The Math and Quiz Game =)

[Survivor]

Topic: Literature history
A certain author has written a story consisting of 6 different parts, dealing with reanimation in the genre of horror. Name both the author and the title of all 6 parts when combined together. As a hint I'll say that the author is most famous for one of his squidlike creations.
Once you get the answer I suggest you read that story. It's a nice one

[kozak6]

HP Lovecraft's "Herbert West: Re-Animator"

[Survivor]

Point goes to kozak6.

[kozak6]

What is the first derivative of the hyperbolic cosecant of x with respect to x?

[demenator]

Is it this?

Althrough I dont understand anything about this formula I googlesd it up. So I could as well be wrong.

[NiTRoX]

Its definately wrong, you need to mention x.

[Bullislander05]

d/dx(cosh(x)) = sinh(x)

Dur.

[kozak6]

Mmmm....

Close.

I asked for d/dx(csch(x))

[Menace13]

Code: [Select]

d/dx(csch(x))
1/x(csch(x))
1/x(1/sinh(x))
that's as far as i got but i am only 12 so what can you expect?

[kozak6]

Not quite.

d/dx refers to taking the derivative of a function.  The d's don't actually cancel.

Taking the derivative is a rather fascinating mathematical operation you can read all about here: http://en.wikipedia.org/wiki/Derivative .

[Survivor]

Derivatives are a very annoying mathematical operation i have to learn all about in university. You do not want to see what they mess us up with >:E

[Troy1]

(d/dx) csch x = -csch x coth x

csch x = 1/sinhx
d/dx 1/sinhx = - coshx/sinhx^2 = - 1/sinhx * coshx/sinhx = - cschx * cothx


=> (d/dx) csch x = -csch x coth x

Used :

(cosh x)' = sinhx

(U/V)' = (U'V - UV')/V^2

(1)' = 0

nothing really hard here.


EDIT :

Pretty sure that the answer is right, if not then Ill delete this edit

Problem :

An infinitely extensible rubber band is attached to a wall at one side and to a car at another one. the initial length of a rubber band is 2 m. There is an ant on the rubber band right near the wall, which moves along the band into the direction of the car.

Simultaneously ant and a car starting to move, speed of ant is 1 m/s speed of the car is 2 m/s. Car is moving away from the wall. When will ant reach the car? (and can it do so? Can it reach it if car moves at 0.5 m/s?)

Take the ant as a dimensionless object. Rubber can be extended infinitely. Car and ant moves at constant speed.

[Bullislander05]

Whups.  Misread it.  I thought it was derivative of the hyperbolic cosine of x.   

d/dx(csch(x)) = -csch(x)coth(x)  Troy got it.

Soo...you're saying that the ant is moving in the direction of the car at 1 m/s while the car is moving 2 m/s, and the car is 2 meters in front?  So the diagram would look like:

|>..........>>

| <-- wall
> <-- ant
.......... <-- represents 2 meters
>> <-- car

Am I correct?  Also, since the rubber band is infinitely extensible, there is no force acting on the ant that would make it move faster or accelerate (Assuming this because you said speeds are constant)?

If that is the case, the ant cannot catch the car and the car will constantly outrun the ant.  If the car moves at .5 m/s, then the problem is changed.  In that case, the ant, with position equation: x = t, and the car, with position x = 2 + .5t, would be at the same position when t = 4 seconds.

Correct?

[Troy1]

W|
A|
L|Ant -> (1 m/s)
L|----------------------Car -> (2 m/s)
=|rubber band 2m
=|
=|

(Ill get normal drawing soon)

Ant's speed is relative to the surface he moves on. So if he is on top of the car his speed is 1 m/s relative to the car and 3 m/s relative to the floor.  There is a sufficient friction force between ant and band to allow ant not to slid on the band.

[Bullislander05]

If that is the case, the ant cannot catch the car and the car will constantly outrun the ant.  If the car moves at .5 m/s, then the problem is changed.  In that case, the ant, with position equation: x = t, and the car, with position x = 2 + .5t, would be at the same position when t = 4 seconds.

^^Is that the correct answer?

[Troy1]

no Band stretches => ant has to crawl longer distance each time, but there is a trick there.

[Nux]

The behavior of the ant with respect to the expanding point on which it stands hasn't been clearly defined. Am I right in saying this infinitely stretching band has a velocity at distance x from the wall of C(x/L), where C is the speed of the car and L is the current length of the band and that the ant has velocity Vr + C(x/L), where Vr is the ants velocity relative to the point on the band on which it stands?

[Troy1]

Good, that is the main idea of the problem. And well, ant is a mathematical point + as I said, there is enough friction for it to not slid = it will grip in to a stretching point, gaining the speed of it + its own speed relative to that point. Good start Nux.

[Nux]

In which case (and please forgive me if I get this terribly wrong) it can be shown that for Vr = 1, C = 2, L = 2 and time t:

V = Vr + C(x/L)
V = 1 + 2(x/2) = 1 + x
x (= Vt) = (1 + x)t
t = x/(1+x)

Where the Cars distance Cx = 2t + 2, for the distances to be equal,

t = 2t+2/(1+2t+2) = 2t+2/2t+3
2t² + 3t = 2t + 2
2t² + t - 2 = 0
t = (-1 + sqrt(1+16))/4 = (-1 + sqrt(17))/4 [or t = (-1 - sqrt(17))/4 but this has no meaning here)]

If the C = 0.5 then,

V = 1 + 0.5(x/2) = 1 + x/4
x = (1 + (x/4))t
t = x/(1+(x/4))

Cx = 0.5t + 2
t = 0.5t + 2/(1+0.5t + 2) = 0.5t+2/0.5t+3
0.5t² + 3t = 0.5t + 2
0.5t² + 1.5t - 2 = 0
t = (-1.5 + sqrt((1.5)^2+4)) = -1.5 + sqrt(25/4) [or -1.5 - sqrt(25/4) but not meaningful here]

[/dev/humancontroller]

~6.3891s
~2.5949s

[Troy1]

In which case (and please forgive me if I get this terribly wrong) it can be shown that for Vr = 1, C = 2, L = 2 and time t:

V = Vr + C(x/L)
V = 1 + 2(x/2) = 1 + x

Wrong. You cannot substitute 2m right away. Say if the car moved and the distance is 4m from the wall at which ant (for example) reaches the car. from your equation his speed is 5 m/s which is not possible, since his maximum speed can be only 3 m/s (sitting on the car). L is a variable, not a constant.

[Nux]

Ah yes. I'd noticed that the first time, but managed to forget it to make things easier for myself =)

When trying to integrate this velocity function, I run into a lot of trouble with these inner functions which are defined in terms of each other. I've tried limits, recursive definitions and pretending variables are constants >.>..

All in all I haven't had enough practice with this sort of problem.

Funnily enough, I made up a new variable Cx which meant exactly the same thing as L. xD

Let's start again:

V (= dx/dt) = Vr + C(x/(Ct+L))

Where L is now the initial length.

As a guess, I'd say this ant has a positive but exponentially decaying acceleration.

[==Troy==]

As a guess, I'd say this ant has a positive but exponentially decaying acceleration.

Relatively to what?

[Nux]

Relative to the same frame as the car's 2m/s (the ground). If I meant relative to the band then that's already covered by the ant's stated velocity of 1m/s.

[TheEternalDretch]

Okay, I'll pick up where Nuxy left off.

V (= dx/dt) = Vr + C(x/(Ct+L)) = Vr + x/(t+L/C)

Writing this out a little differently, we get:  dx/dt - 1/(t+L/C)*x = Vr

Now, I'm going to define a cute little function as: M(t) = 1/(t+L/C)

I multiply our previous equation by my new function: M*dx/dt - (M/(t+L/C))*x = M*Vr

Which happens to be (feel free to check this, but I defined it this way on purpose): M*dx/dt + dM/dt*x = M*Vr

Using a well-known law which I don't know the English name for (product rule! thanks, nux!):
d(M*x)/dt = M*Vr = Vr/(t+L/C)

Integrating both sides (according to t, of course), gives: M*x = Vr*ln(t+L/C)

We then need only divide by M to get: x = Vr*(t+L/C)*ln(t+L/C)

In order to find out when the ant will reach the car, let's equate!

Vr*(t+L/C)*ln(t+L/C) = Ct + L = C*(t+L/C)
Vr*ln(t+L/C) = C
ln(t+L/C) = C/Vr
t + L/C = e^(C/Vr)
t = e^(C/Vr) - L/C

Substituting the given Vr = 1, C = 2, L = 2, we get: t = e^2 - 1 seconds

And for the second part (Vr = 0.5, C = 2, L = 2), we get: t = e^4 - 1 seconds

Yes - so long as the ant is moving, no matter how slow, it will eventually reach the car. Pretty neat. ^_^ So, am I right?

(By the way, I'm mildly disappointed in you, Nux. You said you love math! ^_~)

[Nux]

I do love maths! I really do!

This is good, because it's made me want to practice more.

I understand everything you've put, but only through practice can you do it yourself.

Edit: Oh, and that law you didn't know the egnlish name of is the 'Product Rule'.

[TheEternalDretch]

Well, the one thing I didn't explain is how I got to M, and I can explain that if you wish. But this is easy stuff.... this was the very first sort of problem I learned how to solve in my differential equations course. I can give you a harder problem if I won. ^_^

[==Troy==]

Okay, I'll pick up where Nuxy left off.

V (= dx/dt) = Vr + C(x/(Ct+L)) = Vr + x/(t+L/C)

Writing this out a little differently, we get:  dx/dt - 1/(t+L/C)*x = Vr

Now, I'm going to define a cute little function as: M(t) = 1/(t+L/C)

I multiply our previous equation by my new function: M*dx/dt - (M/(t+L/C))*x = M*Vr

Which happens to be (feel free to check this, but I defined it this way on purpose): M*dx/dt + dM/dt*x = M*Vr

Using a well-known law which I don't know the English name for (product rule! thanks, nux!):
d(M*x)/dt = M*Vr = Vr/(t+L/C)

Integrating both sides (according to t, of course), gives: M*x = Vr*ln(t+L/C)

We then need only divide by M to get: x = Vr*(t+L/C)*ln(t+L/C)

In order to find out when the ant will reach the car, let's equate!

Vr*(t+L/C)*ln(t+L/C) = Ct + L = C*(t+L/C)
Vr*ln(t+L/C) = C
ln(t+L/C) = C/Vr
t + L/C = e^(C/Vr)
t = e^(C/Vr) - L/C

Substituting the given Vr = 1, C = 2, L = 2, we get: t = e^2 - 1 seconds

And for the second part (Vr = 0.5, C = 2, L = 2), we get: t = e^4 - 1 seconds

Yes - so long as the ant is moving, no matter how slow, it will eventually reach the car. Pretty neat. ^_^ So, am I right?

(By the way, I'm mildly disappointed in you, Nux. You said you love math! ^_~)

Bravo! Very well done. Your turn, sir.

just a bit of misunderstanding in the end : the speed of ant is still 1ms and car is now moving at 0.5 ms. But this is not important, since you gave the general formula, which is exactly what I was looking for.

[TheEternalDretch]

Yay! ^_^ Okay, here's a fun one, not too hard. Knowing that x³- x -1 = 0, determine the exact value of: (3x² - 4x)^1/3 + x(2x²+3x+2)^1/4

(This is a really easy problem to solve with a calculator, but it's much cooler/funner to solve algebraically. Don't disappoint me!)

[NiTRoX]

Yay! ^_^ Okay, here's a fun one, not too hard. Knowing that x³- x -1 = 0, determine the exact value of: (3x² - 4x)^1/3 + x(2x²+3x+2)^1/4

(This is a really easy problem to solve with a calculator, but it's much cooler/funner to solve algebraically. Don't disappoint me!)

2x³ + 4x² - 1.55