## Related questions with answers

Familiarize yourself with parametric representations of important surfaces by deriving a representation, by finding the parameter curves (curves u=const and v=const) of the surface and a normal vector N=ru*rv of the surface. Show the details of your work. Helicoid r(u, v)=[u cos v, u sin v, v]. Explain the name.

Solution

VerifiedThe paramater curves of the surface in the $xy$-plane parametrised by

$\pmb{r}(u,v) = [u \cos v, \, u \sin v, \, v ]$

are:

$u=u_0( = const): \,\, r(u_0,v) = [u_0 \cos v, \, u_0 \sin v, \, v ]$

and

$v=v_0( = const): \,\, r(u ,v_0) = [u \cos v_0, \, u \sin v_0, \, v_0 ]$

Notice that the first family of parameter curves is the family $\textbf{helixes}$ whose axis is the $z$-axis and distance from the axis is $|u_0|$. The second family of parameter curves is the family of lines who pass through the points $( \cos v_0, \, \sin v_0, \, v_0)$ and $(0,0,v_0)$. From this information we see that the name is justified since we "construct" the surface by connecting the helixes to the axis (in our case the $z$-axis)

We have

$\pmb{r}_u = [\cos v, \, \sin v, \, 0], \,\, \pmb{r}_v = [-u\sin v, \, u \cos v , \, 1]$

Let us now compute the normal vector:

$\begin{aligned} \pmb{N} & = [\cos v, \, \sin v, \, 0] \times [-u\sin v, \, u \cos v , \, 1] = \\ & = \begin{vmatrix} \pmb{i} & \pmb{j} & \pmb{k} \\ \cos v & \sin v & 0 \\ -u\sin v & u \cos v & 1 \end{vmatrix} = \\ & = \begin{vmatrix} \sin v & 0 \\ u \cos v & 1 \end{vmatrix} \pmb{i} - \begin{vmatrix} \cos v & 0 \\ - u \sin v & 1 \end{vmatrix} \pmb{j} + \begin{vmatrix} \cos v & \sin v \\ -u \sin v & u \cos v \end{vmatrix} \pmb{k} = \\ & = \sin v \cdot \pmb{i} - \cos v \cdot \pmb{j} + (u\cos^2 v + u \sin^2 v) \pmb{k} = \\ & = \sin v \cdot \pmb{i} - \cos v \cdot \pmb{j} + u \cdot \pmb{k} = [\sin v, \, \cos v, \, u ] \end{aligned}$

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