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Find the unit normal vector to the curve $r(t) = (3\sin t)i + (3\cos t)j + 4t k$ at point $( \pi /2, 0,1)$.

The tangent vector of this curve is $(3 \cos t)i -(3 \sin t)j + 4k$ and unit normal vector should be perpendicular to this vector at given point. But I couldn't get the final answer. How to find unit normal vector at the given point$?$

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    $\begingroup$ This point is not on the curve? $\endgroup$ Jan 30, 2019 at 20:39
  • $\begingroup$ This point is on the curve. I couldn't find any value of $t$ for this point $\endgroup$
    – Mathaddict
    Jan 30, 2019 at 20:43
  • $\begingroup$ Therefore it is not on the curve? $\endgroup$ Jan 30, 2019 at 20:45
  • $\begingroup$ So, there's no solution? $\endgroup$
    – Mathaddict
    Jan 30, 2019 at 20:46
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    $\begingroup$ If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found. $\endgroup$ Jan 30, 2019 at 20:47

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Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.

First of all you should parametrize your curve by arclength, which is given by \begin{align} s(t) & = \int_0^t \lVert r'(x) \rVert\,dx \\ & = 5t. \end{align} As a function of $s$, $r$ can be rewritten as $r(s) = \left(3\sin \frac{s}{5},3\cos \frac{s}{5}, \frac{4}{5}s\right)$. The unit tangent is then $$T(s) = \frac{1}{\lVert \frac{dr}{ds}\rVert}\frac{dr}{ds} = \left(\frac{3}{5}\cos \frac{s}{5}, -\frac{3}{5}\sin \frac{s}{5}, \frac{4}{5} \right).$$ The unit normal is just given by the renormalized derivative of the tangent vector: $$N(s)=\frac{T'(s)}{\lVert T'(s)\rVert} = \left(-\sin \frac{s}{5},-\cos \frac{s}{5}, 0\right). $$

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