Recent content by Andrés85

I solved the problem deriving two times the function f(t) = G(tx, ty). Thanks.

I can't increase the degree of the Taylor polynomial because G is C2, so the second degree term is the remainder written in matrix notation. HG(c) is the Hessian matrix of G evaluated at c, where c lies on the segment that goes from (0,0) to (x,y).

Homework Statement Let G:R2\rightarrowR be a C2 function such that G(tx,ty)=t2G(x, y). Show that: 2G(x,y)=(x,y).HG(0,0).(x,y)t The Attempt at a Solution G is C2, so its Taylor expansion is: G(x,y) = G(0,0) + \nablaG(0,0).(x,y) + \frac{1}{2}(x,y).HG(c).(x,y)t, where c lies on...

I solved it that way! Thanks!

if B(x,y) is an open ball with center (x,y) and radius r, i show that if (a,b) belongs to the ball B, then (a,b) belongs to the set, so a^2 + b^2 < 7 |(a,b)| = |(a,b) - (x,y) + (x,y)| = |(a-x, b-y) + (x,y)| ≤ |(a-x, b-y)| + |(x,y)| |(a-x, b-y)| < r, so |(a-x, b-y)| + |(x,y)| < r +...

I think r < min {√7-|(x,y)| ; |(x,y)| - 1} works. I tried to separate the problem in two sets, because the intersection of two open sets is an open set, and proved that x^2+y^2<7 is open, but the method I used with that proof don't work with the set 1 < x^2 + y^2

Homework Statement Show that {(x,y)ℝ^2/1<x^2+y^2<7} is an open set. Homework Equations The Attempt at a Solution =(

i think in D) T3 = m1g + m2g + Mpg - T1 - T2 so T3 is less than m1g + m2g + mpg

i was wrong, i think the work is 0 because the centripetal force is perpendicular to the direction of v

mmmm i think in this case the centripetal force is the one applied by the spring, so (constant)x(4cm) = (m)x(v^2/r) solve and get v, and then work of the disk = (1/2)x(m)x(v^2) + (1/2)x(I)x(W^2) = (1/2)x(m)x(v^2) + (1/2)x(1/2x m xR^2)x(W^2)...