Recent content by Simone Furcas

I'll try answering by smartphone. It is an exercise to show the student remember that first fundamental form is a tensorial product addition. I red in my note my professor 's idea, he showed as you do the tensorial product, furthermore he takes 2 general vector of the space and with an unusual...

I translate it, it's in Italian. I've just solved it by myself. This is first fundamental form##ds^2=cos^2(v)du^2+dv^2## v∈(-##\pi##/2,##\pi##/2). Check it is bilinear, symmetric and positive.

I solved it. I solved the exercise.

How could I proof this ##ds^2=cos^2(v)du^2+dv^2## is bilinear?

##j^T * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}## there was a mistake, it was the transpose not the inverse.

I did a mistake, ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{cosh{y}}##, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{cosh{y}} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 &...

I have ##ds^2=\cos^2(v)du^2 + dv^2## , i take a coordinate transformation x=u and cos(v)=##\frac{1}{(cosh(y))}##, I have to find a new metric with this coordinate transformation and proof it is in agreement with Theorema Egregium. ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{(y^2(1-y^2))}##...

Is it the same with an ellipsoid (a*cos(x)*sen(y),b*sen(x)*sen(y),c*cos(y)) ? N is (a*cos(x)*sen(y),b*sen(x)*sen(y),c*cos(y)) ?

I was thinking it was very difficult... Now I think to be a bit silly! :) thx

Let r:R2 →R3 be given by the formula Compute the second fundamental form with respect to this basis (Hint: There’s a shortcut to computing the unit normal n). I can't find thi shortcut, does anyone help me? I'm solving it with normal vector and first and second derivate, but I obtained...