Results 1 to 2 of 2

Math Help - arc Length of a spiral

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    56

    arc Length of a spiral

    HI,

    I need to find the length of a spiral given by r(t) = pi*(t*cos(pi*t)i - t*sin(pi*t)j +t k) , from t=0 to t=3
    I have been given the formula

    integral (sqrt(a^2 + b^2*t^2)dt = 1/2*t*sqrt(a^2 + b^2*t^2) + (a^2/2b)*ln(b*t + sqrt(a^2 + b^2*t^2).

    I proceeded to find dot product of tangent vectors

    r'(t) . r'(t) = (pi*cos(pi*t) - pi^2*tsin(pi*t))^2 + pi^2(pi*t*cos(pi*t) + sin(pi*t)) +pi^2

    However, now i have come unstuck, where do i go from here?

    I was thinking about letting a^2 = (pi*cos(pi*t) - pi^2*tsin(pi*t))^2,

    b^2= (pi*t*cos(pi*t) + sin(pi*t)) +1, t^2=pi^2

    then substituting into the formula and solving t=3 and t=0

    But i really have no idea where to get the parameters a , b , t from to use in the formula

    Any help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by olski1 View Post
    HI,

    I need to find the length of a spiral given by r(t) = pi*(t*cos(pi*t)i - t*sin(pi*t)j +t k) , from t=0 to t=3
    I have been given the formula

    integral (sqrt(a^2 + b^2*t^2)dt = 1/2*t*sqrt(a^2 + b^2*t^2) + (a^2/2b)*ln(b*t + sqrt(a^2 + b^2*t^2).

    I proceeded to find dot product of tangent vectors

    r'(t) . r'(t) = (pi*cos(pi*t) - pi^2*tsin(pi*t))^2 + pi^2(pi*t*cos(pi*t) + sin(pi*t)) +pi^2

    However, now i have come unstuck, where do i go from here?

    I was thinking about letting a^2 = (pi*cos(pi*t) - pi^2*tsin(pi*t))^2,

    b^2= (pi*t*cos(pi*t) + sin(pi*t)) +1, t^2=pi^2

    then substituting into the formula and solving t=3 and t=0

    But i really have no idea where to get the parameters a , b , t from to use in the formula

    Any help would be greatly appreciated.
    I think you need to check your derivative. I get that

    \mathbf{r}'(t)=\pi[(\cos(\pi t)-t\pi \sin(\pi t))\mathbf{i}+(-\sin(\pi t)-t\pi \cos(\pi t))\mathbf{j}+\mathbf{k}]

    After dotting this with itself and simplifying I get

    \mathbf{r}'(t) \cdot \mathbf{r}'(t)=\pi^2(\pi^2t^2+2)

    So the integral should be

    \displaystyle \pi \int_{0}^{3}\sqrt{\pi^2t^2+2}

    Now use the substitution \displaystyle t= \frac{\sqrt{2}}{\pi}\sinh(x)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Archimedes spiral?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: April 8th 2011, 07:46 PM
  2. Replies: 4
    Last Post: February 13th 2011, 08:49 AM
  3. [SOLVED] Implicit differentiation to find the length of a spiral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 9th 2011, 06:38 AM
  4. Replies: 3
    Last Post: August 30th 2009, 07:51 PM
  5. Logarithmic spiral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 13th 2009, 11:36 AM

/mathhelpforum @mathhelpforum