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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.

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The paramater curves of the surface in the xyxy-plane parametrised by

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

are:

u=u0(=const):r(u0,v)=[u0cosv,u0sinv,v]u=u_0( = const): \,\, r(u_0,v) = [u_0 \cos v, \, u_0 \sin v, \, v ]

and

v=v0(=const):r(u,v0)=[ucosv0,usinv0,v0]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 helixes\textbf{helixes} whose axis is the zz-axis and distance from the axis is u0|u_0|. The second family of parameter curves is the family of lines who pass through the points (cosv0,sinv0,v0)( \cos v_0, \, \sin v_0, \, v_0) and (0,0,v0)(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 zz-axis)

We have

ru=[cosv,sinv,0],rv=[usinv,ucosv,1]\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:

N=[cosv,sinv,0]×[usinv,ucosv,1]==ijkcosvsinv0usinvucosv1==sinv0ucosv1icosv0usinv1j+cosvsinvusinvucosvk==sinvicosvj+(ucos2v+usin2v)k==sinvicosvj+uk=[sinv,cosv,u]\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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