Results 1 to 2 of 2

Math Help - messy vectors

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    154

    messy vectors

     \bold{r}(t) = (\cosh t) \bold{i} + (\sinh t) \bold{j} + t \bold{k}

     \bold{v}(t) = (\sinh t) \bold{i} + (\cosh t) \bold{j} + \bold{k}

    So we want to do the same thing: find the cosine of the angle between  \bold{r} and  \bold{v} .

    So:  \cos \theta = \frac{2e^{2t} + 2e^{-2t} + t}{\sqrt{4e^{4t} + 8 + 2e^{2t} + 4e^{-4t} + 2e^{-2t} + 2t^{2}e^{2t} + 2t^{2}e^{-2t} + t^{2}}}

    But the numerator is always positive, so  \bold{r} is never perpendicular to  v .

    For  \bold{r} to be parallel to  \bold{v} , both the numerator and denominator have to equal each other. This is messy to solve. But I don't think there are values of  t for which  \bold{r} is parallel to  \bold{v} . Is this correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by heathrowjohnny View Post
     \bold{r}(t) = (\cosh t) \bold{i} + (\sinh t) \bold{j} + t \bold{k}

     \bold{v}(t) = (\sinh t) \bold{i} + (\cosh t) \bold{j} + \bold{k}

    So we want to do the same thing: find the cosine of the angle between  \bold{r} and  \bold{v} .

    So:  \cos \theta = \frac{2e^{2t} + 2e^{-2t} + t}{\sqrt{4e^{4t} + 8 + 2e^{2t} + 4e^{-4t} + 2e^{-2t} + 2t^{2}e^{2t} + 2t^{2}e^{-2t} + t^{2}}}

    But the numerator is always positive, so  \bold{r} is never perpendicular to  v .

    For  \bold{r} to be parallel to  \bold{v} , both the numerator and denominator have to equal each other. This is messy to solve. But I don't think there are values of  t for which  \bold{r} is parallel to  \bold{v} . Is this correct?
    Use the idea from my other post - the scalar business:

    \cosh t = \lambda \sinh t .... (1)

    \sinh t = \lambda \cosh t .... (2)

    t = \lambda ... (3)

    Sub (3) into (1) and (2):

    \cosh t = t \sinh t .... (1')

    \sinh t = t \cosh t .... (2')

    Sub (2') into (1'):

    \cosh t = t^2 \cosh t which has t = \pm 1 as its only solutions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Messy differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 18th 2010, 11:07 AM
  2. Messy Integral!
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 5th 2010, 06:22 AM
  3. a messy looking ODE system
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: December 16th 2009, 05:09 AM
  4. Messy Integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 10th 2008, 02:26 PM
  5. messy integral (?)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 9th 2008, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum