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Math Help - Finding the Curvature K

  1. #1
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    Finding the Curvature K

    Hi
    Need help on the following question:

    A curve has vector equation r=(3t-t^3)i+(3t^2)j+(3t++t^3)k Find its curvature K

    i know that K = \frac{|v * a|}{|v^3|}

    This is what i have done.

    v= (3-3t^2)i+6tj+(3++3t^2)k
    a= -6ti+6j+6tk

    |v * a| = \begin {bmatrix} 3-t^2 & 6t & 3+3t^2 || -6t & 6 & 6t \end{bmatrix}

    i(36t^2-18+18t^2)-j(18t-18^3+18t+18t^3)+k(18-18t^2+36t^2)
    i(18t^2-18)-j(36t)+k(18-18t^2)
    18t^2-18-36t+18-18t^2
    |v*a|=-36t

    |v| = 3-3t^2+6t+3+3t^2

    This is where i am not sure what to do

    how do you work out |v|^3 ???
    P.S
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  2. #2
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    Dear Paymemoney,

    I hope the correct vector equation must be, r=(3t-t^3)i+(3t^2)j+(3t+t^3)k

    If you want to find the curvature first find \underline{T}=\frac{dr}{ds}=\frac{r'(t)}{\mid{r'(t  )}\mid}; where \underline{T} is the unit tangent vector.

    K=\mid\frac{dT}{ds}\mid=\mid\frac{dT}{dt}\times\fr  ac{dt}{ds}\mid=\mid\frac{1}{\mid{r'(t)}\mid}\frac{  dT}{dt}\mid

    Hope you could continue.
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  3. #3
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    oh yeh, the first part i found the unit tangent so i get this:

    K =\frac{|-36t|}{|\sqrt{2}(1+t^2)|^3}

    what would i do next?
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  4. #4
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    Dear Paymemoney,

    That is the answer. Since the denominator is positive, K=\frac{36\mid{t}\mid}{(\sqrt{2}(1+t^2))^3}
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  5. #5
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    the answer in the book is K=\frac{1}{3(1+t^2)^2} unless this is incorrect
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    the answer in the book is K=\frac{1}{3(1+t^2)^2} unless this is incorrect
    Dear Paymemoney,

    I went all through this problem and found out K=\frac{1}{3(1+t^2)^2} is the answer. Probably you have done something wrong in your calculations. Try to solve the problem according to the method I have given and if you have any problem please dont hesitate to ask.
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  7. #7
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    in relation to this question the book had an example on these types of question and i didn't understand how they got

    r^{'}(t) * r^{''}(t) = \begin{matrix} i & j & k \\ 1 & 2t & 3t^2 \\ 0 & 2 & t \end{bmatrix} = 6^2i-6tj+2k

    to this:

    Hence |r^{'}(t) * r^{''} (t)| = \sqrt{4+36^2+36t^4}=2\sqrt{1+9t^2+9t^4}

    why is the values inside 4+36&2+36^4?
    Last edited by Paymemoney; October 28th 2010 at 05:42 PM.
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  8. #8
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    Quote Originally Posted by Paymemoney View Post
    in relation to this question the book had an example on these types of question and i didn't understand how they got

    r^{'}(t) * r^{''}(t) = \begin{matrix} i & j & k \\ 1 & 2t & 3t^2 \\ 0 & 2 & t \end{bmatrix} = 6^2i-6tj+2k

    to this:

    Hence |r^{'}(t) * r^{''} (t)| = \sqrt{4+36^2+36t^4}=2\sqrt{1+9t^2+9t^4}

    why is the values inside 4+36&2+36^4?
    Dear Paymemoney,

    Cross product of two vectors can be written in matrix notation. That is, if A=a_{1}i+a_{2}j+a_{3}k~and~B=b_{1}i+b_{2}j+b_{3}k~  then~A\times{B}=\left(\begin{matrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix}\right)

    Does this answer your question?
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  9. #9
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    Quote Originally Posted by Sudharaka View Post
    Dear Paymemoney,

    Cross product of two vectors can be written in matrix notation. That is, if A=a_{1}i+a_{2}j+a_{3}k~and~B=b_{1}i+b_{2}j+b_{3}k~  then~A\times{B}=\left(\begin{matrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix}\right)

    Does this answer your question?

    not really, i understand that part, its the part where you square root i don't understand
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  10. #10
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    Quote Originally Posted by Paymemoney View Post
    not really, i understand that part, its the part where you square root i don't understand
    Dear Paymemoney,

    The magnitude of a vector \underline{r} is denoted by \mid\underline{r}\mid. Suppose you have a vector \underline{r}=xi+yj+zk. Then the magnitude of this vector is given by, \mid{\underline{r}\mid=\sqrt{x^2+y^2+z^2}.

    Hope this helps.
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  11. #11
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    yep i understand now, but as for the original question i still cannot get the answer would you show me your solutions to it?
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  12. #12
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    Quote Originally Posted by Paymemoney View Post
    yep i understand now, but as for the original question i still cannot get the answer would you show me your solutions to it?
    Dear Paymemoney,

    Okay. What did you get for the unit tangent vector? It should be, \underline{T}=\frac{(1-t^2)i+2tj+(1+t^2)k}{\sqrt{2}(1+t^2)}

    Now differentiate this with respect to 's' and you would end up with, \frac{dT}{ds}=\frac{1}{3(1+t^2)}\left[\frac{-2t}{(1+t^2)^2}i+\frac{1-t^2}{(1+t^2)^2}j\right]

    Therefore, K=\mid\frac{dT}{ds}\mid=\frac{1}{3(1+t^2)^3}\mid(-2t)i+(1-t^2)j\mid

    Hope you can continue.
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