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Thread: More Unit Tangent and Unit Normal Vectors

  1. #1
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    More Unit Tangent and Unit Normal Vectors

    Find the unit tangent vector.

    r (t) = 4t^2i

    r'(t) = 8ti

    ||r'(t)|| = root {(8t)^2} = root {64t^2} = 8t

    So, T (t) = 8ti/8t = i.

    Yes? No?

    I now need the unit normal vector.

    I am stuck here again.

    I know that T (t) = i (see above).

    I need to find T'(t) and ||T'(t)||.
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  2. #2
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    Re: More Unit Tangent and Unit Normal Vectors

    you cannot learn math like this.

    Math within a given subject domain has to be learned in order.

    You cannot start playing the tangent and normal vectors until you understand how to differentiate a vector valued function.

    If you can't immediately see that $\dfrac{d}{dt} (1i + 0j + 0k )= 0i + 0j + 0k$

    you're just not ready to proceed further.
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    Re: More Unit Tangent and Unit Normal Vectors

    According to the textbook, my answer for T (t) is correct.
    So, T (t) = i is correct.

    I need N (t) = T'(t)/||T'(t)||

    I found T'(t) = 0. Then the magnitude is root {0^2} = 0.

    I concluded that N (t) = 0/0 = 0 or is it really zero?

    According to the textbook, "N (t) is undefined. The path is a line and the speed is a variable."

    Why is N (t) undefined? Can we say that 0/0 is undefined?
    Last edited by USNAVY; Jan 6th 2017 at 01:39 PM.
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    Re: More Unit Tangent and Unit Normal Vectors

    Quote Originally Posted by USNAVY View Post
    According to the textbook, my answer for T (t) is correct.
    So, T (t) = i is correct.

    I need N (t) = T'(t)/||T'(t)||

    I found T'(t) = 0. Then the magnitude is root {0^2} = 0.

    I concluded that N (t) = 0/0 = 0 or is it really zero?

    According to the textbook, "N (t) is undefined. The path is a line and the speed is a variable."

    Why is N (t) undefined? Can we say that 0/0 is undefined?
    it's undefined because there is no way you can express the 0 vector as a unit vector.

    or as you said.. $\dfrac{T^\prime}{\|T^\prime\|} = \dfrac {(0,0,0)} 0 \text{ which is undefined.}$
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    Re: More Unit Tangent and Unit Normal Vectors

    Quote Originally Posted by romsek View Post
    it's undefined because there is no way you can express the 0 vector as a unit vector.

    or as you said.. $\dfrac{T^\prime}{\|T^\prime\|} = \dfrac {(0,0,0)} 0 \text{ which is undefined.}$
    Cool.
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    Re: More Unit Tangent and Unit Normal Vectors

    Quote Originally Posted by romsek View Post
    you cannot learn math like this.

    Math within a given subject domain has to be learned in order.

    You cannot start playing the tangent and normal vectors until you understand how to differentiate a vector valued function.

    If you can't immediately see that $\dfrac{d}{dt} (1i + 0j + 0k )= 0i + 0j + 0k$

    you're just not ready to proceed further.
    You appear to have misread the OP's question Romsek. The question is to find the unit tangent vector to $\displaystyle \begin{align*} 4\,t^2\,\mathbf{i} \end{align*}$, which is $\displaystyle \begin{align*} \mathbf{i} \end{align*}$, not to find the unit tangent vector to $\displaystyle \begin{align*} \mathbf{i} \end{align*}$.

    What the OP posted is correct.
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    Re: More Unit Tangent and Unit Normal Vectors

    Surely you realized that this "curve" is a straight line, the x- axis? Its unit tangent vector is the unit vector in the direction of the x- axis, i. And there is no single unit normal vector- every vector in the yz-plane is normal to it.
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