How to find a unit normal vector to the given curve?

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$$?$$

• This point is not on the curve? Jan 30, 2019 at 20:39
• This point is on the curve. I couldn't find any value of $t$ for this point Jan 30, 2019 at 20:43
• Therefore it is not on the curve? Jan 30, 2019 at 20:45
• So, there's no solution? Jan 30, 2019 at 20:46
• 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. Jan 30, 2019 at 20:47

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: $$N(s)=\frac{T'(s)}{\lVert T'(s)\rVert} = \left(-\sin \frac{s}{5},-\cos \frac{s}{5}, 0\right).$$