a = 6, b = 2 are the values

Let **a****xyz – ****a****y^3** **+xz^2 =*** bw^3* be a homogenous polynomial in

1. Show the view of V in affine patches *Ux, Uy, Uz, Uw. *when x=1,y=1, z=1, w=1.

2. What is the dimension of V?

3. Is V an irreducible variety?

4. Find all singular points.

5. Give the ideal of **V**. Is it prime? Is your variety irreducible? Describe the ring k(**V**) = O(V) of polynomials (regular functions) on **V**.

6. Calculate curvature at (at least two) smooth points.

Curvature_surfaces _definition.docx

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7. Describe the symmetries of your surface **V. Is it ** bounded or unbounded?

8. Can you find a line on your surface?

Plane curve C = (*x*(t),*y*(t)): k =1/R of the circle of best fit. k = | *x’*(t)* y’’*(t) – *x’’*(t)* y’*(t)| /(*x’*(t)2+*y’*(t)2)3/2

OR the tangent unit vector T(t) = (*x’*(t),*y’*(t))/|| T(t) || , where the norm || T(t) ||T= (*x’*(t)2+*y’*(t)2)1/2 .

k(t) = || T’(t) ||= (*x’*(t)2+*y’*(t)2)1/2

For a surface **V** defined by the equation *F (x,y,z,w)* = 0 in **P3**, the Gaussian curvature at a given point with coordinates point (*x,y,z,* 1) is given by the formula below.

1. Example:

Let *F (x,y,z,w)* = *x+y+z+w+* 1 = 0 in **P3** . Then **V** is a plane, and take the point (1,-1,-1,1) on the plane. Now consider **R3 **when *w*= 1. Then derivatives *Fx*=1, *Fy*=1, *Fz*=1, hence all not zero, i.e. the denominator equals to 9. Since all second derivatives are equal to zero, the numerator is equal to zero. Hence curvature of the plane at the given points 0. But the calculations do not depend on the point we have started with. Therefore the curvature is zero at every point on the plane.

2. Calculate the curvature of all quadratic surfaces.

3. Calculate the curvature of your surface at several non singular points and analyze the results for a pattern (note you may have to permute the coordinates to make sure that the denominator is not zero!) How does the curvature change on your surface? Is it always positive or negative?

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