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Thread: messy vectors

  1. #1
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    messy vectors

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

    $\displaystyle \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 $\displaystyle \bold{r} $ and $\displaystyle \bold{v} $.

    So: $\displaystyle \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 $\displaystyle \bold{r} $ is never perpendicular to $\displaystyle v $.

    For $\displaystyle \bold{r} $ to be parallel to $\displaystyle \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 $\displaystyle t $ for which $\displaystyle \bold{r} $ is parallel to $\displaystyle \bold{v} $. Is this correct?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    $\displaystyle \bold{r}(t) = (\cosh t) \bold{i} + (\sinh t) \bold{j} + t \bold{k} $

    $\displaystyle \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 $\displaystyle \bold{r} $ and $\displaystyle \bold{v} $.

    So: $\displaystyle \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 $\displaystyle \bold{r} $ is never perpendicular to $\displaystyle v $.

    For $\displaystyle \bold{r} $ to be parallel to $\displaystyle \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 $\displaystyle t $ for which $\displaystyle \bold{r} $ is parallel to $\displaystyle \bold{v} $. Is this correct?
    Use the idea from my other post - the scalar business:

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

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

    $\displaystyle t = \lambda$ ... (3)

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

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

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

    Sub (2') into (1'):

    $\displaystyle \cosh t = t^2 \cosh t$ which has $\displaystyle t = \pm 1$ as its only solutions.
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