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Math Help - Simplification of the Unit Normal Vector Proof Help

  1. #1
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    Simplification of the Unit Normal Vector Proof Help

    This is what my book says:

    If C is given by the vector equation \vec{r}(t)=x(t)\vec{i}+y(t)\vec{j} a \leq t \leq b

    then the unit tangent vector is \vec{T}(t)= \frac{x'(t)}{|\vec{r}'(t)|}\vec{i} + \frac{y'(t)}{|\vec{r}'(t)|}\vec{j}

    You can verify that the outward unit normal vector to C is given by  \vec{n}(t) = \frac{y'(t)}{|\vec{r}'(t)|}\vec{i} - \frac{x'(t)}{|\vec{r}'(t)|}\vec{j}

    I don't get how that's true. I tried proving it using the fact that \vec{n}(t)=\frac{\vec{T}'(t)}{|\vec{T}'(t)|} and all I got was this:

    \vec{n}(t)=\frac{\frac{d}{dt}\frac{x'(t)}{|\vec{r}  '(t)|} + \frac{d}{dt}\frac{y'(t)}{|\vec{r}'(t)|}}{\sqrt{(\f  rac{d}{dt}\frac{x'(t)}{|\vec{r}'(t)|})^2+ \frac{d}{dt}\frac{y'(t)}{|\vec{r}'(t)|}^2}}

    and I don't know how to simplify that to  \vec{n}(t) = \frac{y'(t)}{|\vec{r}'(t)|}\vec{i} - \frac{x'(t)}{|\vec{r}'(t)|}\vec{j}

    Thanks in advance!
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  2. #2
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    It is easy to check that n has unit length, and the dot product of n and T is 0.
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  3. #3
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    That makes sense, but how was \vec{n} derived?
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  4. #4
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    Apply the quotient rule to each of \frac{x'}{\left(x'^2+ y'^2\right)^2}.
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