Find angles, length of edges, and area of curvilinear triangle

Consider the helicoid S given by the parametrization x(u,v)=(vcosu,vsinu,u). Let T be the curvilinear triangle on S which is the image under x of the triangle {(u,v): 0<=u<=a, o<=v<=sinhu} for some a>0. Find the angles of T, the length of its edges, and its area.

Re: Find angles, length of edges, and area of curvilinear triangle

Hey julietteeden.

I don't know the terminology used in common differential geometry but the arc-length formalization should give the length and a tensor formulation to translate the vector identities between the curved space and R^3 should give the area (or you could use the vector calculus result by calculating the normal of the surface and calculating a surface integral):

Surface integral - Wikipedia, the free encyclopedia

The angles should have a tensor formulation with regards to the inner product between co-ordinate spaces.

Re: Find angles, length of edges, and area of curvilinear triangle

I don't know what curvilinear triangle is either. It is not in my textbook and my professor has never mentioned it.

Re: Find angles, length of edges, and area of curvilinear triangle

Take a look at this:

Bézier triangle - Wikipedia, the free encyclopedia

That is one way of generating a curvilinear triangle, but my best guess is that it refers a triangle on either a sphere or a saddle point and not something as complex as the normal bezier triangle.

Re: Find angles, length of edges, and area of curvilinear triangle

I'm not sure what a tensor formulation is.

Re: Find angles, length of edges, and area of curvilinear triangle

Tensor formulation is just the mathematics of co-ordinate system invariant mathematics.

So typically we have calculus and geometry in R^n and tensors allow us to do this symbolically and also numerically in other systems that have the appropriate bijective properties between systems (which is given with the metric tensor and bijectivity of points between systems).

Once you have this then there are co-ordinate system invariant ways to do calculus in any system and this is why this approach is used in for example general relativity because you have all the physics developed in R^n (or at least R^3/R^4) but curved geometry is a different system where everything behaves different, so the co-ordinate invariant approach was developed and tools were made to deal with arbitary systems.

Since we know the results for calculus (including differentiation and integration for the geometry of R^n with vector results including inner and outer products), then we can use this as a bridge to do calculus in other co-ordinate systems and this is exactly what physicists do (sometimes other scientists as well but physics and engineering are the bigs ones apart from applied and pure mathematics).