Mathematics Problems Thread

Originally posted by AbnormalButSane
Actually the integral is (1/3)x^3 + C

Originally posted by Mr. Bacon
yes, yes it is, stupid C. mmm

I did a problem this morning:

k(del^2 u/ del x^2)=(del u/del t)
where 0<x<L, t>0

with the conditions
u(0,t)=0, u(L,t)=0, t>0
u(x,0)=f(x), 0<x<L

Hint: Assume u(x,t) = X(x)T(t)

God, I love DE. doped

Originally posted by AbnormalButSane
You were close though. petpet

I did a problem this morning:

k(del^2 u/ del x^2)=(del u/del t)
where 0<x<L, t>0

with the conditions
u(0,t)=0, u(L,t)=0, t>0
u(x,0)=f(x), 0<x<L

Hint: Assume u(x,t) = X(x)T(t)

God, I love DE. doped

and yea, **** that 😐

Originally posted by JacopeX
This is thread will die quickly.

People hate math.

But people love spam

[QUOTE=11969548]Originally posted by retturnnerr
[B]Self-explanatory.

E.g. you ask someone:

Jack Sparrow had 1200 galleons. If Jack Sparrow sold 500 galleons to Davy Jones. But Davy Jones then destroyed another 500 galleons owned by Jack Sparrow.

How many galleons does Sparrow have left?

1200 - 500 = 700
700 - 500 = 200

ANSWER 200

1+1= WINDOWS

Dominic jump roped. He jump roped 450 times.

He weighs exactly 100 KG. (I just weighed myself and It's almost exactly 100Kg.)

Dominic averages .25 meters on the maximum height of each jump.

Each jump also expends an extra .2 joules by lifting both feet up, unnecessarily, as humans tend to do that when jump roping. (If we want to get even more accurate, I could give an average distance the feet are raised during the jump, to actually give people the chance to calculate exactly how many joules are used to pull those feet up...it would require the weight of the feet, distance between knee and bottom of the foot, weight distribution of the leg, i.e. the top part weighs more, but the bottom part is the furthest out. etc. But that would make it far too complicated.)

Likewise, it requires Dominic .08 joules to spin the rope one complete turn, on average.

Calculate the number of joules required for Dominic to complete one jump with the jump rope. After finding that number, discover the total energy expenditure, in joules, of Dominic's jump rope adventure.

mgh + .2 + .08

where m is mass, h is height jumped, and g is acceleration of gravity 9.81m/s^2

? It's late oh well.

Originally posted by AbnormalButSane
mgh + .2 + .08

where m is mass, h is height jumped, and g is acceleration of gravity 9.81m/s^2

? It's late oh well.

M = 100 kg

H = .25 meters.

Yeah, I'm much too lazy to use math with numbers.

Originally posted by AbnormalButSane
Yeah, I'm much too lazy to use math with numbers.

Word problems are pretty much what every single physics problem I had, were.

My least favorite was fluid dynamics. The pressure...flow rates.

AAAAHH! I hated it. 🙁