sqrt(1+4(sin∅)^2)


    or does this one look correct?

    integral (0 to 1) sqrt(1+1/(4*(4-x^2))


    either way, they both look hard to integrate.
     
  7. Mar 8, 2013 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi whatlifeforme! :smile:

    i haven't followed what you've done

    start again, writing everything out carefully :wink:
     
  8. Mar 8, 2013 #7

    Zondrina

    User Avatar
    Homework Helper

    Use the formula I gave you.... it really shouldn't be to difficult to compute this.
     
  9. Mar 8, 2013 #8
    so starting over:
    find the length of the curve from x=0 to x=1.

    y=sqrt(4-x^2)

    arc length formula.
    [itex]\displaystyle\int_0^1 {\sqrt{1+F'(x)^2} dx} [/itex]



    [itex]\frac{dy}{dx} = \frac{1}{2\sqrt{4-x^2}} * -2x [/itex]

    [itex]\frac{dy^2}{dx^2} = \frac{x^2}{4-x^2}[/itex]




    [itex]\displaystyle\int_0^1 {\sqrt{1+[(x^2)/(4-x^2)]} dx}[/itex]


    does this look correct so far, any tips to go from here?
     
    Last edited: Mar 8, 2013
  10. Mar 8, 2013 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Sure. What are you waiting for? Simplify it.
     
  11. Mar 8, 2013 #10
    how? common denominator?

    [itex]\displaystyle\int_0^1 {\sqrt{1+[(x^2)/(4-x^2)]} dx}[/itex]


    [itex]\sqrt{\frac{4-x^2+x^2}{4-x^2}}[/itex]

    [itex]\sqrt{\frac{4}{4-x^2}}[/itex]

    [itex]\frac{2}{\sqrt{4-x^2}}[/itex]

    [itex]2\displaystyle\int_0^1 {\frac{dx}{\sqrt{4-x^2}}} [/itex]

    2arcsin(x/2)][itex]^{1}_{0}[/itex]

    2(∏/6) - 2(0) = ∏/3
     
    Last edited: Mar 8, 2013
  12. Mar 8, 2013 #11
    Use the trig substitution x=2sinθ when you get [itex]\int \frac{4}{\sqrt{4-x^2}}dx[/itex]
     
  13. Mar 8, 2013 #12

    that would be the same as recognizing that it is the derivative of arcsin. however, if you were to do the substitution you would still end with the values i have, correct?
     
  14. Mar 8, 2013 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, but 2(∏/6) - 2(0) isn't equal to ∏/2. Typo?
     
  15. Mar 8, 2013 #14

    Bacle2

    User Avatar
    Science Advisor

    How about squaring both sides of the original :

    y=[itex]{\sqrt{4-x^2}}[/itex]


    to see what curve y is ?
     
  16. Mar 8, 2013 #15
    sorry. fixed. it's ∏/3.
     
  17. Mar 8, 2013 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Then it's correct. Nice TeX by the way.
     
  18. Mar 9, 2013 #17
    thanks. i'm trying to get the hang of it.
     
  19. Mar 9, 2013 #18

    Bacle2

    User Avatar
    Science Advisor

    Well, since you already did the full problem, my idea was to use the fact that y is a circle of radius 2, centered at (0,0) , and that (1,√3) is a point corresponding to π/3 radians (easier to see after normalizing by dividing each term by 2 ), and (0,2) corresponds to π/2 radians (there is some ambiguity on wether for x=0 we choose y=2 , or y=-2 ). The length of an arc of (π/2 -π/3) radians in a circle of radius 2 is 2(π/2 -π/3) = π/3 .
     
  20. Mar 9, 2013 #19

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Length of Curve
  1. Length of a Curve (Replies: 5)

  2. Length of a curve (Replies: 1)

  3. Length of a curve? (Replies: 15)

  4. Length of a curve (Replies: 10)

  5. Length of A Curve (Replies: 32)

Loading...