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Math Help - Finding a unit normal vector

  1. #1
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    Finding a unit normal vector

    I wanted to make sure i got the right answer
    T=cost*cos(sint) i - cost*sin(sint) j- sint k
    T'=[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk
    |T'|=\sqrt{(-sint*cos(sint)-sin(sint)cos^2t)^2 + (sint*sin(sint)-cos^2+cos(sint))^2 - cos^2t}
    I got the answer for |T'| to be cos^2t
    \frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk}{cos^2t}
    \frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j}{cos^2t} - sectk

    I don't know if I should simplify more.
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  2. #2
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    Quote Originally Posted by alexgo View Post
    I wanted to make sure i got the right answer
    T=cost*cos(sint) i - cost*sin(sint) j- sint k
    T'=[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk
    |T'|=\sqrt{(-sint*cos(sint)-sin(sint)cos^2t)^2 + (sint*sin(sint)-cos^2+cos(sint))^2 - cos^2t}
    I got the answer for |T'| to be cos^2t
    \frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk}{cos^2t}
    \frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j}{cos^2t} - sectk
    You are making hard on yourself.

    \displaystyle T' = \frac{{R' \times (R'' \times R')}}<br />
{{\left\| {R'} \right\|^3 }}
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  3. #3
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    Quote Originally Posted by Plato View Post
    You are making hard on yourself.

    \displaystyle T' = \frac{{R' \times (R'' \times R')}}<br />
{{\left\| {R'} \right\|^3 }}
    ha that is a pain too.
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  4. #4
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    Quote Originally Posted by alexgo View Post
    ha that is a pain too.
    Well it is all a pain if you ask me.
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