Differential Geometry of Curves and Surfaces: Difficult Problems with Solutions

Solution:
The coordinate [tex]y(u)=\frac{u^3}{3a^2}[/tex] is strictly increasing, so each plane meets the curve exactly once:
[tex]\frac{u^3}{3a^2}=\frac{a}{3}\Rightarrow u^3=a^3\Rightarrow u=a,\qquad \frac{u^3}{3a^2}=9a\Rightarrow u^3=27a^3\Rightarrow u=3a[/tex]
Hence we need the arc with [tex]a\le u\le 3a[/tex]. We have
[tex]\vec{r}\,{}'(u)=\left(1,\ \frac{u^2}{a^2},\ -\frac{a^2}{2u^2}\right),\qquad \left|\vec{r}\,{}'(u)\right|^2=1+\frac{u^4}{a^4}+\frac{a^4}{4u^4}=\left(\frac{u^2}{a^2}+\frac{a^2}{2u^2}\right)^2[/tex]
because the doubled product in the square is [tex]2\times \frac{u^2}{a^2}\times \frac{a^2}{2u^2}=1[/tex]. So [tex]\left|\vec{r}\,{}'(u)\right|=\frac{u^2}{a^2}+\frac{a^2}{2u^2}[/tex] (a positive quantity) and
[tex]L=\int_a^{3a}\left(\frac{u^2}{a^2}+\frac{a^2}{2u^2}\right)du=\left[\frac{u^3}{3a^2}-\frac{a^2}{2u}\right]_a^{3a}=\left(9a-\frac{a}{6}\right)-\left(\frac{a}{3}-\frac{a}{2}\right)=9a-\frac{a}{6}+\frac{a}{6}=9a[/tex]

Solution:
We have
[tex]\vec{r}\,{}'=\left(1,\ u,\ u^2\right),\qquad \vec{r}\,{}''=(0,\ 1,\ 2u),\qquad \vec{r}\,{}'×\vec{r}\,{}''=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&u&u^2\\0&1&2u\end{vmatrix}=\left(u^2,\ -2u,\ 1\right)[/tex]
This vector is never zero, so the osculating plane exists at every point, and [tex]\vec{r}\,{}'×\vec{r}\,{}''[/tex] is its normal vector. The osculating plane at the point [tex]\vec{r}(u)[/tex] is
[tex]u^2(x-u)-2u\left(y-\frac{u^2}{2}\right)+\left(z-\frac{u^3}{3}\right)=0\ \Leftrightarrow\ u^2x-2uy+z-\frac{u^3}{3}=0[/tex]
It passes through [tex]A(0, 0, 9)[/tex] exactly when
[tex]9-\frac{u^3}{3}=0\ \Leftrightarrow\ u^3=27\ \Leftrightarrow\ u=3[/tex]
([tex]u=3[/tex] is the only real root). Substituting [tex]u=3[/tex] gives the unique such plane; it touches the curve at [tex]\left(3, \frac{9}{2}, 9\right)[/tex]:
[tex]9x-6y+z-9=0[/tex]

Solution:
We use the formula
[tex]\tau=\frac{\left(\vec{r}\,{}',\vec{r}\,{}'',\vec{r}\,{}'''\right)}{\left|\vec{r}\,{}'×\vec{r}\,{}''\right|^2}[/tex]
The derivatives and their values at [tex]u=1[/tex] are
[tex]\vec{r}\,{}'=\left(3-3u^2,\ 6u,\ 3+3u^2\right)=(0,\ 6,\ 6),\qquad \vec{r}\,{}''=(-6u,\ 6,\ 6u)=(-6,\ 6,\ 6),\qquad \vec{r}\,{}'''=(-6,\ 0,\ 6)[/tex]
Then
[tex]\vec{r}\,{}'×\vec{r}\,{}''=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0&6&6\\-6&6&6\end{vmatrix}=(0,\ -36,\ 36),\qquad \left|\vec{r}\,{}'×\vec{r}\,{}''\right|^2=2592\ne 0[/tex]
so the torsion is defined. The scalar triple product is
[tex]\left(\vec{r}\,{}',\vec{r}\,{}'',\vec{r}\,{}'''\right)=\left(\vec{r}\,{}'×\vec{r}\,{}''\right)\centerdot\vec{r}\,{}'''=0+0+36\times 6=216[/tex]
Hence
[tex]\tau(1)=\frac{216}{2592}=\frac{1}{12}[/tex]

Solution:
The curves meet where [tex]u-v=0[/tex] and [tex]u+v=2[/tex], i.e. at [tex]u=v=1[/tex]. For the first fundamental form we compute
[tex]\vec{r}_u=(\cos v,\ \sin v,\ 2u),\qquad \vec{r}_v=(-u\sin v,\ u\cos v,\ 0)[/tex]
[tex]E=\vec{r}_u\centerdot\vec{r}_u=1+4u^2,\qquad F=\vec{r}_u\centerdot\vec{r}_v=0,\qquad G=\vec{r}_v\centerdot\vec{r}_v=u^2[/tex]
so at [tex]u=1[/tex] we get [tex]E=5[/tex], [tex]F=0[/tex], [tex]G=1[/tex]. Along [tex]c_1[/tex] we have [tex]du=dv[/tex], so we take the direction [tex](du,dv)=(1,1)[/tex]; along [tex]c_2[/tex] we have [tex]\delta u=-\delta v[/tex], so we take [tex](\delta u,\delta v)=(1,-1)[/tex]. The angle between two directions on the surface is given by
[tex]\cos\varphi=\frac{E\,du\,\delta u+F\left(du\,\delta v+dv\,\delta u\right)+G\,dv\,\delta v}{\sqrt{E\,du^2+2F\,du\,dv+G\,dv^2}\,\sqrt{E\,\delta u^2+2F\,\delta u\,\delta v+G\,\delta v^2}}=\frac{5-1}{\sqrt{6}\times \sqrt{6}}=\frac{2}{3}[/tex]
Since [tex]\cos\varphi\gt 0[/tex], this is the acute angle between the curves:
[tex]\varphi=\arccos\frac{2}{3}[/tex]

Solution:
The first derivatives:
[tex]\vec{r}_u=\left(\cos 2v , \sin 2v , 0\right),\quad \vec{r}_v=\left(-2u\sin 2v , 2u\cos 2v , -\frac{1}{v^2}\right)[/tex]
Coefficients of the first fundamental form:
[tex]E=\vec{r}_u\centerdot\vec{r}_u=1,\quad F=\vec{r}_u\centerdot\vec{r}_v=-2u\sin 2v\cos 2v+2u\sin 2v\cos 2v=0,\quad G=\vec{r}_v\centerdot\vec{r}_v=4u^2+\frac{1}{v^4}[/tex]
The vector product [tex]\vec{m}[/tex] of [tex]\vec{r}_u[/tex] and [tex]\vec{r}_v[/tex] is a normal vector; its length is [tex]\sqrt{EG-F^2}=\sqrt{G}[/tex]:
[tex]\vec{m}=\left(-\frac{\sin 2v}{v^2} , \frac{\cos 2v}{v^2} , 2u\right),\quad |\vec{m}|=\sqrt{\frac{1}{v^4}+4u^2}=\sqrt{G},\quad \vec{n}=\frac{\vec{m}}{\sqrt{G}}[/tex]
Second derivatives and coefficients of the second fundamental form (the coefficient [tex]N[/tex] is not needed, because [tex]L=0[/tex]):
[tex]\vec{r}_{uu}=\vec{0},\quad \vec{r}_{uv}=\left(-2\sin 2v , 2\cos 2v , 0\right)[/tex]
[tex]L=\vec{r}_{uu}\centerdot\vec{n}=0,\quad M=\vec{r}_{uv}\centerdot\vec{n}=\frac{1}{\sqrt{G}}\left(\frac{2\sin ^2 2v}{v^2}+\frac{2\cos^2 2v}{v^2}\right)=\frac{2}{v^2\sqrt{G}}[/tex]
Therefore
[tex]K=\frac{LN-M^2}{EG-F^2}=-\frac{M^2}{G}=-\frac{4}{v^4G^2}=-\frac{4}{v^4\left(4u^2+\frac{1}{v^4}\right)^2}=-\frac{4v^4}{\left(1+4u^2v^4\right)^2}[/tex]
This agrees with the fact that [tex]S[/tex] is a ruled surface (the [tex]u[/tex]-lines are straight lines), for which [tex]K\le 0[/tex].

Solution:
The first derivatives and their vector product [tex]\vec{m}[/tex]:
[tex]\vec{r}_u=\left(2u , 2u , v\right),\quad \vec{r}_v=\left(2v , -2v , u\right),\quad \vec{m}=\left(2u^2+2v^2 , 2v^2-2u^2 , -8uv\right)[/tex]
Here [tex]\vec{m}=\vec{0}[/tex] only for [tex]u=v=0[/tex], so all points of [tex]S[/tex] are regular. The second derivatives:
[tex]\vec{r}_{uu}=\left(2 , 2 , 0\right),\quad \vec{r}_{uv}=\left(0 , 0 , 1\right),\quad \vec{r}_{vv}=\left(2 , -2 , 0\right)[/tex]
With the unit normal [tex]\vec{n}=\frac{\vec{m}}{|\vec{m}|}[/tex] the coefficients of the second fundamental form are
[tex]L=\frac{8v^2}{|\vec{m}|},\quad M=-\frac{8uv}{|\vec{m}|},\quad N=\frac{8u^2}{|\vec{m}|}[/tex]
[tex]II=\frac{8}{|\vec{m}|}\left(v^2\,du^2-2uv\,du\,dv+u^2\,dv^2\right)=\frac{8}{|\vec{m}|}\left(v\,du-u\,dv\right)^2[/tex]
Asymptotic directions satisfy [tex]II=0[/tex], i.e. [tex]v\,du-u\,dv=0[/tex]. The form is a perfect square and is not identically zero, so at every point there is exactly one (double) asymptotic direction: all points are parabolic ([tex]LN-M^2=0[/tex], [tex]K=0[/tex]).
Integrating [tex]v\,du-u\,dv=0[/tex] gives [tex]u=Cv[/tex] (and [tex]v=0[/tex]); along [tex]u=Cv[/tex]
[tex]\vec{r}=v^2\left(C^2+1 , C^2-1 , C\right)[/tex]
which is a ray from the origin. Indeed [tex]x^2-y^2=4u^2v^2=4z^2[/tex], so [tex]S[/tex] lies on the cone [tex]x^2-y^2=4z^2[/tex] and its asymptotic lines are the generators of the cone. Hence exactly one asymptotic line passes through each point of [tex]S[/tex].

Solution:
Write the family as [tex]F(x,y,x_0)=0[/tex], [tex]x_0\gt 0[/tex], where
[tex]F(x,y,x_0)=y-\frac{2}{x_0}-x\left(1-\frac{1}{x_0^2}\right)[/tex]
The envelope is obtained from the system [tex]F=0,\ \frac{\partial F}{\partial x_0}=0[/tex]:
[tex]\frac{\partial F}{\partial x_0}=\frac{2}{x_0^2}-\frac{2x}{x_0^3}=\frac{2}{x_0^3}\left(x_0-x\right)=0\iff x=x_0[/tex]
Substituting into [tex]F=0[/tex]:
[tex]y=\frac{2}{x_0}+x_0-\frac{1}{x_0}=x_0+\frac{1}{x_0}[/tex]
So the envelope is the curve [tex]y=x+\frac{1}{x},\ x\gt 0[/tex]. Check: at [tex]x=x_0[/tex] its slope is [tex]y'=1-\frac{1}{x_0^2}[/tex], the same as the slope of the line with parameter [tex]x_0[/tex], and this line passes through [tex]\left(x_0 , x_0+\frac{1}{x_0}\right)[/tex]; hence every line of the family is tangent to the curve.

Solution:
Let [tex]P\left(x_0 , y_0 , x_0y_0\right)[/tex] be any point of the surface and consider a line through [tex]P[/tex] with direction vector [tex]\left(p , q , r\right)\ne\vec{0}[/tex]:
[tex]x=x_0+pt,\quad y=y_0+qt,\quad z=x_0y_0+rt,\quad t\in\mathbb{R}[/tex]
It lies on the surface if and only if for all [tex]t[/tex]
[tex]x_0y_0+rt=\left(x_0+pt\right)\left(y_0+qt\right)\iff rt=\left(py_0+qx_0\right)t+pq\,t^2[/tex]
Two polynomials in [tex]t[/tex] coincide for all [tex]t[/tex] only if their coefficients are equal:
[tex]pq=0,\quad r=py_0+qx_0[/tex]
If [tex]p=0[/tex], then [tex]q\ne 0[/tex] (otherwise [tex]r=0[/tex] too and the vector would be zero), and the direction is [tex]\left(0 , q , qx_0\right)\parallel\left(0 , 1 , x_0\right)[/tex]: this is the line [tex]\ell_1:\ x=x_0,\ z=x_0y[/tex].
If [tex]q=0[/tex], then [tex]p\ne 0[/tex] and the direction is [tex]\left(p , 0 , py_0\right)\parallel\left(1 , 0 , y_0\right)[/tex]: this is the line [tex]\ell_2:\ y=y_0,\ z=y_0x[/tex].
Both lines really lie on the surface (on [tex]\ell_1[/tex] we have [tex]z=x_0y=xy[/tex], on [tex]\ell_2[/tex] we have [tex]z=y_0x=xy[/tex]); their directions are not parallel, so they are different, and by the computation above there are no other lines through [tex]P[/tex] on the surface.
Hence exactly two lines of the surface pass through every point, and [tex]z=xy[/tex] is a ruled surface, for example the union of the lines [tex]x=c,\ z=cy[/tex], [tex]c\in\mathbb{R}[/tex].