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Math Help - Parametric Arc Length

  1. #1
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    Parametric Arc Length

    I am given a curve defined by:

    x=a\cos(t)+at\sin(t)

    y=a\sin(t)-at\cos(t)

    "The a is a positive constant"

    How do I handle this? Am I supposed to factor the a out of the original expression in order to get my \frac{dy}{dt} and \frac{dx}{dt} ?
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  2. #2
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    What is the question?

    Can you find \dfrac{dx}{dt}~\&~\dfrac{dy}{dt}~?
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  3. #3
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    Quote Originally Posted by MechEng View Post
    I am given a curve defined by:

    x=a\cos(t)+at\sin(t)

    y=a\sin(t)-at\cos(t)

    "The a is a positive constant"

    How do I handle this? Am I supposed to factor the a out of the original expression in order to get my \frac{dy}{dt} and \frac{dx}{dt} ?
    x=a(cost+tsint)

    y=a(sint-tcost)

    \displaystyle\huge\frac{dx}{dt}=a(....)

    \displaystyle\huge\frac{dy}{dt}=a(,,,,)
    Last edited by Archie Meade; August 9th 2010 at 04:32 PM.
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    Plato,

    I am asked to find the length of curve, for 0<t<pi/2, defined by x and y above. If I supposed to pull a out of expression, I can find dx/dt and dy/dt... as follows:

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

    \frac{dy}{dt}=a(t \sin (t))
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  5. #5
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    So, should my formula for arc length look like this?


    \displaystyle s=\int_{0}^{\pi/2}{\sqrt{a(t\cos t)^2+a(t\sin t)^}}dt

    Or is this supposed to say...


    \displaystyle s=a\int_{0}^{\pi/2}{\sqrt{(t\cos t)^2+(t\sin t)^}}dt
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  6. #6
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    Quote Originally Posted by MechEng View Post
    Plato,

    I am asked to find the length of curve, for 0<t<pi/2, defined by x and y above. If I supposed to pull a out of expression, I can find dx/dt and dy/dt... as follows:

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

    \frac{dy}{dt}=a(t \sin (t))
    Then the arc length is

    \displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}\sqrt{  \left(\frac{dy}{dt}\right)^2+\left(\frac{dx}{dt}\r  ight)^2}\ dt

    =\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}\sqrt  {\left(atcost\right)^2+\left(atsint\right)^2}\ dt

    =\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}\sqrt  {a^2t^2\left(cos^2t\right)+a^2t^2\left(sin^2t\righ  t)}\ dt

    =\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}at\sq  rt{\left(cos^2t\right)+\left(sin^2t\right)}\ dt

    =\displaystyle\huge\ a\int_{0}^{\frac{{\pi}}{2}}t\ dt
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  7. #7
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    Thank you!

    So I should wind up with:

    \displaystyle s=\frac{a\pi^2}{8}
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  8. #8
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    Quote Originally Posted by MechEng View Post
    Thank you!

    So I should wind up with:

    \displaystyle s=\frac{a\pi^2}{8}
    Yes,

    in cases like that, you can keep the constant with the other terms,
    until you notice that there is a common factor within the square root.

    \displaystyle\huge\int{\sqrt{a^2t^2cos^2t+a^2t^2si  n^2t}}\ dt=\int{\sqrt{(at)^2\left(cos^2t+sin^2t\right)}\ dt

    =\displaystyle\huge\int{\sqrt{(at)^2}\sqrt{cos^2t+  sin^2t}\ dt=\int{at\sqrt{1}\ dt=a\int{t}\ dt
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  9. #9
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    the next one I am working on is:


    x=e^t \cos t


    x=e^t \sin t


    So far I have:


    \displaystyle\frac{dx}{dt}=te^t \cos t - e^t \sin t


    \displaystyle\frac{dx}{dt}=te^t \sin t - e^t \cos t


    I am getting stuck on the square root portion as follows:


    \displaystyle\sqrt{(te^t cos t - e^t sin t)^2+(te^t sin t + e^t cos t)^2}


    \displaystyle\sqrt{e^{2t}(t cos t - sin t)^2+e^{2t}(t sin t + cos t)^2}


    \displaystyle e^t \sqrt{(t cos t - sin t)^2+(t sin t + cos t)^2}


    Do you have a hint or a direction from this point?
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  10. #10
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    Quote Originally Posted by MechEng View Post
    the next one I am working on is:


    x=e^t \cos t


    y=e^t \sin t


    So far I have:


    \displaystyle\frac{dx}{dt}=e^t \cos t - e^t \sin t


    \displaystyle\frac{dy}{dt}=e^t \sin t + e^t \cos t


    I am getting stuck on the square root portion as follows:


    \displaystyle\sqrt{(e^t cos t - e^t sin t)^2+(e^t sin t + e^t cos t)^2}


    \displaystyle\sqrt{e^{2t}(cos t - sin t)^2+e^{2t}(sin t + cos t)^2}


    \displaystyle e^t \sqrt{(cos t - sin t)^2+(sin t + cos t)^2}


    Do you have a hint or a direction from this point?
    You had a stray "t" and a small typo on the declared function for "y".
    It's been edited above.

    Next, expand the expression under the square root and simplify..

    (cost-sint)(cost-sint)=cos^2t-2costsint+sin^2t

    (cost+sint)(cost+sint)=cos^2t+2sintcost+sin^2t

    Bring those together...
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  11. #11
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    Thank you once again. I know exactly what I did wrong... for some reason, I treated e^t as e^tn when I took the derivative. That gave me te^t. Wrong either way, but at least I know why I was wrong.

    Anyway... finishing this one out produces...

    s=\int_{0}^{\pi}{e^t\sqrt{2}}

    So...

    s=(e^{\pi}-1)\sqrt{2}
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  12. #12
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    Roots are really causing me some grief tonight.

    Same scenario... given the following:

    0 \leq t \leq 2\pi

    x= a \cos^3 t

    y= a \sin^3 t

    \frac{dx}{dt}= a 3 cos^2 t sin t

    \frac{dy}{dt}= a 3 sin^2 t cos t

    I get...

    \sqrt{(a 3 cos^2 t sin t)^2 + (a 3 sin^2 t cos t)^2}

    \sqrt{a3}\sqrt{(cos^2 t sin t)^2 + (sin^2 t cos t)^2}

    \sqrt{a3}\sqrt{cos^4 t sin^2 t + sin^4 t cos^2 t}

    I keep running myself into these corners. I can plug this into my calculator and come up with:

    \sqrt{a3}|sin^2 t cos^2 t|

    But I'm not exactly sure how to get there. In addition, this does not look particularly useful for the integration. So, I must have made some sort of error along the way.
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  13. #13
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    Ok... Let me try that again...


    0 \leq t \leq 2\pi


    x= a \cos^3 t


    y= a \sin^3 t


    \frac{dx}{dt}= a 3 cos^2 t sin t


    \frac{dy}{dt}= a 3 sin^2 t cos t


    I get...


    \sqrt{(a 3 cos^2 t sin t)^2 + (a 3 sin^2 t cos t)^2}


    a\sqrt{(3 cos^2 t sin t)^2 + (3 sin^2 t cos t)^2}


    a\sqrt{9 cos^4 t sin^2 t + 9 sin^4 t cos^2 t}


    This is where I ran into a dead-end last time...


    a3|sin^2 t cos^2 t|


    Much more usable this time around...


    Now, plugging this into my integral...


    \displaystyle s= \int_{0}^{2\pi}{3a|\sin t \cos t|} dt


    \displaystyle s= 3a\int_{0}^{2\pi}{|\sin t \cos t|} dt


    let u= \cos t


    let du= -\sin t


    \displaystyle s= 3a\int_{\cos 0}^{\cos 2\pi}{|-u|} du


    s= 3a \frac{u^2}{2}|_{1}^{1}


    Wait... how did my terminals of integration become 1 to 1?
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  14. #14
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    Quote Originally Posted by MechEng View Post
    Ok... Let me try that again...


    0 \leq t \leq 2\pi


    x= a \cos^3 t


    y= a \sin^3 t


    \frac{dx}{dt}= a 3 cos^2 t(- sin t)


    \frac{dy}{dt}= a 3 sin^2 t cos t


    I get...


    \sqrt{(a 3 cos^2 t(- sin t))^2 + (a 3 sin^2 t cos t)^2}


    \sqrt{(3a)^2( cos^2 t (-sin t))^2 + (3a)^2( sin^2 t cos t)^2}


    \sqrt{(3a)^2(cos^4 t sin^2 t +  sin^4 t cos^2 t)}


    This is where I ran into a dead-end last time...


    a3|\sqrt{sin^2 t cos^2 t}|


    Much more usable this time around...


    Now, plugging this into my integral...


    \displaystyle s= \int_{0}^{2\pi}{3a|\sin t \cos t|} dt


    \displaystyle s= 3a\int_{0}^{2\pi}{|\sin t \cos t|} dt


    let u= \cos t ..........No! it's too early to do this! You need to examine the shape of this graph


    let du= -\sin t\ dt


    \displaystyle s= 3a\int_{\cos 0}^{\cos 2\pi}{|-u|} du


    s= 3a \frac{u^2}{2}|_{1}^{1}


    Wait... how did my terminals of integration become 1 to 1?
    Some editing above....

    If you continue from

    \displaystyle\huge\int_{0}^{2{\pi}}\sqrt{(3a)^2cos  ^4tsin^2t+(3a)^2sin^4tcos^2t}\ dt

    then under the square root can be factored to

    \displaystyle\huge\int_{0}^{2{\pi}}\sqrt{(3a)^2\le  ft(cos^2tsin^2t\right)\left(cos^2t+sin^2t\right)}\ dt

    =\displaystyle\huge\int_{0}^{2{\pi}}\sqrt{(3a)^2}\  sqrt{(costsint)^2}(1)\ dt

    =\displaystyle\huge\ 3a\int_{0}^{2{\pi}}costsint\ dt

    Now, the graph alternates from positive to negative, so be careful with your limits,
    to avoid getting a result of zero, or subtracting any part of the area.
    You are correct to look for the integral of the modulus.
    However, you need to locate the x-axis crossing points of your graph
    in order to evaluate the area between the resulting function and the x-axis
    and if it is symmetrical.

    u=sint,\ du=costdt

    t=0,\ u=0

    t=\frac{{\pi}}{2},\ u=1

    \displaystyle\huge\ (4)3a\int_{0}^1u\ du

    gives the area between the curve and the x-axis.

    You may also express

    costsint=\frac{1}{2}sin2t
    Last edited by Archie Meade; August 10th 2010 at 05:12 AM.
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  15. #15
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    Quote Originally Posted by Archie Meade View Post

    Now, the graph alternates from positive to negative, so be careful with your limits,
    to avoid getting a result of zero, or subtracting any part of the area.
    You are correct to look for the integral of the modulus.
    However, you need to locate the x-axis crossing points of your graph
    in order to evaluate the area between the resulting function and the x-axis
    and if it is symmetrical.

    u=sint,\ du=costdt

    t=0,\ u=0

    t=\frac{{\pi}}{2},\ u=1

    What does this notation mean?

    \displaystyle\huge\ (4)3a\int_{0}^1u\ du

    gives the area between the curve and the x-axis.

    You may also express

    costsint=\frac{1}{2}sin2t
    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?

    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?

    I think a value of \frac{\pi}{2} would accomplish what I want.
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