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This is not a homework problem, but a problem from an old exam, published online, which I am trying to solve.

Consider a regular parameterized surface $X:U\to\mathcal{R}^3$ with Gauss map $N:X(U) \to S^2$ and principal curvatures $\kappa_1 = 1/r_1$ and $\kappa_2 = 1/r_2$. Let $r\in \mathcal{R}$ and define a regular parametrized surface $X^r:U\to\mathcal{R}^3$ by $$X^r(u,v) = X(u,v)+rN(u,v).$$ Show that the principal curvatures of $X^r$ are $k_1(r)=1/(r_1-r)$ and $k_2(r)=1/(r_2-r)$, respectively.

In principle what we need to do is to determine the second and first fundamental form of $X^r$ in terms of the corresponding forms of $X$. We could then calculate the Gaussian curvature $K$, the mean curvature $K$, and calculate the principal curvatures $\kappa = H \pm \sqrt{H^2-K}$. However, the calculations for this approach seem to get very messy, so I suspect it should be done differently. Any suggestions? Using $\langle X_u,N \rangle = \langle X_v,N \rangle = 0$, one can easily see that $$X^r_u \times X^r_v = X_u \times X_v + r^2N_u \times N_v.$$ If we choose the $+$ sign in calculating the normal from the cross product of the tangent basis vectors, and using the fact that $N_u \times N_v = -N_{uv} \times N \perp X_u \times X_v$, we get the Gauss map $$N^r = \frac{|X_u\times X_v|N+r^2N_u\times N_v}{\sqrt{|X_u\times X_v|^2 + r^4|N_u\times N_v|^2 }}.$$ From this, I see no reasonably simple way to get at the second fundamental form of $X^r$ in terms of the second (and first) fundamental form of $X$.

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Apparently this question was asked before, see Differential geometry - proving an expression for the principal curvature. Nevertheless, I will write an answer in my own words below. Feel free to delete the question if it's deemed appropriate to do so.

First of all, we note that the Gauss map of $X^r$ is also $N$. Indeed, $$\langle X^r_u,N \rangle = \langle X_u+rN_u,N\rangle = r\langle N_u,N \rangle = (r/2)\langle N,N \rangle_u = 0$$ as $\langle N,N \rangle = 1$. Similarly, $\langle X^r_v,N\rangle = 0$.

Now, let $\gamma$ be a curve in $X(U)$ such that $\gamma'(0)$ is an eigenvector of the shape operator $S_{\gamma(0)}=-dN_{\gamma(0)}$. Then $S_{\gamma(0)}(\gamma'(0))=-dN_{\gamma(0)}(\gamma'(0))=\kappa\gamma'(0)$, where $\kappa = \kappa_1$ or $\kappa = \kappa_2$.

Consider the curve $\gamma_r$ in $X^r(U)$ defined by $\gamma_r(t)= \gamma(t) + rN(\gamma(t))$. We have $$\gamma_r'(0) = \gamma'(0)+rdN_{\gamma(0)}(\gamma'(0))=(1-r\kappa)\gamma'(0)$$ Furthermore, we have $N(\gamma_r(t))=N(\gamma(t))$, which implies that $dN_{\gamma_r(0)}(\gamma_r'(0)) = dN_{\gamma(0)}(\gamma'(0))$. Thus it follows that $$S_{\gamma_r(0)}(\gamma_r'(0)) = - dN_{\gamma_r(0)}(\gamma_r'(0)) = -dN_{\gamma(0)}(\gamma'(0)) = S_{\gamma(0)}(\gamma'(0))=\kappa\gamma'(0) = \frac{\kappa}{1-r\kappa}\gamma_r'(0).$$ Hence $\gamma_r'(0)$ is an eigenvector of the shape operator, so it follows that $\gamma_r'(0)$ is a principal direction of curvature. The eigenvalue $\kappa/(1-r\kappa)$ is the corresponding principal curvature (maximum or minimum). But $\kappa = 1/r_i$, where $i=1$ or $i=2$, which shows that the principal curvatures of $X^r$ are $1/(r_1-r)$ and $1/(r_2-r)$, as claimed.

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