Recent content by mrguru34

Ok guys! Thats treally helpful! Thanks :D One other thing how do you differentiate a vector...do you just differentiate each term by say t...or am i just being dumb! Thanks for being patient with me! :)

How do you then prove the point xi? And to be honest i don't really know how to do any of the rest! And pointers/tips to get me going in the right direction

A curve in space is specied by the one parameter set of vectors x(t). Also given is a surface in space parameterised by x(u, v): x(t)= <2+t, -t, 1+3t2> x(u,v)=(u2 - v + u, u+5, v-2> A) Show that the curve intersects the surface in exactly two points. Show that xi = <4 - \frac{\sqrt{46}}{2}...

so using the sine addition formula you get sin(\frac{n*Pi*x}{a})*cos(\frac{n*pi}{2})+sin(\frac{n*pi}{2})*cos(\frac{n*Pi*x}{a})

So basically simplifying it you get \sqrt{\frac{2}{a}}* sin * (\frac{n*Pi*(2x+a)}{2a} So then how do you do incorporate the odd an even?

that a on the end of the bottom equation is suposed to be on the bottom where the bracket is

Ok so for basically part iii) i got \Psi(x) = \sqrt{\frac{2}{a}} sin \frac{n*Pi*x}{a} \Phi(x) = \sqrt{\frac{2}{a}} sin \frac{n*Pi*(x+\frac{a}{2}}){a}

Vela! That really helps! :) Thanks! :D The only problem i have now is i don't understand part iv) Could you help with that?

Hiya mate! Na I've got literally no where WHat bout you ? Howd you manage to do the 1st and 2nd bits?

I see ok! So basically the graph of the box has move a/2 to the left

I have a problem: Let V0(x) denote the potential corresponding to the innite square well (`box') extending from x = 0 to x = a . Let us replace it by the potential V (x) = 0 ; |x| \leq a/2 \infty; |x| \geq a/2 i want to find the solutions of the Schrodinger equation associated...