Principal curvature of a surface defined by an equation

Question

Let be $S$ a set of points $ (x,y,z)\in\mathbb{R^3}$ that satisfy the equation $x^3+y^2+z^2=1$. Calculate the main curvatures and Gaussian curvature at the points $p_1=(1,0,0)$, $p_2=(0,1,0)$ e $p_3=(0,0,1)$.

I want to know if this is correct:

We have that the tangent plane in $p_1$ is generated by $(0,1,0)$ and $(0,0,1)$ (from the Kernel of $dg_{p_1}$ where $g=x^3+y^2+z^2-1$), then we have that $k_1= <N(p_1),\gamma_1''(0)>$ and $k_2= <N(p_1),\gamma_2''(0)>$ (where $k_1$ and $k_2$ are the two main curvatures and $N$ is the Gauss Map, in this case $N=\frac{dg}{||dg||}$) where $\gamma_1,\gamma_2$ are two curves in $S$ that $\gamma_1(0)=\gamma_2(0)=p_1$ and $\gamma'_1(0)=(0,1,0)$,$\gamma'_2(0)=(0,0,1)$. We have $N(p_1)=(1,0,0)$, if we choose $\gamma_1(t)=((1-t^2)^{1/3},t,0)$ and $\gamma_2(t)=((1-t^2)^{1/3},0,t)$, we obtained $k_1=k_2=-2/3$, then Gaussian curvature is $4/9$.

Doing the same with $p_2$ we obtained $k_1=0$ and $k_2=-1$, so $K(p_2)=0$ and doing the same with $p_3$ we obtained $k_1=0$ and $k_2=-1$, so $K(p_3)=0$.

Answer 1

We start with $p_1=(1,0,0)$ and consider the parametrization $\varphi_1: \mathbb{R}^2/S^1 \to S$ defined as $\varphi_1(y,z)=((1-y^2-z^2)^{1/3},y,z)$. Notice that $\varphi_1(0,0)=p_1$, and moreover, $\dfrac{\partial\varphi_1}{\partial y}(0,0)=(0,1,0)$ and $\dfrac{\partial\varphi_1}{\partial z}(0,0)=(0,0,1)$ are the principal directions in $p_1$. Then, continue the reasoning as you did to find the principal curvatures at $p_1$.

For $p_2=(0,1,0)$ consider the parametrization $\varphi_2: \mathbb{R}^2/{(x,z)|x^3+z^2=1} \to S$ defined as $\varphi_2(x,z)=(x,\sqrt(1-x^3-z^2),z)$. Notice that $\varphi_2(0,0)=p_2$, and moreover, $\dfrac{\partial\varphi_2}{\partial x}(0,0)=(1,0,0)$ and $\dfrac{\partial\varphi_2}{\partial y}(0,0)=(0,0,1)$ are the principal directions in $p_2$. Then, continue the reasoning as you did to find the principal curvatures at $p_2$.

In the end $p_3=(0,0,1)$ consider the parametrization $\varphi_3: \mathbb{R}^2/{(x,z)|x^3+y^2=1} \to S$ defined as $\varphi_3(x,y)=(x,y,\sqrt(1-x^3-y^2))$. Notice that $\varphi_3(0,0)=p_3$, and moreover, $\dfrac{\partial\varphi_3}{\partial x}(0,0)=(1,0,0)$ and $\dfrac{\partial\varphi_3}{\partial y}(0,0)=(0,1,0)$ are the principal directions in $p_3$. Then, continue the reasoning as you did to find the principal curvatures at $p_3$.

Answer 2

Guys i think i undestand what do you mean:

for example if we consider the surfaces that is the image of the parametrizzazion $\varphi: (0, +\infty) \times (0, +\infty) \to \mathbb{R}^3$ defined as $\varphi(u,v)=(\frac{1}{2}u^2,\frac{1}{2}v^2,u-v)$, if we want to calculate the principal direction in $\varphi(1,1)=(\frac{1}{2},\frac{1}{2},0)$ we have that $w_1=\dfrac{\partial\varphi}{\partial u}(1,1)=(1,0,1)$ and $w_2=\dfrac{\partial\varphi}{\partial v}(1,1)=(0,1,-1)$ and the representation matrix of $dN_p$ compared to this basis is $$ \begin{pmatrix} \frac{2}{3 \sqrt3} & \frac{1}{3 \sqrt3} \\ -\frac{1}{3 \sqrt3} & -\frac{2}{3 \sqrt3} \end{pmatrix}. $$ that is not diagonal, so $\dfrac{\partial\varphi}{\partial u}(1,1)$ and $\dfrac{\partial\varphi}{\partial v}(1,1)$ are not the principal direction in this case. But i can diagonalize the matrix, infact consider $$ H=\begin{pmatrix} 1 & 1 \\ \sqrt3-2 & -\sqrt3-2 \end{pmatrix}. $$ we have $$ \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & -\frac{1}{3} \end{pmatrix}= H^{-1} \begin{pmatrix} \frac{2}{3 \sqrt3} & \frac{1}{3 \sqrt3} \\ -\frac{1}{3 \sqrt3} & -\frac{2}{3 \sqrt3} \end{pmatrix} H $$ and so the principal directions are $w_1+(\sqrt3-2)w_2=(1,\sqrt3-2,3-\sqrt3)$ and $w_1+(-\sqrt3-2)w_2=(1,-\sqrt3-2,3+\sqrt3)$.