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Math Help - Difficulty Calculating the Principal Unit Normal Vector.

  1. #1
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    Difficulty Calculating the Principal Unit Normal Vector.

    Hi,

    I am having difficulties find the principal unit normal vector N(t) for the vector function

    \\r(t)=(t,4-t^2,0).

    I have worked up to this point:

    \\r(t)=(t,4-t^2,0); -2\leq t\leq 2
    \\r'(t)=(1,-2t,0)
    \\r'(t) \right \|=\sqrt(1+4t^2)
    \\T(t)=<\frac{1}{\sqrt(1+4t^2)},\frac{-2t}{\sqrt(1+4t^2)},0>

    I know that I need to take the derivative of this, and devide it by its length to find the princiipal normal vector; however, I am not sure how to do this!

    Thanks!
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  2. #2
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    Re: Difficulty Calculating the Principal Unit Normal Vector.

    I have either made progress or gone the wrong direction. Either way, here is the work that I have done since I made this question:
    I first factored the (1+4t^2)^(-1/2), leaving <1,-2t,0> as the vector. Then, to multiply the two, I set (1+4t^2)^(-1/2) as u and <1,-2t,0> as v to use the product rule. I found the derivative of u to be  frac {-(1+4t^2)^{-3/2}}{2}8t , which further simplifies to   frac {-4t(1+4t^2)^{-3/2}} , which I multiplied against the scalar. I did the same with taking the derivative of the vector, and multiplying it by the (1+4t^2)^(-1/2).
    At the end, I know have:
    <-4t(1+4t^2)\tfrac{-3}{2},8t^3,0> + <0,-2(1+4t^2)^{^{1/2}},0>\
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  3. #3
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    Re: Difficulty Calculating the Principal Unit Normal Vector.

    Quote Originally Posted by SC313 View Post
    Hi,

    I am having difficulties find the principal unit normal vector N(t) for the vector function

    \\r(t)=(t,4-t^2,0).

    I have worked up to this point:

    \\r(t)=(t,4-t^2,0); -2\leq t\leq 2
    \\r'(t)=(1,-2t,0)
    \\r'(t) \right \|=\sqrt(1+4t^2)
    \\T(t)=<\frac{1}{\sqrt(1+4t^2)},\frac{-2t}{\sqrt(1+4t^2)},0>

    I know that I need to take the derivative of this, and devide it by its length to find the princiipal normal vector; however, I am not sure how to do this!

    Thanks!
    \frac{dT}{dt}=\left<\frac{-4 t}{(4 t^2+1)^{3/2}}\,,\ \frac{-2}{(4 t^2+1)^{3/2}}\,,\ 0\right>

    Divide this by it's magnitude.
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  4. #4
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    Re: Difficulty Calculating the Principal Unit Normal Vector.

    Quote Originally Posted by SammyS View Post
    \frac{dT}{dt}=\left<\frac{-4 t}{(4 t^2+1)^{3/2}}\,,\ \frac{-2}{(4 t^2+1)^{3/2}}\,,\ 0\right>

    Divide this by it's magnitude.
    Where do you get this from? Thanks!
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  5. #5
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    Re: Difficulty Calculating the Principal Unit Normal Vector.

    Quote Originally Posted by SC313 View Post
    Where do you get this from? Thanks!
    I took the derivative of T(t) that you had in your OP.
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  6. #6
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    Re: Difficulty Calculating the Principal Unit Normal Vector.

    Okay, I see now. And how do I find the magnitude of this..., and divide the number by this? It's so confusing. thanks by the way
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  7. #7
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    Re: Difficulty Calculating the Principal Unit Normal Vector.

    Magnitude: Square root of the sum of the squares of the components.

    At least they have a common denominator.
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