Differential Geometry of Curves and Surfaces: Very Difficult Problems with Solutions

Solution:
We have
[tex]\vec{r}^{\,\prime}=\left(2 , \frac{1}{u} , 2u\right),\quad |\vec{r}^{\,\prime}|=\sqrt{4u^2+4+\frac{1}{u^2}}=2u+\frac{1}{u}=\frac{2u^2+1}{u}[/tex]
so the unit tangent vector is
[tex]\vec{t}=\frac{1}{2u^2+1}\left(2u , 1 , 2u^2\right)[/tex]
We look for a fixed vector [tex]\vec{e}=\left(\alpha , \beta , \gamma\right)\ne\vec{0}[/tex] with [tex]\vec{t}\centerdot\vec{e}=k[/tex] constant, i.e. for all [tex]u\gt 0[/tex]
[tex]2\gamma u^2+2\alpha u+\beta=k\left(2u^2+1\right)\Rightarrow \alpha=0,\ \beta=\gamma=k[/tex]
So [tex]\vec{e}\parallel\left(0 , 1 , 1\right)[/tex], and every tangent forms an angle of [tex]45^\circ[/tex] with this direction:
[tex]\vec{t}\centerdot\frac{\left(0 , 1 , 1\right)}{\sqrt{2}}=\frac{1+2u^2}{\sqrt{2}\left(2u^2+1\right)}=\frac{1}{\sqrt{2}}[/tex]
(This agrees with Lancret's theorem: [tex]\kappa=\frac{2u}{\left(2u^2+1\right)^2}[/tex] and [tex]\tau=-\frac{2u}{\left(2u^2+1\right)^2}[/tex], so [tex]\frac{\tau}{\kappa}=-1[/tex] is constant.)
A point [tex]\left(x , y , z\right)[/tex] lies on the cylinder if and only if for some [tex]\lambda[/tex] the point [tex]\left(x , y-\lambda , z-\lambda\right)[/tex] lies on [tex]c[/tex]:
[tex]x=2u,\quad y-\lambda=\ln u,\quad z-\lambda=u^2[/tex]
Eliminating [tex]\lambda[/tex] and substituting [tex]u=\frac{x}{2}[/tex]:
[tex]z-y=u^2-\ln u=\frac{x^2}{4}-\ln\frac{x}{2},\quad x\gt 0[/tex]
Check: the left side does not change under [tex]\left(y , z\right)\to\left(y+\lambda , z+\lambda\right)[/tex], and the points of [tex]c[/tex] satisfy the equation.

Solution:
Take [tex]x[/tex] as the parameter:
[tex]\vec{r}(x)=\left(x , x^n , f(x)\right),\quad \vec{r}^{\,\prime}=\left(1 , nx^{n-1} , f'\right),\quad \vec{r}^{\,\prime\prime}=\left(0 , n(n-1)x^{n-2} , f''\right)[/tex]
The vector product [tex]\vec{B}[/tex] of [tex]\vec{r}^{\,\prime}[/tex] and [tex]\vec{r}^{\,\prime\prime}[/tex] is a normal vector of the osculating plane:
[tex]\vec{B}=\left(nx^{n-1}f''-n(n-1)x^{n-2}f' , -f'' , n(n-1)x^{n-2}\right)[/tex]
Its third coordinate is non-zero for [tex]x\gt 0[/tex], so the osculating plane exists at every point. The projection of [tex]M=\vec{r}(x)[/tex] onto [tex]Oy[/tex] is [tex]Q\left(0 , x^n , 0\right)[/tex] and [tex]\overrightarrow{MQ}=\left(-x , 0 , -f\right)[/tex]. The condition [tex]\overrightarrow{MQ}\centerdot\vec{B}=0[/tex] reads
[tex]-x\left(nx^{n-1}f''-n(n-1)x^{n-2}f'\right)-n(n-1)x^{n-2}f=0[/tex]
Dividing by [tex]-nx^{n-2}\ne 0[/tex] we get an Euler equation:
[tex]x^2f''-(n-1)xf'+(n-1)f=0[/tex]
Substituting [tex]f=x^m[/tex]:
[tex]m(m-1)-(n-1)m+(n-1)=m^2-nm+n-1=(m-1)(m-n+1)=0\Rightarrow m=1\text{ or }m=n-1[/tex]
Since [tex]n\ge 3[/tex], the roots are different ([tex]n-1\ge 2[/tex]), so [tex]x[/tex] and [tex]x^{n-1}[/tex] are two linearly independent solutions of this linear second-order equation on [tex]x\gt 0[/tex] (after division by [tex]x^2[/tex] its coefficients are continuous there). Hence all solutions are
[tex]f(x)=C_1x+C_2x^{n-1},\quad C_1,C_2\in\mathbb{R}[/tex]

Solution:
Let [tex]\vec{x}(s)[/tex] be the natural parametrization of [tex]C[/tex] with Frenet frame [tex]\vec{t},\ \vec{n},\ \vec{b}[/tex]; the prime denotes [tex]\frac{d}{ds}[/tex]. The Frenet formulas are
[tex]\vec{t}^{\,\prime}=\kappa\vec{n},\quad \vec{n}^{\,\prime}=-\kappa\vec{t}+\tau\vec{b},\quad \vec{b}^{\,\prime}=-\tau\vec{n}[/tex]
The center of the osculating circle lies on the principal normal at distance [tex]\frac{1}{\kappa}[/tex], so [tex]C^*:\ \vec{x}^*(s)=\vec{x}(s)+\frac{1}{\kappa}\vec{n}(s)[/tex]. Since [tex]\kappa[/tex] is constant,
[tex]\frac{d\vec{x}^*}{ds}=\vec{t}+\frac{1}{\kappa}\left(-\kappa\vec{t}+\tau\vec{b}\right)=\frac{\tau}{\kappa}\vec{b}[/tex]
As [tex]\tau\ne 0[/tex], the curve [tex]C^*[/tex] is regular and its arc length [tex]s^*[/tex] satisfies [tex]\frac{ds^*}{ds}=\frac{|\tau|}{\kappa}[/tex]. Let [tex]\varepsilon=\frac{\tau}{|\tau|}[/tex]; it is a constant ([tex]1[/tex] or [tex]-1[/tex]), because [tex]\tau[/tex] is continuous and never zero. The unit tangent of [tex]C^*[/tex] is [tex]\vec{t}^*=\varepsilon\vec{b}[/tex], and
[tex]\frac{d\vec{t}^*}{ds^*}=\varepsilon\,\vec{b}^{\,\prime}\,\frac{ds}{ds^*}=\varepsilon\left(-\tau\vec{n}\right)\frac{\kappa}{|\tau|}=-\kappa\vec{n}[/tex]
Hence the curvature of [tex]C^*[/tex] is [tex]\kappa^*=\left|\frac{d\vec{t}^*}{ds^*}\right|=\kappa[/tex], a constant, and its principal normal is [tex]\vec{n}^*=-\vec{n}[/tex].
The binormal [tex]\vec{b}^*[/tex] completes [tex]\vec{t}^*,\ \vec{n}^*[/tex] to a right-handed orthonormal frame. The frame [tex]\vec{b},\ -\vec{n},\ \vec{t}[/tex] is right-handed (it is obtained from [tex]\vec{t},\ \vec{n},\ \vec{b}[/tex] by swapping [tex]\vec{t}[/tex] and [tex]\vec{b}[/tex] and changing the sign of [tex]\vec{n}[/tex]), and so is [tex]\varepsilon\vec{b},\ -\vec{n},\ \varepsilon\vec{t}[/tex]; hence [tex]\vec{b}^*=\varepsilon\vec{t}[/tex].
For the torsion we differentiate [tex]\vec{n}^*[/tex]:
[tex]\frac{d\vec{n}^*}{ds^*}=-\vec{n}^{\,\prime}\,\frac{\kappa}{|\tau|}=\left(\kappa\vec{t}-\tau\vec{b}\right)\frac{\kappa}{|\tau|}=\frac{\kappa^2}{|\tau|}\vec{t}-\varepsilon\kappa\vec{b}[/tex]
On the other hand, by the Frenet formula for [tex]C^*[/tex]
[tex]\frac{d\vec{n}^*}{ds^*}=-\kappa^*\vec{t}^*+\tau^*\vec{b}^*=-\varepsilon\kappa\vec{b}+\varepsilon\tau^*\vec{t}[/tex]
The coefficients of [tex]\vec{b}[/tex] agree; comparing the coefficients of [tex]\vec{t}[/tex] we get
[tex]\varepsilon\tau^*=\frac{\kappa^2}{|\tau|}\Rightarrow \tau^*=\frac{\kappa^2}{\varepsilon|\tau|}=\frac{\kappa^2}{\tau}[/tex]
This proves both claims.

Solution:
Take the axis of rotation as [tex]Oz[/tex] with [tex]O[/tex] at the origin, and put [tex]c=\sqrt{a^2-b^2}\gt 0[/tex]. Then
[tex]T:\ \left(\sqrt{x^2+y^2}-a\right)^2+z^2=b^2[/tex]
Position of [tex]\Pi[/tex]. For [tex]P\in T[/tex] let [tex]Q[/tex] be the center of the meridian circle through [tex]P[/tex] ([tex]Q[/tex] lies on the circle [tex]x^2+y^2=a^2,\ z=0[/tex]). The tangents to the meridian and to the parallel through [tex]P[/tex] are both perpendicular to [tex]QP[/tex], so [tex]QP[/tex] is the normal of [tex]T[/tex] at [tex]P[/tex], and [tex]|QP|=b[/tex].
The plane [tex]z=0[/tex] is not tangent to [tex]T[/tex] ([tex]QP[/tex] is vertical only at the highest and lowest points of the meridians, which are not in this plane). So [tex]\Pi[/tex] meets [tex]z=0[/tex] in a line through [tex]O[/tex]; after a rotation about [tex]Oz[/tex] (which maps [tex]T[/tex] onto itself) this line is [tex]Oy[/tex], and
[tex]\Pi:\ \alpha x+\gamma z=0,\quad \alpha\ne 0[/tex]
If [tex]\Pi[/tex] touches [tex]T[/tex] at [tex]P[/tex], then [tex]QP\perp\Pi[/tex], i.e. [tex]\overrightarrow{QP}\parallel\left(\alpha , 0 , \gamma\right)[/tex]. But [tex]\overrightarrow{QP}[/tex] lies in the meridian plane through [tex]Q=\left(a\cos\varphi , a\sin \varphi , 0\right)[/tex], spanned by [tex]\left(\cos\varphi , \sin \varphi , 0\right)[/tex] and [tex]\left(0 , 0 , 1\right)[/tex]; since [tex]\alpha\ne 0[/tex], this forces [tex]\sin \varphi=0[/tex], i.e. [tex]Q=\left(\pm a , 0 , 0\right)[/tex]. Moreover, the distance from [tex]Q[/tex] to [tex]\Pi[/tex] equals [tex]|QP|=b[/tex]:
[tex]\frac{|\alpha|a}{\sqrt{\alpha^2+\gamma^2}}=b\iff \gamma^2b^2=\alpha^2\left(a^2-b^2\right)\iff \gamma=\pm\frac{c}{b}\alpha[/tex]
So [tex]\Pi:\ bx\mp cz=0[/tex]; since the reflection [tex]z\to -z[/tex] maps [tex]T[/tex] onto itself, we may assume [tex]\Pi:\ bx-cz=0[/tex].
Equation of [tex]T[/tex]. Expanding and using [tex]a^2-b^2=c^2[/tex] (squaring is reversible, since the left side is positive):
[tex]T:\ x^2+y^2+z^2+c^2=2a\sqrt{x^2+y^2}\iff\left(x^2+y^2+z^2+c^2\right)^2=4a^2\left(x^2+y^2\right)[/tex]
Factorization. Let [tex]\rho=x^2+y^2+z^2[/tex]. Using [tex]a^2=b^2+c^2[/tex]:
[tex]\left(x^2+(y+b)^2+z^2-a^2\right)\left(x^2+(y-b)^2+z^2-a^2\right)=\left(\rho-c^2+2by\right)\left(\rho-c^2-2by\right)=\left(\rho-c^2\right)^2-4b^2y^2[/tex]
and a direct check (the difference of the two sides is [tex]4c^2\rho-4c^2\left(x^2+y^2\right)-4c^2z^2=0[/tex]) gives
[tex]\left(\rho+c^2\right)^2-4a^2\left(x^2+y^2\right)=\left(\rho-c^2\right)^2-4b^2y^2-4\left(b^2x^2-c^2z^2\right)[/tex]
[tex]\left(\rho+c^2\right)^2-4a^2\left(x^2+y^2\right)=\left(x^2+(y+b)^2+z^2-a^2\right)\left(x^2+(y-b)^2+z^2-a^2\right)-4(bx-cz)(bx+cz)[/tex]
For the points of [tex]\Pi[/tex] we have [tex]bx-cz=0[/tex], so such a point lies on [tex]T[/tex] if and only if it lies on one of the spheres
[tex]S_1:\ x^2+(y+b)^2+z^2=a^2,\qquad S_2:\ x^2+(y-b)^2+z^2=a^2[/tex]
Thus [tex]T\cap\Pi=\left(S_1\cap\Pi\right)\cup\left(S_2\cap\Pi\right)[/tex]. The centers [tex]\left(0 , \mp b , 0\right)[/tex] of the spheres lie in [tex]\Pi[/tex], so each intersection is a great circle of radius [tex]a[/tex]. Therefore the section of the torus by [tex]\Pi[/tex] consists of two circles of radius [tex]a[/tex] (the Villarceau circles).