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Thread: The curvature for a vector

  1. #1
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    The curvature for a vector

    Find the t-value where \kappa(t)=0.0001m^{-1}
    for this vectorfunction:
    \overrightarrow{r}(t)=\binom{450t}{1200-30t^2-7t^3},   t\in[-5,5]

    I know that I must use this formula:

    \kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^\frac{3}{2}}

    It's just that I don't know where to begin and what to do.
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  2. #2
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    Quote Originally Posted by No Logic Sense View Post
    Find the t-value where \kappa(t)=0.0001m^{-1}
    for this vectorfunction:
    \overrightarrow{r}(t)=\binom{450t}{1200-30t^2-7t^3},   t\in[-5,5]

    I know that I must use this formula:

    \kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^\frac{3}{2}}

    It's just that I don't know where to begin and what to do.
    Hi

    x(t) = 450t and y(t) = 1200-30t^2-7t^3

    You can substitute in the formula but it is going to be a hard job
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  3. #3
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    Oh, I get it now!

    I can just insert x(t) and y(t) in the respective places and then solve the fraction with the solve button on my calculator, right?
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  4. #4
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    That's it !
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  5. #5
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    Thanks!
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