1
$\begingroup$

In Pressley's Elementary Differential geometry, Q6.4.7 p148 stomps me.

Question: Let $\sigma(u,v)$ be a surface patch with standard unit normal $N$. Show that $$N \times \sigma_u = \frac{E\sigma_v - F\sigma_u}{(EG-F^2)^{1/2}}$$ with $E = \langle\sigma_u , \sigma_u \rangle , F = \langle \sigma_u , \sigma_v \rangle$ and $ G = \langle \sigma_v , \sigma_v \rangle$.

Answer: $N \times \sigma_u = \alpha \sigma_u+\beta \sigma_v$ and $(N \times \sigma_u) \cdot\sigma_u = 0 , (N \times \sigma_u) \cdot \sigma_v = (EG-F^2)^{1/2}$ (makes sense)
This implies that $\alpha E+\beta F = 0 , \alpha F + \beta G = (EG-F^2)^{1/2}$ (does not make sense) How are the equalities deduced? any hint is appreciated!

$\endgroup$
2
$\begingroup$

Taking the dot product of $N \times \sigma_u$ with $\sigma_u$ we get

$$ 0 = (N \times \sigma_u) \cdot \sigma_u = (\alpha \sigma_u + \beta \sigma_v) \cdot \sigma_u = \alpha (\sigma_u \cdot \sigma_u) + \beta ( \sigma_v \cdot \sigma_u) = \alpha E + \beta F. $$

Taking the dot product of $N \times \sigma_u$ with $\sigma_v$ we get

$$ (EG - F^2)^{1/2} = (N \times \sigma_u) \cdot \sigma_v = (\alpha \sigma_u + \beta \sigma_v) \cdot \sigma_v = \alpha (\sigma_u \cdot \sigma_v) + \beta ( \sigma_v \cdot \sigma_v) = \alpha F + \beta G. $$

$\endgroup$
  • $\begingroup$ thank you! Silly of me not to have thought of that! $\endgroup$ – rannoudanames Mar 25 '17 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.