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Calculating Distance Traveled (Arc Length)

  1. Oct 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Given y = 156 - (x - 40)2/60. x = 0 and x = 85 Find distance traveled

    2. Relevant equations

    Arc Length S = integral of square root of ( y' )2

    3. The attempt at a solution

    Doing this I get the trig sub tan(t) = (x - 40)/ 30 (Told by teacher to use this instead of -(x - 40)/30 )
    so x = 30tan(t) + 40
    and dx = 30sec2(t)

    So then under the radical I get 1 + ( - (tan (t) )2
    So is this equal to the square root of sec2(t)?

    Assuming it is, I get the integral of sec2(t) times 30sec2(t) or sec3(t)

    I know how to do this (I think) but I'm really not sure (I can't use what's written in the back of my book).

    Doing the work I get:

    S(x) = 15sec(t)tan(t) + (1/2) ln[ (sec(t) + tan(t) ]
    Substitution x back in for t gives

    tan(t) = (x-40)/30
    and
    sec(t) = square root [ 1 + ( ( -x + 40)/(30) )2

    But evaluating this from 0 to 85 gives me an arc length of ~ 75.05, which I know can't be right, but I can't figure out where I went wrong.
     
  2. jcsd
  3. Oct 12, 2010 #2

    cronxeh

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    Isn't it S= integral (sqrt( 1+ (y')^2)) ?

    y' = (40-x)/30
    (y')^2 = ((40-x)^2)/900
    integral sqrt(1+((40-x)^2)/900), x=0..85 ~ 108.296

    As a matter of fact you could've estimated this distance by taking mid-way point (in this case its an inflection point (x-40)/30 = 0 -> x=40

    at x=40, y=156
    at x=0, y=129.3
    at x=85, y=122.25

    As you can see you can draw 2 lines from x=0 to x=40, and from x=40 to x=85, then add up the distances to get your approximate arc length:

    sqrt((0-40)^2+(156-129.3)^2) + sqrt((85-40)^2 + (156-122.25)^2) ~ 104.34
     
    Last edited: Oct 12, 2010
  4. Oct 12, 2010 #3
    What equation did you use to evaluate s(x)?

    Once I redid/rechecked everything I got

    s(x) = 15 [ sqrt [ 1 + ( ( -x + 40)/(30 ) )2 ] * ( ( x - 40 ) / (30) ) + ln [ sqrt [ 1 + ( ( -x + 40)/(30 ) )2 + ( ( x - 40 ) / (30) ) ]
     
  5. Oct 12, 2010 #4

    cronxeh

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    [tex]S = \int^{85}_{0} \sqrt{1+\frac{(40-x)^{2}}{900}} dx[/tex]
     
  6. Oct 12, 2010 #5
    I mean after all the trig sub, the final equation that you substitute 85 and 0 into.
     
  7. Oct 12, 2010 #6

    cronxeh

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    1/60 sqrt((x-40)^2+900) (x-40)+15 sinh^(-1)((x-40)/30)
     
  8. Oct 12, 2010 #7
    We haven't learned about hyperbolic functions yet. Is there a way to do it without using sinh, i.e. by using trig sub and integration by parts.
     
  9. Oct 12, 2010 #8

    cronxeh

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    Sure, its just what you did but with the correct limits, which I think you should double check

    Let u=40-x, du=-dx, now do the limits here

    integral -sqrt(1+(u^2)/900) du

    then substitute u=30*tan(s), du=30*sec^2(s)ds

    = integral -sqrt(tan^2(s)+1) =
    where s=arcan(u/30)

    integral -30*sec^3(s)ds =
    -15*tan(s)sec(s) - 15*integral(sec(s)ds)
    = -15tan(s)sec(s) - 15*ln(tan(s)+sec(s))
    sub back s=arctan(u/30) and u=40-x:

    1/60*(x-40)*sqrt(x^2-80*x+2500) - 15*ln(1/30*(sqrt(x^2-80*x+2500)-x+40))
    =58.4839-(-49.8125) = 108.2964
     
  10. Oct 13, 2010 #9
    For some reason when I plug in the numbers I get an arc length of 118.693625.

    Also, where did the 1/60 and 1/30 come from? It seems like you were plugging in ( 40 - x) / 900 for tan(s) instead of ( 40 - x ) /30
     
  11. Oct 13, 2010 #10

    cronxeh

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    = -15tan(s)sec(s) - 15*ln(tan(s)+sec(s))
    sub back s=arctan(u/30) and u=40-x:

    If you agree with that line, then watch the magic:

    tan(arctan(x)) = x
    cos(arctan(x)) = 1/sqrt(x^2+1)
    sec(arctan(x)) = sqrt(x^2+1)
    s=arctan((40-x)/30)

    -15tan(s)sec(s) - 15*ln(tan(s)+sec(s)) =
    -15*ln(tan(s) + 1/cos(s)) - (15*tan(s))/cos(s) =
    -15*((40-x)/30)*sqrt(((40-x)/30)^2+1) - 15*ln(((40-x)/30)+sqrt(((40-x)/30)^2+1)) =
    (x/2 - 20)*((x/30 - 4/3)^2 + 1)^(1/2) - 15*ln(((x/30 - 4/3)^2 + 1)^(1/2) - x/30 + 4/3)
    Plug in your limits x=85, x=0:

    58.4839 - (-49.8125) = 108.2964

    If you are not a believer in the answer, consider that we forget calculus for a second and just approximate it:

    Given y = 156-(x-40)^2/60, lets do 10 intervals from x=0 to x=85:
    distance travelled is then ( in matlab ):
    Code (Text):

    x=0:85
    d=0;
    for i=1:85
    d=d+sqrt(((156-(x(i)-40)^2/60) - (156-(x(i+1)-40)^2/60))^2 + (x(i)-x(i+1))^2);
    end
    d
     
    d =

    108.2942

    If you want me to go more accurate,

    Code (Text):

    x=0:0.1:85
    d=0;
    for i=1:850
    d=d+sqrt(((156-(x(i)-40)^2/60) - (156-(x(i+1)-40)^2/60))^2 + (x(i)-x(i+1))^2);
    end
    d
     
    d =

    108.2964
     
    Last edited: Oct 13, 2010
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