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Homework Help: Calculus 3 - Length of Curve

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the length of the parabolic segment r = [itex]\frac{12}{1+cos(θ))}[/itex], 0[itex]\leq[/itex]θ[itex]\leq[/itex][itex]\frac{∏}{2}[/itex].

    2. Relevant equations

    L = ∫[itex]\sqrt{f(θ)^{2}+f^{'}(θ)^{2}}[/itex]dθ

    3. The attempt at a solution

    I got up to here and didn't know whta to do

    L = 12 [itex]∫^{0}_{\frac{∏}{2}}[/itex] [itex]\frac{1}{1+cos(θ)}[/itex][itex]\sqrt{1+(\frac{sin(θ)}{1+cos(θ)})^{2}}[/itex]dθ

    Wolfram alpha doesn't know neither

    [itex]f(θ)^{2}[/itex] = [itex](\frac{12}{1+cos(θ))})^{2}[/itex]
    [itex]f^{'}(θ)^{2}[/itex] = [itex](\frac{12sin(θ)}{(1+cos(θ))^{2}})^{2}[/itex]
    [itex]\sqrt{f(θ)^{2}+f^{'}(θ)^{2}}[/itex] = [itex]\sqrt{(\frac{12}{1+cos(θ))})^{2}+(\frac{12sin(θ)}{(1+cos(θ))^{2}})^{2}}[/itex]=[itex]\frac{12}{1+cos(θ)}\sqrt{1+(\frac{sin(θ)}{1+cos(θ)})^{2}}[/itex]
     
    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Mod note: Fixed incorrect = in the identity below.
    First, simplify the integrand using trig identities such as [itex] \sin^2(\theta) + \cos^2(\theta) = 1. [/itex] After that, Maple manages to do the integral, but I won't report the answer here: you can try again after simplification, substitution, etc. BTW: to get [itex] \int_{0}^{\pi/2}[/itex] you need to type \int_{0}^{\pi/2}, not what you typed.

    RGV
     
    Last edited by a moderator: Feb 6, 2012
  4. Feb 3, 2012 #3
    oh hm ok thanks i got to go ill be back later
     
  5. Feb 4, 2012 #4
    well i tired to simplify it further but keep on getting stuck

    12 square root (2) * integral[0,pi/2] d theta/(sin(theta)+sin(2*theta))

    is as far as i got
     
  6. Feb 4, 2012 #5
    I entered a approximation into the answer box and it told me i was wrong so i guess it wants a exact answer

    here is what it said (see attachment) I'm not sure it really helps

    also note that I tried to evaluate
    12 square root (2) * integral[0,pi/2] d theta/(sin(theta)+sin(2*theta))
    using trig substitution with angle alpha
    cot(alpha)^2/2 = cos(theta)
    I would have to solve for theta
    and then find d theta/ d alpha
    solve for d alpha
    and use the fact that
    cos(alpha)^2/sin(theta) = 1/(sin(theta)+sin(2theta))
    but I'm not sure if this will all work

    i believe that it's equivalent
    that is 12 square root (2) * integral[0,pi/2] d theta/(sin(theta)+sin(2*theta))
     

    Attached Files:

    Last edited: Feb 4, 2012
  7. Feb 6, 2012 #6
    So here is were I'm at now

    L = 12 [itex]∫^{\frac{∏}{2}}_{0}[/itex] [itex]\frac{1}{1+cos(θ)}[/itex][itex]\sqrt{1+(\frac{sin(θ)}{1+cos(θ)})^{2}}[/itex]dθ

    from here I used the fact that
    1 + cos(θ) = 2[itex]cos^{2}[/itex]([itex]\frac{θ}{2}[/itex])
    so
    [itex]\frac{sec^{2}(\frac{θ}{2})}{2}[/itex] = [itex]\frac{1}{1+cos(θ)}[/itex]
    similarly
    [itex]\sqrt{1+(\frac{sin(θ)}{1+cos(θ)})^{2}}[/itex] = [itex]\sqrt{1+\frac{sec^{4}(\frac{θ}{2})sin^{2}(θ)}{4}}[/itex]
    So I know have the following

    L = 12 [itex]∫^{\frac{∏}{2}}_{0}[/itex] [itex]\frac{1}{1+cos(θ)}[/itex][itex]\sqrt{1+(\frac{sin(θ)}{1+cos(θ)})^{2}}[/itex]dθ = 12 [itex]∫^{\frac{∏}{2}}_{0}[/itex] [itex]\frac{sec^{2}(\frac{θ}{2})}{2}[/itex] [itex]\sqrt{1+\frac{sec^{4}(\frac{θ}{2})sin^{2}(θ)}{4}}[/itex]dθ
    from here I use the fact that
    [itex]sin^{2}[/itex](θ)=4[itex]sin^{2}[/itex]([itex]\frac{θ}{2}[/itex])[itex]cos^{2}[/itex]([itex]\frac{θ}{2}[/itex])
    wolfram alpha confirm it's true
    http://www.wolframalpha.com/input/?i=sin(x)^2=4sin(x/2)^2*cos(x/2)^2
    plugging this in I get

    12 [itex]∫^{\frac{∏}{2}}_{0}[/itex] [itex]\frac{sec^{2}(\frac{θ}{2})}{2}[/itex] [itex]\sqrt{1+sec^{4}(\frac{θ}{2})sin^{2}(\frac{θ}{2})cos^{2}(\frac{θ}{2})}[/itex]dθ=6 [itex]∫^{\frac{∏}{2}}_{0}[/itex] [itex]sec^{2}(\frac{θ}{2})[/itex][itex]\sqrt{1+sec^{2}(\frac{θ}{2})sin^{2}(\frac{θ}{2})}[/itex]dθ, u = [itex]\frac{θ}{2}[/itex], [itex]\frac{du}{dθ}[/itex]=[itex]\frac{1}{2}[/itex],du=[itex]\frac{dθ}{2}[/itex],dθ=2du,L=12 [itex]∫^{θ=\frac{∏}{2}}_{θ=0}[/itex] [itex]sec^{2}(u)[/itex][itex]\sqrt{1+sec^{2}(u)sin^{2}(u)}[/itex]du=12 [itex]∫^{θ=\frac{∏}{2}}_{θ=0}[/itex] [itex]sec^{2}(u)[/itex][itex]\sqrt{1+tan^{2}(u)}[/itex]du
    then from here I use
    [itex]sin^{2}(θ)[/itex])+[itex]cos^{2}(θ)[/itex]=1
    that if I divide through by [itex]cos^{2}(θ)[/itex] I get
    [itex]tan^{2}(θ)[/itex])+1=[itex]sec^{2}(θ)[/itex]
    L=12 [itex]∫^{θ=\frac{∏}{2}}_{θ=0}[/itex] [itex]sec^{2}(u)[/itex][itex]\sqrt{sec^{2}(u)}[/itex]du=12[itex]∫^{θ=\frac{∏}{2}}_{θ=0}[/itex] [itex]sec^{3}(u)[/itex]du
    then because
    ∫[itex]sec^{n}(x)dx = \frac{sec^{n-2}(x)tan(x)}{n-1}+\frac{n-2}{n-1}∫sec^{n-2}(x)dx[/itex]
    we got
    12[itex]∫^{θ=\frac{∏}{2}}_{θ=0}[/itex] [itex]sec^{3}(u)du=12[\frac{sec(u)tan(u)}{2}+\frac{1}{2}∫sec(u)du]|^{θ=\frac{∏}{2}}_{θ=0}[/itex]
    and because
    ∫sec(x)dx=ln|sec(x)+tan(x)|+c
    we got
    [itex]12[\frac{sec(u)tan(u)}{2}+\frac{1}{2}∫sec(u)du]|^{θ=\frac{∏}{2}}_{θ=0}=12[\frac{sec(u)tan(u)}{2}+\frac{1}{2}ln(|sec(u)+tan(u)|)]|^{θ=\frac{∏}{2}}_{θ=0}[/itex]
    plugging back in
    u = [itex]\frac{θ}{2}[/itex]
    we got
    [itex]12[\frac{sec(u)tan(u)}{2}+\frac{1}{2}ln(|sec(u)+tan(u)|)]|^{θ=\frac{∏}{2}}_{θ=0}=12[\frac{sec(\frac{θ}{2})tan(\frac{θ}{2})+ln|sec(θ/2)+tan(\frac{θ}{2})|}{2}]|^{θ=\frac{∏}{2}}_{θ=0}=12[\frac{sec(\frac{∏}{4})tan(\frac{∏}{4})+ln(|sec(∏/4)+tan(\frac{∏}{4})|)}{2}-\frac{sec(0)tan(0)+ln(|sec(0)+tan(0)|}{2})=12[\frac{\sqrt{2}+ln(|\sqrt{2}+1|)}{2}-0][/itex]=6([itex]\sqrt{2}[/itex]+ln([itex]\sqrt{2}+1)[/itex])
     
    Last edited: Feb 6, 2012
  8. Feb 6, 2012 #7

    Mark44

    Staff: Mentor

    I haven't worked the problem through, but here is what I got.

    r2 + (r')2 = [tex]\frac{288}{(1 + cosθ)^3}[/tex]

    You still need to take the square root, and then integrate that result. No doubt another trig identity will come in useful.
     
  9. Feb 6, 2012 #8
    I have no idea how you got that. I get

    [itex]\frac{144}{(1+cos(θ))^{2}}[/itex] + [itex]\frac{144(1-cos^{2}(θ))}{(1+cos(θ))^{4}}[/itex] = [itex]\frac{144}{(1+cos(θ))^{2}}(1+[/itex][itex]\frac{1-cos^{2}(θ)}{(1+cos(θ))^{2}}[/itex])

    when I try to expand it
     
    Last edited: Feb 6, 2012
  10. Feb 6, 2012 #9
    well i found one mistake in my work i forgot to square the two lets see...
     
  11. Feb 6, 2012 #10

    Mark44

    Staff: Mentor

    In the first line above, 1 - cos2θ = sin2θ, right?
     
  12. Feb 6, 2012 #11
    yes...
    if you see I believe I had the right answer above... I actually got an answer which is a improvement...
    I got the integral of sec^3(x) dx
    but I get the wrong answer did I use the wrong reduction formula or something =( I think it's the right reduction formula but I'm not sure
     
  13. Feb 6, 2012 #12
    ok well i did use the right reduction formula
    at first i had tan(pi/4)=sqrt(2)/2 but it's one i fixed that and updated my answer above
    I don't see what I've done wrong but apparently it is
    my calculator gives me 13.77735
    and when i put in the answer i got i get 7.4787
     
  14. Feb 6, 2012 #13

    Mark44

    Staff: Mentor

    I checked what I had in wolframalpha, and got 13.7735 (two 7's, not 3), which sort of agrees with your calculator answer.

    You have a ton of stuff in post #6, too much to wade through to find an error.

    The integral should look like this, if my work is correct:
    [tex]12\sqrt{2}\int_0^{\pi/2} \frac{d\theta}{(1 + cos\theta)^{3/2}}[/tex]

    There's a half-angle identity that might be useful here: cosθ + 1 = 2 cos2(θ/2). This turns the integral above into
    [tex]12\sqrt{2}\int_0^{\pi/2} \frac{d\theta}{(2^{3/2}cos^3(\theta/2))}[/tex]

    Then it should be a relatively simple matter of integrating sec3 of something. There's a wikepedia page on this integral.
     
  15. Feb 6, 2012 #14
    ya the only problem is that i don't think there is an error in my work i have checked over it a lot, it's not as long as it seems it's just that I showed each step and my reasoning and any formulas I used, I included the reduction formula for integrating secant
    I got it down to
    12 integral sec^3(u) du

    similar to what you got
    6*sqrt(2) integral sec^3(x)
    interesting enough when i enter this into my calculator with limits i get divide by zero error :O when i try to put in what you got 12*sqrt(2)*integral[0,pi/2] sec^3(x)dx
    and wolfram alpha
    http://www.wolframalpha.com/input/?...t(2)*integral[0,pi/2]+sec^3(x)dx&incTime=true
    couldn't give an approximation or exact answer
    hmm...
     
  16. Feb 6, 2012 #15
    it told me that the correct answer is
    6*(sqrt(2)+ln(1+sqrt(2))
    yay i got it right and found my error yippeeee
     
    Last edited: Feb 6, 2012
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