Solutions to Exercises

Exercise 2.1 $\vec{u}$ and $\vec{v}$ are perpendicular to each other.

Exercise 2.2 It suffices to prove the hint because $\sin^2\theta+\cos^2\theta=1$ .

$\begin{align*} \|\vec{u}\times\vec{v}\|^2+\langle\vec{u}, \vec{v}\rangle^2&=(y_1z_2-y_2z_1)^2+(x_2z_1-x_1z_2)^2+(x_1y_2-x_2y_1)^2+(x_1x_2+y_1y_2+z_1z_2)^2\\ &=y_1^2z_2^2+y_2^2z_1^2+x_2^2z_1^2+x_1^2y_2^2+x_2^2y_1^2+x_1^2x_2^2+y_1^2y_2^2+z_1^2z_2^2\\ &=(x_1^2+y_1^2+z_1^2)(x_2^2+y_2^2+z_2^2)\\ &=\|\vec{u}\|^2\|\vec{v}\|^2 \end{align*}$

Exercise 2.4 Let $\alpha(t)=(x(t), y(t), z(t))$ and $\beta(t)=(u(t), v(t), w(t))$ be the smooth parametrizations of $\alpha$ and $\beta$ . Then

$\begin{align*} \frac{d}{dt}\langle\alpha(t), \beta(t)\rangle&=\frac{d}{dt}(x(t)u(t)+y(t)v(t)+z(t)w(t))\\ &=(x'(t)u(t)+x(t)u'(t))+(y'(t)v(t)+y(t)v'(t))+(z'(t)w(t)+z(t)w'(t))\\ &=(x'(t)u(t)+y'(t)v(t)+z'(t)w(t))+(x(t)u'(t)+y(t)v'(t)+z(t)w'(t))\\ &=\langle\alpha'(t), \beta(t)\rangle+\langle\alpha(t), \beta'(t)\rangle \end{align*}$ If $\alpha$ is a smooth parametrized curve lying on a sphere of radius $R$ centered at the origin in $\mathbb{R}^3$ , then $\langle\alpha(t), \alpha(t)\rangle=R^2$ . Differentiating both sides with respect to $t$ and using the above result, we have $\langle\alpha'(t), \alpha(t)\rangle+\langle\alpha(t), \alpha'(t)\rangle=0\Rightarrow\langle\alpha'(t), \alpha(t)\rangle=0$

Exercise 2.5 By Exercise 2.4, $\langle\alpha'(t), \alpha''(t)\rangle=0$ . By the hint of Exercise 2.2,

$\|\alpha'(t)\times\alpha''(t)\|=\|\alpha'(t)\|\|\alpha''(t)\|=\|\alpha''(t)\|$ The result follows.

Exercise 2.6

$\begin{align*} \kappa(t)&=\frac{\|(-R\sin t, R\cos t, b)\times(-R\cos t, -R\sin t, 0)\|}{(R^2+b^2)^{\frac{3}{2}}}\\ &=\frac{\|(bR\sin t, -bR\cos t, R^2)\|}{(R^2+b^2)^{\frac{3}{2}}}\\ &=\frac{R(b^2+R^2)^{\frac{1}{2}}}{(R^2+b^2)^{\frac{3}{2}}}\\ &=\frac{R}{R^2+b^2} \end{align*}$

Exercise 2.7 Note that $\alpha'(t)= (3t^2, 6t^5, 9t^8)$ and $\alpha'(0)=\vec{0}$ . So $\alpha(t)$ is not a regular parametrized curve. This curve can be reparametrized as

$\widetilde{\alpha}(t)=(t, t^2, t^3)$ This parametrization is regular. So $\begin{align*} \widetilde{\alpha}'(t)&=(1, 2t, 3t^2)\\ \widetilde{\alpha}''(t)&=(0, 2, 6t)\\ \kappa(t)&=\frac{\|(1, 2t, 3t^2)\times(0, 2, 3t^2)\|}{\|(0, 2, 6t)\|^3}\\ &=\frac{\|(6t^3-6t^2, -3t^2, 2)\|}{(4+36t^2)^{\frac{3}{2}}}\\ &=\frac{(4+36t^4(t-1)^2+9t^4)^{\frac{1}{2}}}{(4+36t^2)^{\frac{3}{2}}} \end{align*}$ So $\kappa(0)=\frac{1}{4}$ .

Exercise 2.8.1 By Exercise 2.2, $\vec{u}\times\vec{v}=\vec{0}$ iff $\sin\theta=0$ , where $\theta$ is the included angle of $\vec{u}$ and $\vec{v}$ , iff $\theta=0$ or $\pi$ , iff $\vec{u}$ and $\vec{v}$ are multiple of each other.

Exercise 2.10 Observe that $\langle N, \vec{x}_u\rangle=0$ , because the normal vector is defined to be perpendicular to the tangent vector. Hence

$\begin{align*} \langle N, \vec{x}_u\rangle&=0\\ \frac{d}{du}\langle N, \vec{x}_u\rangle&=0\\ \left\langle\frac{dN}{du}, \vec{x}_u\right\rangle+\langle N, \vec{x}_{uu}\rangle&=0\\ \langle N, \vec{x}_{uu}\rangle&=\left\langle -\frac{dN}{du}, \vec{x}_u\right\rangle\\ &=e \end{align*}$ The other cases are similar.

Exercise 3.1 By the hint of Exercise 2.2,

$\begin{align*} \|\vec{x}_u\times\vec{x}_v\|&=\sqrt{\|\vec{x_u}\|^2\|\vec{x}_v\|^2-\langle\vec{x}_u, \vec{x}_v\rangle^2}\\ &=\sqrt{EG-F^2} \end{align*}$

Exercise 3.2

$\begin{align*} \int_0^{2\pi}\int_0^{2\pi}\|\vec{x}_u\times\vec{x}_v\|dudv&=\int_0^{2\pi}\int_0^{2\pi}r(R+r\cos u)dudv\\ &=\int_0^{2\pi}2\pi rRdv\\ &=4\pi^2rR \end{align*}$

Exercise 3.3

$\begin{align*} \vec{x}_u&=(-av\sin u, av\cos u, b)\\ \vec{x}_v&=(a\cos u, a\sin u, 0)\\ \vec{x}_u\times\vec{x}_v&=(-ab\sin u, ab\cos u, -a^2v)\\ \|\vec{x}_u\times\vec{x}_v\|&=a\sqrt{b^2+v^2}\\ \vec{x}_{uu}&=(-av\cos u, -av\sin u, 0)\\ \vec{x}_{uv}&=(-a\sin u, a\cos u, 0)\\ \vec{x}_{vv}&=\vec{0}\\ N&=\left(-\frac{b\sin u}{\sqrt{b^2+v^2}}, \frac{b\cos u}{\sqrt{b^2+v^2}}, -\frac{av}{\sqrt{b^2+v^2}}\right)\\ \end{align*}$ $E=a^2v^2+b^2, F=0, G=a^2$ $e=g=0, f=\frac{ab}{\sqrt{b^2+v^2}}$ Hence $H=0$ and helicoid is indeed a minimal surface.

Exercise 4.1 We shall first show that the above parametrization is conformal:

$\begin{align*} \vec{x}_u&=(-\cosh u\sin v, \cosh u\cos v, 0)\\ \vec{x}_v&=(-\sinh u\cos v, -\sinh u\sin v, -1) \end{align*} \[E=G=\cosh^2 u, F=0\]$ Observe that $\begin{align*} \left(\frac{\partial^2}{\partial u^2}+\frac{\partial^2}{\partial v^2}\right)(-\sinh u\sin v)&=-\sinh u\sin v+\sinh u\sin v\\ &=0\\ \left(\frac{\partial^2}{\partial u^2}+\frac{\partial^2}{\partial v^2}\right)(\sinh u\cos v)&=\sinh u\cos v-\sinh u\cos v\\ &=0\\ \left(\frac{\partial^2}{\partial u^2}+\frac{\partial^2}{\partial v^2}\right)(-v)&=0 \end{align*}$