Recent content by psi^

hmmm the formula is right, but are you sure that sin function receives the input in degrees? Because i used the a sin function that receives the value of the angle in radians and gave it a input in degrees and the graph result is almost the same. At t=1,2s gives me 1,544584079m.

y it could be in \ R^2 my bad :rolleyes:

The vector input in f is linear combination of the basis (and so is the output vector because its an endomorphism). think f like this: f(x_1, x_2, x_3) = (x_2 + x_3, x_1 + x_3, x_2 - x_1) Hints: a) multiply A by (x_1, x_2, x_3) and verify results b) if Av = v then v has to be a combination of...

2x-z=4 represents the family of points in the form (x, 0, 2x-4) and the points (0,0,4) and (1,0,-2) are part of that family, with them we can form the vector (1,0,2) So we get the family of vectores parallel to the line by x.(1,0,2) Notice from your solution its easy to see that they...

yeah :) i am new to TeX too :'( next time i'll do it on paper first! LoL

d^2 = x^2 + y^2 = v_{0x}^2 t^2 + v_{0y}^2 t^2 - v_{0y} g t^3 + \frac{1}{4}g^2t^4 \Rightarrow v_0^2 t^2 + \frac{1}{4} g^2 t^4 > v_{0y} g t^3 \Leftrightarrow \frac{v_0^2}{t}+ \frac{1}{4} g^2 t > v_0 cos(\alpha) g \Leftrightarrow \frac{v_0}{g t}+ \frac{g t}{4v_0} > cos(\alpha) \Leftrightarrow...

Humm did you notice that v0y is negative in this case right?

All you have to do is to compute the change in frequency due to Doppler's Effect, you have the speed of the source and the speed of the wave. Check the general section on the wiki for more details on the formulas :) http://en.wikipedia.org/wiki/Doppler_effect

y = y0 + v0y*t - (1/2)*g*t^2 If you consider your referential center at the floor: y0 = 14, the graph is going to decrease from y = 14 to y = 0. If you consider your referential center at the edge of the roof: y0 = 0, the graph is going to decrease from y = 0 to y = -14. You only need to...

If f is a isometry, u.v=f(u).f(v) holds. So the vector norm (u.u)^1/2 and distance stays the same.

But in the case of a reflection the transformation of a midpoint is still a midpoint, no?

:) Just notice, you can find c when t = 0, v = v0: ln(g + kv) = -kt + c => ln(g + kv0) = -k.0 + c <=> c = ln(g + kv0) So, g + kv = e^{-kt + c} <=> g + kv = e^{-kt + ln(g + kv0)} <=> g + kv = e^{-kt}.(g + kv0)