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HAHA!!! Math funnies..

aggrorider

Monkey
Sep 20, 2005
209
0
this one's corny, but in my calculus class a korean sudent said this, "I want to be your derivative, so I can be a tangent to your curve."

...she was not impressed
 

Westy

the teste
Nov 22, 2002
54,486
20,287
Sleazattle
I have notoriously bad handwriting. I remember taking a differential equations quiz where I had to prove that one formula was equal to another. I simply wrote the first formula, wrote it again really messy, wrote the first half of the first formula really messy along with the second half and so on until I had the second formula. Dumba ass TA who was grading it gave me half credit.
 

BurlyShirley

Rex Grossman Will Rise Again
Jul 4, 2002
19,180
17
TN
I have notoriously bad handwriting. I remember taking a differential equations quiz where I had to prove that one formula was equal to another. I simply wrote the first formula, wrote it again really messy, wrote the first half of the first formula really messy along with the second half and so on until I had the second formula. Dumba ass TA who was grading it gave me half credit.
Oh yeah, that's one of my signature moves... and "d" can look alot like an "a" on an answer sheet if you're sly enough:biggrin:
 

kizzi77

Monkey
Aug 11, 2005
564
0
nashvegas
Hahahahahaha!!!!!!!

:clapping:

that was funny! I heart math and math funnies...i even sent it to my mom who loves math as much as me, maybe more because she taught it forever!
 

Jay Gatz

Monkey
Aug 14, 2004
169
0
NE for college, CO when i can
A constant function and e^x are walking on down the street. Then suddenly the constant function sees a differential operator approaching and runs away.

So e^x runs after the constant function and asks why the hurry. "Well, you see, there's this differential operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!"

"Ah," says e^x, "that mother****er won't bother ME, I'm e to the x!" and heads back.

Of course he meets the differential operator after a short distance and says with a sneer, "Hi, I'm e^x."

The differential operator replies, "Hi, I'm d/dy."


i bet most of you wont get that.
 

kinghami3

Future Turbo Monkey
Jun 1, 2004
2,239
0
Ballard 4 life.
A constant function and e^x are walking on down the street. Then suddenly the constant function sees a differential operator approaching and runs away.

So e^x runs after the constant function and asks why the hurry. "Well, you see, there's this differential operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!"

"Ah," says e^x, "that mother****er won't bother ME, I'm e to the x!" and heads back.

Of course he meets the differential operator after a short distance and says with a sneer, "Hi, I'm e^x."

The differential operator replies, "Hi, I'm d/dy."


i bet most of you wont get that.
If it was two years ago, I would have gotten it. I get why the constant is running, and I'm sure the punchline has to do with d/dx, but other than that I'm lost.
 

DamienC

Turbo Monkey
Jun 6, 2002
1,165
0
DC
Of course he meets the differential operator after a short distance and says with a sneer, "Hi, I'm e^x."

The differential operator replies, "Hi, I'm d/dy."
:biggrin: Everyone likes a good differentiation joke.
 

urbaindk

The Real Dr. Science
Jul 12, 2004
4,819
0
Sleepy Hollar
You guys... LOL. Differentiating e^x with respect to x is equal to e^x, yes. However differentiating e^x with respect to y is equal to 0. Therein lies the joke. Duh!

edit: ninja'ed by n8.
 

N8 v2.0

Not the sharpest tool in the shed
Oct 18, 2002
11,003
149
The Cleft of Venus
Math Problem
Advanced Mathematics: Curve Superelevation


What is the superelevation of the outer rail above the inner rail of a 2° railroad curve with a 70 mph design speed if the friction factor is neglected?


a. 8 ½ in
b. 7 ¼ in
c. 6 ½ in
d. 5 ¼ in
 

skinny mike

Turbo Monkey
Jan 24, 2005
6,415
0
You guys... LOL. Differentiating e^x with respect to x is equal to e^x, yes. However differentiating e^x with respect to y is equal to 0. Therein lies the joke. Duh!

edit: ninja'ed by n8.
i feel stupid not being able to get this. after all, this is what we are currently covering in ap calculus. :banghead: but then again, we are focusing more on derivatives than we are on differentials currently.
 

sneakysnake

Monkey
Apr 2, 2006
875
1
NC
You guys... LOL. Differentiating e^x with respect to x is equal to e^x, yes. However differentiating e^x with respect to y is equal to 0. Therein lies the joke. Duh!

edit: ninja'ed by n8.

This is the reason I'm in ap statistics.

stats = :banana:

calc = :rant: or teh suck
 
You guys... LOL. Differentiating e^x with respect to x is equal to e^x, yes. However differentiating e^x with respect to y is equal to 0. Therein lies the joke. Duh!

edit: ninja'ed by n8.
No, N8 was dead wrong:

"i really don't think d/dy is gonna do anything to e to the x". It's going to turn e^x to nothing, which, as you observed, is the point of the joke...
 

urbaindk

The Real Dr. Science
Jul 12, 2004
4,819
0
Sleepy Hollar
Math Problem
Advanced Mathematics: Curve Superelevation


What is the superelevation of the outer rail above the inner rail of a 2° railroad curve with a 70 mph design speed if the friction factor is neglected?


a. 8 ½ in
b. 7 ¼ in
c. 6 ½ in
d. 5 ¼ in
I'm going to make a guess and say that the super elevation is small at that speed and radius so I'll go with answer d. I'm too lazy to work it out (don't feel like doing the unit conversions) but here is the equation: h = V^2*s/gR

V = velocity, 70 mph
s = track guage (not given??? n8, you're slipping.)
g = acceleration due to gravity 32.2 ft/sec^2
R = radius of curve (you have to convert 2° to a radius. I believe that the railroad defines this as the angle between the tanget of the curve and a chord length of the arc, how they define the chord length I don't know)
 

N8 v2.0

Not the sharpest tool in the shed
Oct 18, 2002
11,003
149
The Cleft of Venus
I'm going to make a guess and say that the super elevation is small at that speed and radius so I'll go with answer d. I'm too lazy to work it out (don't feel like doing the unit conversions) but here is the equation: h = V^2*s/gR

V = velocity, 70 mph
s = track guage (not given??? n8, you're slipping.)
g = acceleration due to gravity 32.2 ft/sec^2
R = radius of curve (you have to convert 2° to a radius. I believe that the railroad defines this as the angle between the tanget of the curve and a chord length of the arc, how they define the chord length I don't know)


Solution:

What's the weight of the train? The inertial force on the exterior railing would be directly proportional to the lateral sway of each individual boxcar. If the lead boxcar (A) was 1/3 heavier than the next boxcar (B), it would create a greater pivotal momentum about the rotational plane of the longitudinal axis along the arc of the rails. However, if boxcar B weighs less than A, then just the opposite would occur. In other words, it would be pushing the lead car (A) which if the rail is not high enough, it could have a tendency to push it off the edge of the outer rail. However, if we take each individual sprocket of the boxcar and multiplied the instantaneous center of rotation (ICOR) of each wheel times the velocity(not speed, we need a vector quantity and only velocity has that), each would create an individual value that when added would be the sum total of the all the tangential accelerations--this would give us a value that will alleviate the lesser degree of superelevation.

Here's the kicker, if A weighs more than B, then D would be the answer. If B weighs more than A, answer B would be correct due to the outside rail would be substantially higher than the inner rail. If all boxcars were assumed to be the same, then the answer should be "C".
 

urbaindk

The Real Dr. Science
Jul 12, 2004
4,819
0
Sleepy Hollar
Solution:

What's the weight of the train? The inertial force on the exterior railing would be directly proportional to the lateral sway of each individual boxcar. If the lead boxcar (A) was 1/3 heavier than the next boxcar (B), it would create a greater pivotal momentum about the rotational plane of the longitudinal axis along the arc of the rails. However, if boxcar B weighs less than A, then just the opposite would occur. In other words, it would be pushing the lead car (A) which if the rail is not high enough, it could have a tendency to push it off the edge of the outer rail. However, if we take each individual sprocket of the boxcar and multiplied the instantaneous center of rotation (ICOR) of each wheel times the velocity(not speed, we need a vector quantity and only velocity has that), each would create an individual value that when added would be the sum total of the all the tangential accelerations--this would give us a value that will alleviate the lesser degree of superelevation.

Here's the kicker, if A weighs more than B, then D would be the answer. If B weighs more than A, answer B would be correct due to the outside rail would be substantially higher than the inner rail. If all boxcars were assumed to be the same, then the answer should be "C".

Ok??? That makes very little sense, I think I need to see a picture. What is the equation for the solution? That's not a very 'real world' answer, because a you don't know the weigh of the cars a priori and you can't realistically change to superelevation to match the train as it passes. But I think it (the heavier cars pushing the lighter cars through the turn) is why they tend to put the empties at the back of the train.