bloodhound

I would say its more spherical, as spherical is defined in precise mathematical terms, whereas the concept of colour is different to everyone. i.e if an alien came about, it would not be seeing the same colour, but it would see the same shape.

ahhhh, ur one step ahead of me in everything.. have a module on that next term

ahhhh, ur one step ahead of me in everything.. have a module on that next term

same here, altough i just plugged the equation in maple and made it solve it.

same here, altough i just plugged the equation in maple and made it solve it.

We just started sequences of functions in our analysis module. We should be doing this stuff in a short while

We just started sequences of functions in our analysis module. We should be doing this stuff in a short while

oh rite... maybe its equivalent to a "proper" subset. cheers .. ill have a go at doing this question.

oh rite... maybe its equivalent to a "proper" subset. cheers .. ill have a go at doing this question.

What is a "proper" subgroup? never came across that term before. If G is a group, and you have a [math]g \in G[/math] such that [math]G=<g>[/math], then G is called a cyclic group and g is called the generator of G Here, [math]<g>=\{g^k:k \in \mathbb{Z}\}[/math] or additively [math]<g>=\{mg:m \in \mathbb{Z}\}[/math] Well basically its just saying that, if u have a group G, then its a cyclic group if there exists an element in G such that if u take all the integer powers of that element, you get the group G. A nice property is that a subgroup of a cyclic group is also cyclic. Cyclic groups are also abelian by the way.

What is a "proper" subgroup? never came across that term before. If G is a group, and you have a [math]g \in G[/math] such that [math]G=<g>[/math], then G is called a cyclic group and g is called the generator of G Here, [math]<g>=\{g^k:k \in \mathbb{Z}\}[/math] or additively [math]<g>=\{mg:m \in \mathbb{Z}\}[/math] Well basically its just saying that, if u have a group G, then its a cyclic group if there exists an element in G such that if u take all the integer powers of that element, you get the group G. A nice property is that a subgroup of a cyclic group is also cyclic. Cyclic groups are also abelian by the way.

oops sorry. got confused with my reaction forces and normal forces.

oops sorry. got confused with my reaction forces and normal forces.

Argghhhh I Wanna See The Rest If The Pics

Argghhhh I Wanna See The Rest If The Pics

the reaction force will always be perpendicular to the slope.

one word "SUCK IT" ARGGHHHHH YES IM CLEVAAAAAA

with fan base like that , you can make a living out of it

Ok . this is how it goes. using P(0)=0 and P'(0)=0 you get c=d=0 and then 1)[math]P(l)=al^3+bl^2=h[/math] and 2)[math]P'(l)=3al^2+2bl=0[/math] from 2) [math](3)a=-\frac{2b}{3l}[/math] substituting a into 1) we get [math]h=-\frac{2b}{3l}l^3+bl^2[/math] [math]=\frac{-2bl^2+3bl^2}{3}[/math] [math]=\frac{bl^2}{3}[/math] Therefore, [math]b=\frac{3h}{l^2}[/math] put that back into (3) we have [math]a=-\frac{2b}{3l}[/math] [math]=\frac{\frac{-6h}{l^2}}{3l}[/math] [math]=-\frac{2h}{l^3}[/math]

thats impossible.

I dont if the above solution is rite. can anyone check.... as i am not able to complete the second part of the question using P(x) from above [edit: oops seems fine now.. sorted ] [edit no.2] oops no fine at all.. redoing it again[/edit] [edit no.3] everything is good now [/edit]

The gerenal cubic has 4 arbitary constants. To find the particular soultion you thus require 4 initial (boundary) conditions. The first two are obvious. 1)at distance x = l, the plane must be at height p(l)=h 2)at x=0 , the plane must be height 0 The second two are the bit subtle The flight of the plane will have this function f(x)= 0 for x<=0 f(x)=P(x) for 0<=x<=l f(x)=h for x=>l and since , the motion of the place has to be smooth (i.e f(x) must be differentiable everywhere ) it follows that f'(0)=P'(0)=0=f'(l)=P'(l) [The above is a bit long winded, and the results should be clear just by intuition] Once, you put in those conditions, you get two equations for two unknowns {a,b} solve it, and you get [math]a=\frac{-2h}{l^3}[/math] and [math]b=\frac{3h}{l^2}[/math] thus the particural solution for P(x) is [math]P(x)=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2[/math]

just look at his avatar for gods sake

is the pc version coming out???? but them ill have to wait till i can get a computer with decent graphiocs

Well, like someone said, talking about it is not the same thing as doing it. if they did happen to do it, they they would be in a big hole. i.e sanctions etc.