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Math Help - Parametric Arc Length

  1. #16
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    Quote Originally Posted by MechEng View Post
    Is this something specific to parametric equations? I don't think I have seen this sort of shift before. I have another problem that is giving me the same result. I start off with:

    x=a(t-\sin t)

    y=a(1-\cos t)

    \frac{dx}{dt}=a(1-\cos t)

    \frac{dy}{dt}=a(\sin t)

    \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=a\sqrt{  2-2\cos t}

    Plugging that into my surface are formula yields:

    \displaystyle s=2\pi\int_{0}^{2\pi}{a^2(1-\cos t)|\sqrt{2-2\cos t}|dt

    s=2\sqrt{2}\pi a^2\int_{0}^{2\pi}{(1-\cos t)^{3/2}}dt

    Now, to finish this up I would like to substitute:

    u=1-\cos t

    du=\sin t dt

    If I understand this correctly... I want to change my terminals of integration to ensure that I do not try to "add" a negative portion of the curve to a positive portion. How does the logic work for this?

    You must identify x-axis (t-axis) "crossing points", not just where the graph is zero.
    Remember, integration "sums", hence it will sum negatives to an overall negative.
    Thus we need to change the sign of the integration for parts of the graph under the horizontal axis.

    2\pi is one complete revolution. During this revolution the function \cos t crosses the x-axis twice. Is there some rule of thumb for ensuring that I am always dealing with a positive value?

    Either graphically or by calculation, find the zeros of the function being integrated
    and discover if the function crosses the horizontal axis there.
    You need to locate where the function changes sign.


    I think a value of \frac{\pi}{2} would accomplish what I want.
    That substitution gives you a sint to contend with.

    Instead.. 2sin^2\left(\frac{t}{2}\right)=1-cost

    Then you can use u=\frac{t}{2}

    This graph does not cross the x-axis as the maximum value of cost is 1.


    The notation you asked about....
    That graph crosses the x-axis every 90 degrees,
    however it is symmetrical about the x-axis.
    As you are integrating from 0 to 360 degrees,
    we calculate 4 times the integral from 0 to 90 degrees.

    sintcost=0\ for\ t=0^o,\ 90^0,\ 180^o,\ 270^o,\ 360^o

    Also we have negative f(t) for \frac{{\pi}}{2}<t<{\pi} and \frac{3{\pi}}{2}<t<2{\pi}

    f(-t)=-f(t)
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  2. #17
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    Ok, so if I set (1-\cos t)=0 and solve for t, I should be able to see where y=0, therefore where the graph is either equal to zero or crossing the x-axis. From there, I would need to examine two points on either side of the x value that produces a y=0.

    Is that the general idea? I see that you made a point of saying "not just where the graph is zero", so I assume that it would not hurt to include those points in my evaluation. Is there a faster method to examine this?

    Quote Originally Posted by Archie Meade View Post
    That substitution gives you a sint to contend with.

    Instead.. 2sin^2\left(\frac{t}{2}\right)=1-cost

    Then you can use u=\frac{t}{2}

    This graph does not cross the x-axis as the maximum value of cost is 1.


    The notation you asked about....
    That graph crosses the x-axis every 90 degrees,
    however it is symmetrical about the x-axis.
    As you are integrating from 0 to 360 degrees,
    we calculate 4 times the integral from 0 to 90 degrees.

    sintcost=0\ for\ t=0^o,\ 90^0,\ 180^o,\ 270^o,\ 360^o

    Also we have negative f(t) for \frac{{\pi}}{2}<t<{\pi} and \frac{3{\pi}}{2}<t<2{\pi}

    f(-t)=-f(t)
    If I use the substitution:

    2\sin^2(\frac{t}{2})=(1-\cos t) from \sin^2 t = \frac{1-\cos 2t}{2}

    And then...

    u=\frac{t}{2}

    du=\frac{1}{2}dt

    I find that I get the following (coefficients omitted):

    \displaystyle\int_{0}^{\pi}\frac{(2 \sin^2 u)^{3/2}}{2}}dt

    What do I do with this? If my 2 \sin^2 u was not being raised to the 3/2 power, it would be pretty easy. But I'm not sure where this is supposed to go...

    Did I make a larger mistake that has me way off target?
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  3. #18
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    There is a note that says that these curves were chosen because they produced integrals that are easy to evaluate. I am really hung up on this one, so I plugged the indefinte integral into WolframAlpha to get some sort of idea...

    \int (1-cos(t))^{3/2} dt = \frac{1}{3} \sqrt{1-cos(t)} (cos(\frac{3 t}{2})-9 cos(\frac{t}{2})) csc(\frac{t}{2})+constant

    That doesn't seem particularly user friendly. I am trying to get my last couple of problems finished up today so that I can take a week off before classes start... I need a break.
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  4. #19
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    Quote Originally Posted by MechEng View Post
    There is a note that says that these curves were chosen because they produced integrals that are easy to evaluate. I am really hung up on this one, so I plugged the indefinte integral into WolframAlpha to get some sort of idea...

    \int (1-cos(t))^{3/2} dt = \frac{1}{3} \sqrt{1-cos(t)} (cos(\frac{3 t}{2})-9 cos(\frac{t}{2})) csc(\frac{t}{2})+constant

    That doesn't seem particularly user friendly. I am trying to get my last couple of problems finished up today so that I can take a week off before classes start... I need a break.
    Depending on how you do it,
    you can end up with many alternative expressions,
    giving the same result!

    There may be more convenient ways, but here's one....

    \displaystyle\huge\int_{0}^{2{\pi}}\left(1-cost\right)^{\frac{3}{2}}dt=\int_{0}^{2{\pi}}\left[2sin^2\left(\frac{t}{2}\right)\right]^{\frac{3}{2}}dt

    =\displaystyle\huge\left[\sqrt{2}\right]^3\int_{0}^{2{\pi}}sin^3\left(\frac{t}{2}\right)dt

    u=\displaystyle\huge\frac{t}{2}\ \Rightarrow\ 2u=t\ \Rightarrow\ dt=2du

    gives

    \displaystyle\huge4\sqrt{2}\int_{0}^{{\pi}}sin^3ud  u=4\sqrt{2}\int_{0}^{{\pi}}(sinu)sin^2udu=4\sqrt{2  }\int_{0}^{{\pi}}sinu\left(1-cos^2u\right)du

    x=cosu\ \Rightarrow\ -dx=sinu\ du

    giving

    \displaystyle\huge4\sqrt{2}\int_{1}^{-1}\left(1-x^2\right)(-dx)=4\sqrt{2}\int_{-1}^{1}\left(1-x^2\right)\ dx

    \displaystyle\huge=4\sqrt{2}\left[\left(1-\frac{1}{3}\right)-\left(-1-\frac{-1}{3}\right)\right]

    \displaystyle\huge=4\sqrt{2}\left[\frac{2}{3}-\left(-\frac{2}{3}\right)\right]

    =\displaystyle\huge4\sqrt{2}\left(\frac{4}{3}\righ  t)=\frac{16}{3}\sqrt{2}

    The constant applies when we are looking for an "antiderivative" (returned by Wolfram).
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  5. #20
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    Just to be difficult... and to show that I have at least been trying to finish these...

    Recall that my coeffiecients, 2\pi a^2\sqrt{2} have been set aside for clarity... I will bring them back at the end.

    So, we have...

    \displaystyle\int_{0}^{2\pi}(1-cos(t))^{\frac{3}{2}}dt  =  \int_{0}^{2\pi}(2\sin^2(\frac{t}{2}))^{\frac{3}{2}  }dt

    Because of identity...

    \diplaystyle\sin^2(x)=\frac{1-cos2(x)}{2}

    So...

    \displaystyle\int_{0}^{2\pi}(2\sin^2(\frac{t}{2}))  ^{\frac{3}{2}}dt


    \displaystyle\int_{0}^{2\pi}2^{\frac{3}{2}}\sin^3(  \frac{t}{2})dt


    \displaystyle2\sqrt{2}\int_{0}^{2\pi}\sin^3(\frac{  t}{2})dt

    Substituting...

    2u=t
    2du=dt

    \displaystyle2\sqrt{2}\int_{0}^{\pi}\sin^3udu

    Now, I'm going to take a shortcut here due to the identity...

    \displaystyle\int\sin^3u du=\frac{1}{3}(2+\sin^2u)\cos u + c

    So...

    \displaystyle\frac{1}{3}(2+\sin^2u)\cos u|_{0}^{\pi}


    \displaystyle\frac{1}{3}(2+\sin^2\pi)\cos \pi)-\frac{1}{3}(2+\sin^20)\cos 0)


    \displaystyle\frac{1}{3}(2+0)1)-\frac{1}{3}(2+0)1)


    \displaystyle\frac{(-2)-2}{3}


    \displaystyle\frac{-4}{3}

    Bringing back the coefficients from above...

    \displaystyle\frac{-64\pi a^2}{3}

    Why am I negative here?
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  6. #21
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    Quote Originally Posted by MechEng View Post
    Just to be difficult... and to show that I have at least been trying to finish these...

    Recall that my coeffiecients, 2\pi a^2\sqrt{2} have been set aside for clarity... I will bring them back at the end.

    So, we have...

    \displaystyle\int_{0}^{2\pi}(1-cos(t))^{\frac{3}{2}}dt  =  \int_{0}^{2\pi}\left(2\sin^2\left(\frac{t}{2}\righ  t)\right)^{\frac{3}{2}}dt

    Because of identity...

    \diplaystyle\sin^2(x)=\frac{1-cos2(x)}{2}

    So...

    \displaystyle\int_{0}^{2\pi}\left(2\sin^2\left(\fr  ac{t}{2}\right)\right)^{\frac{3}{2}}dt


    \displaystyle\int_{0}^{2\pi}2^{\frac{3}{2}}\sin^3\  left(\frac{t}{2}\right)dt


    \displaystyle2\sqrt{2}\int_{0}^{2\pi}\sin^3\left(\  frac{t}{2}\right)dt

    Substituting...

    2u=t

    2du=dt

    \displaystyle(2)2\sqrt{2}\int_{0}^{\pi}\sin^3udu

    Remember dt=2du

    Now, I'm going to take a shortcut here due to the identity...

    \displaystyle\int\sin^3u du=-\frac{1}{3}(2+\sin^2u)\cos u + c

    the identity has a negative sign

    So...

    -\displaystyle\frac{1}{3}(2+\sin^2u)\cos u|_{0}^{\pi}


    -\displaystyle\frac{1}{3}(2+\sin^2\pi)(\cos \pi)--\frac{1}{3}(2+\sin^20)(\cos 0)


    -\displaystyle\frac{1}{3}(2+0)(-1)+\frac{1}{3}(2+0)(1)

    ********** [cos({\pi})=-1] *********


    \displaystyle\frac{2+2}{3}


    \displaystyle\frac{4}{3}

    Bringing back the coefficients from above...

    4\sqrt{2}\frac{4}{3}=\frac{16}{3}\sqrt{2}

    2{\pi}a^2\sqrt{2}\frac{16}{3}\sqrt{2}=\frac{64}{3}  {\pi}a^2

    \displaystyle\frac{64\pi a^2}{3}
    Editing done...

    You may not ordinarily have access to that identity, however.
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