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A parametric surface is defined as $$X=140u+20v-40uv-20, \ \ \ Y=80-80v \ \ \ \ Z=50-10u-50v+10uv$$ Where, $0\le u,v\le1$

Find out the maximum principal curvature of given surface.

My Try:

Parametric surface: $P(u,v)=(140u+20v-40uv-20, 80-80v,50-10u-50v+10uv)$ $$P_u=\frac{\partial P}{\partial u}=(140-40v,0,-10+10v), \ \ P_v=\frac{\partial P}{\partial v}=(20-40u,-80,-50+10u)$$ $$\text{Normal vector to the surface}, \vec n=P_u \times P_v=(800(w-1), 200(34-5u-9v), -1600(7-2v)) $$ $$\hat n =\frac{\vec n}{|\vec n|}=\frac{(800(w-1), 200(34-5u-9v), -1600(7-2v))}{\sqrt{25u^2+353v^2+90uv-340u-2130v+4308}}$$ $$P_{uu}=\frac{\partial^2 P}{\partial u^2}=(0,0,0), P_{uw}=\frac{\partial^2 P}{\partial u\partial v}=(-40,0,10), \ \ \ P_{vv}=\frac{\partial^2 P}{\partial v^2}=(0,0,0),$$ $$L=\hat n\cdot P_{uu}=0$$$$ M=\hat n\cdot P_{uv}=\frac{-400}{\sqrt{25u^2+353v^2+90uv-340u-2130v+4308}}$$ $$N=\hat n\cdot P_{vv}=0$$ $$E=P_u\cdot P_u=100(17v^2-114v+197)$$ $$F=P_u\cdot P_v=100(17uv-57u-13v+33)$$ $$G=P_v\cdot P_v=100(17u^2-27u+93)$$ Gauss curvature (K), $$K=\frac{LN-M^2}{EG-F^2}$$$$K=\frac{-16}{(25u^2+353v^2+90uv-340u-2130v+4308)(100u^2+1412v^2+921uv-1360u-9744v+16671)}$$ Mean Curvature (H), $$H=\frac{EN+GL-2FM}{2(EG-F^2)}$$ $k_{max}=H+\sqrt{H^2-K}$

I want to compute maximum principle curvature but the calculation becomes very difficult. Is there is any other easy method to find the maximum principal curvature of this parametric surface. Someone please help me. Thanks

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1 Answer 1

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The principal curvatures are the eigenvalues of the shape operator, whose matrix is given by $$ S = \begin{bmatrix} L & M \\ M & N \end{bmatrix} = \begin{bmatrix} 0 & M \\ M & 0 \end{bmatrix}. $$

The maximum principal eigenvalue is $|M|$. The value $|M|$ is maximal if the expression under the square root in the denominator is minimal. I believe you have made a mistake and that it should be $$ 25 u^2 + 353 v^2 + 90 uv - 340 u - 2436 v + 4308. $$ The global minimum of this function is $(\frac{13}{17}, \frac{57}{17})$, which is outside the square $0 \leq u, v \leq 1$. This means the minimum will lie on the boundary of the square.

Restrict the function to the 4 lines $u = 0$, $u = 1$, $v = 0$ and $v=1$, and find the minimum on that lines. You should only keep the minima that are on the line segments, i.e. check whether $0\leq u, v\leq 1$ is satisfied . Lastly you also calculate the value of the function in the four corners $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$. Then, by comparing all values, you can select the minimum value on the square.

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  • $\begingroup$ Thank you very much for your answer. I found $u=1.7772$ & $v=2.7904$ as the point of minima & $f(u,v)$ has minimum value $1034.055147$. Hence $M=400/\sqrt{1034.055147}=12.43$ But sir these values are out of range $0\le u,v \le 1$. Is the answer correct or is there is any mistake. Please help sir $\endgroup$
    – user766881
    Commented Apr 24, 2020 at 0:23
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    $\begingroup$ There is a mistake in your calculation, but I also made a typo (wrote $335v^2$ instead of $353v^2$). But even with corrections, I believe the global minimum of the expression lies outside the square. This means one must search for a minimum on the boundary of the square. The answer is edited and should help. $\endgroup$
    – Ernie060
    Commented Apr 24, 2020 at 16:56
  • $\begingroup$ Sir again there is a typo in the function in your answer. It should be $f(u,v)=25u^2+353v^2+90uv-340u-2130v+4308$. I have cross checked the global minima occurs at $u=2417/1360=1.7772$ & $v=3795/1360=2.7904$ $\endgroup$
    – user766881
    Commented Apr 24, 2020 at 17:39
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    $\begingroup$ And I typed in the parametrization $P$ in Mathematica and I get $2436$ as coefficient with $v$. Anyhow, apart from typos and calculations errors, the approach is correct. $\endgroup$
    – Ernie060
    Commented Apr 24, 2020 at 17:47
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    $\begingroup$ Thank you sir my problem is resolved. Many many thanks to you. $\endgroup$
    – user766881
    Commented Apr 24, 2020 at 18:38

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