/sci/ - Science & Math

>>5493385
That person will be dead in 50 years, and you can only hope their offspring aren't as stupid.

>>5493385
fossilized cooked hams in 15 years. this is my new favorite argument.

Please /sci/, please send educational messages to this poor stupid fellow.

>>5493404
>implying these filthy psuedo-intellectual subhuman scum aren't incorrigible

poes law

>>5493385
>I am so fucking done.

>Throwing in the towel whenever you encounter 1 retard.

Spoiler: There's a lot more than one retard on the planet, son. Just accept it and move on with your life already.

>>5493412
Actually, I was implying harassment.

>>5493385

We shall use an actual historical event to illustrate how a differential equation arose, how a relationship was then established between the two variables involved, and finally how from the relationship, the answer to a very interesting problem was determined.

In the year 1940, a group of boys was hiking in the vicinity of a town in France named Lascaux. They suddenly became aware that their dog had disappeared. In the ensuing search he was found in a deep hole from which he was unable to climb out. When one of the boys lowered himself into the hole to help extricate the dog, he made a startling discovery. The hole was once part of the roof of an ancient cave that had become covered with brush. On the walls of the cave there were marvellous paintings of stags, wild horses, cattle, and of a fierce-looking black beast which resembled our bull. This accidental discovery, as you may guess, created a sensation. In addition to the wall paintings and other articles of archaeological interest, there were also found the charcoal remains of a fire. The problem we wish to solve is the following: determine from the charcoal remains how long ago the cave dwellers lived.

>>5493385
>>5493523

It is well known that charcoal is burnt wood and that with time certain changes take place in all dead organic matter. It is also known that all living organism contain two isotopes of carbon, namely <span class="math">C^{12}[/spoiler] and <span class="math">C^{14}[/spoiler]. The first element is stable; the second is radioactive. Furthermore the ratio of the amounts of each present in any macroscopic piece of living organism remains constant. However from the moment the organism dies, the <span class="math">C^{14}[/spoiler] that is lost because of radiation, is no longer replaced. Hence the amount of the unstable <span class="math">C^{14}[/spoiler] present in a dead organism, as well as its ratio to the stabled <span class="math">C^{12}[/spoiler], changes with time. The changing entities in this problem are therefore the element <span class="math">C^{14}[/spoiler] and time. If the law which tells us how one of these changing entities is related to the other cannot be expressed without involving their derivative, then a differential equation will result.

>>5493385
>>5493523
>>5493530

Let 't' represent the elapsed time since the tree from which the charcoal came, died, and let 'x' represent the amount of <span class="math">C^{14}[/spoiler] present in the dead tree at any time 't'. Then the instantaneous rate at which the element <span class="math">C^{14}[/spoiler] decomposes is expressed in mathematical symbols as

<div class="math">\frac{dx}{dt}\cdot </div> <span class="math"> (1.1) [/spoiler]

We now make the assumption that this rate of decomposition of <span class="math">C^{14}[/spoiler] varies as the first power of 'x' (remember 'x' is the amount of <span class="math">C^{14}[/spoiler] present at any time 't'). Then the equation which expresses this assumption is

<div class="math">\frac{dx}{dt} = -kx</div> <span class="math"> (1.11) [/spoiler]

where k > 0 is a proportionality constant, and the negative sign is used to indicate that 'x', the quantity of <span class="math">C^{14}[/spoiler] present, is decreasing. Equation (1.11) is a differential equation.

>>5493530
The applicable range of radiocarbon dating is only tens of thousands of years, other methods (or at least other isotopes of other elements) are needed for dating materials presumed to be significantly older than that.

>>5493536

It states that the instantaneous rate of decomposition of <span class="math">C^{14}[/spoiler] present at a moment of time. For example, if 'k' = 0.01 and 't' is measured in years, then when 'x' = 200 units at a moment in time, (1.11) tells us that the rate of decomposition of <span class="math">C^{14}[/spoiler] at that moment if 1/100 of 200 or at the rate of 2 units per year. If, at another moment of time, x = 50 units, then (1.11) tells us that the rate of decomposition of <span class="math">C^{14}[/spoiler] at that moment is 1/100 of 50 or at the rate of <span class="math">\frac{1}{2}[/spoiler] unit per year.

Our next task is to try to determine from (1.11) a law that will express the relationship between the variable 'x' (which, remember, is the amount of <span class="math">C^{14}[/spoiler] present at any time 't') and the time 't'/ To do this, we multiply (1.11) by <span class="math">\frac{dt}{x}<span class="math"> and obtain
<div class="math">\frac{dx}{x}=-k dt</div> 1.12

Integration of (1.12) gives
<div class="math">logx = -kt + c,</div> (1.13)
where 'c' is an arbitrary constant. By the definition of the logarithm, we can write (1.13) as
<div class="math">x = e^{-kt+c} = e^{c}e^{-kt} = Ae^{-kt},</div> (1.14)
where we have replaced the constant e^{c} by a new constant 'A'.

Although (1.14) is an equation which expresses the relationship between the variable 'x' and the variable 't', it will not give us the answer we seek until we know the values of 'A' and 'k'. For this purpose, we fall back on other available information which as yet we have not used.[/spoiler][/spoiler]

>>5493536

>>5493536

It states that the instantaneous rate of decomposition of <span class="math">C^{14}[/spoiler] present at a moment of time. For example, if 'k' = 0.01 and 't' is measured in years, then when 'x' = 200 units at a moment in time, (1.11) tells us that the rate of decomposition of <span class="math">C^{14}[/spoiler] at that moment if 1/100 of 200 or at the rate of 2 units per year. If, at another moment of time, x = 50 units, then (1.11) tells us that the rate of decomposition of <span class="math">C^{14}[/spoiler] at that moment is 1/100 of 50 or at the rate of <span class="math">\frac{1}{2}[/spoiler] unit per year.

Our next task is to try to determine from (1.11) a law that will express the relationship between the variable 'x' (which, remember, is the amount of <span class="math">C^{14}[/spoiler] present at any time 't') and the time 't'/ To do this, we multiply (1.11) by <span class="math">\frac{dt}{x}[/spoiler] and obtain
<div class="math">\frac{dx}{x}=-k dt</div> <span class="math">1.12[/spoiler]

Integration of (1.12) gives
<div class="math">logx = -kt + c,</div> <span class="math">(1.13)
where 'c' is an arbitrary constant. By the definition of the logarithm, we can write (1.13) as
<div class="math">x = e^{-kt+c} = e^{c}e^{-kt} = Ae^{-kt},</div> <span class="math">(1.14)[/spoiler]
where we have replaced the constant <span class="math">e^{c}[/spoiler] by a new constant 'A'.

Although (1.14) is an equation which expresses the relationship between the variable 'x' and the variable 't', it will not give us the answer we seek until we know the values of 'A' and 'k'. For this purpose, we fall back on other available information which as yet we have not used.[/spoiler]

>>5493536

It states that the instantaneous rate of decomposition of <span class="math">C^{14}[/spoiler] present at a moment of time. For example, if 'k' = 0.01 and 't' is measured in years, then when 'x' = 200 units at a moment in time, (1.11) tells us that the rate of decomposition of <span class="math">C^{14}[/spoiler] at that moment if 1/100 of 200 or at the rate of 2 units per year. If, at another moment of time, x = 50 units, then (1.11) tells us that the rate of decomposition of <span class="math">C^{14}[/spoiler] at that moment is 1/100 of 50 or at the rate of <span class="math">\frac{1}{2}[/spoiler] unit per year.

Our next task is to try to determine from (1.11) a law that will express the relationship between the variable 'x' (which, remember, is the amount of <span class="math">C^{14}[/spoiler] present at any time 't') and the time 't'/ To do this, we multiply (1.11) by <span class="math">\frac{dt}{x}[/spoiler] and obtain
<div class="math">\frac{dx}{x}=-k dt</div> <span class="math">1.12[/spoiler]

Integration of (1.12) gives
<div class="math">logx = -kt + c,</div> <span class="math">(1.13)[/spoiler]
where 'c' is an arbitrary constant. By the definition of the logarithm, we can write (1.13) as
<div class="math">x = e^{-kt+c} = e^{c}e^{-kt} = Ae^{-kt},</div> <span class="math">(1.14)[/spoiler]
where we have replaced the constant <span class="math">e^{c}[/spoiler] by a new constant 'A'.

Although (1.14) is an equation which expresses the relationship between the variable 'x' and the variable 't', it will not give us the answer we seek until we know the values of 'A' and 'k'. For this purpose, we fall back on other available information which as yet we have not used.

>>5493556

Since time is being measured from the moment the tree died, i.e., 't' = 0 at death, we learn from (1.14) by substituting 't' = 0 in it, that 'x' = 'A'. Hence we now know, since 'x' is the amount of <span class="math">C^{14}[/spoiler] present at any time 't', that 'A' units of <span class="math">C^{14}[/spoiler] were present when the tree, from which the charcoal came, died.
From the chemist we learn that approximately 99.876 percent of <span class="math">C^{14}[/spoiler] present at death will remain in dead wood after 10 years and that the assumption made after (1.1) is correct. Mathematically this means that when 't' = 10, 'x' = 0.99876'A'. Substituting these values of 'x' and 't' in (1.14), we obtain

<div class="math"> 0.99876A = Ae^{-10k}, 0.99876 = e^{-10k}. </div> <span class="math">(1.15)[/spoiler]

We can now find the value of 'k' in either of two ways. There are tables which tell us for what value of -10k, <span class="math">e^{-10k}[/spoiler] = 0.99876. Division of this value by -10 will then give us the value of 'k'. Or if we take the natural logarithm of both sides of (1.15) there results
<div class="math">log0.99876 = -10k</div><span class="math">(1.2)[/spoiler]

From a table of natural logarithms, we find

<div class="math">-0.00124 = -10k, k = 0.000124 </div> <span class="math"> (1.21) [/spoiler]

approximately. Equation (1.14) now becomes
<div class="math"> x = Ae^{-0.000124t} </div> <span class="math">(1.22)[/spoiler]
where 'A' is the amount of <span class="math">C^{14}[/spoiler] present at the moment the tree died.

>>5493567

Equation (1.22) expresses the relationship between the variable quantity 'x' and the variable time 't'. We are therefore at last in a position to answer the original question: How long ago did the cave dwellers live? By a chemical analysis of the charcoal, the chemist was able to determine the ratio of the amounts of <span class="math">C^{14}[/spoiler] to <span class="math">C^{12}[/spoiler] present at the time of the discovery of the cave. A comparison of this ratio with the fixed ratio of these two carbons in living trees disclosed that 85.5 percent of the amount of <span class="math">C^{14}[/spoiler] present at death had decomposed. Hence 0.145'A' units of <span class="math">C^{14}[/spoiler] remained. Substituting this value for 'x' in (1.22), we obtain

<div class="math">0.145 = Ae^{-0.000124t}
0.145 = e^{-0.000124t}
log0.145 = -0.000124t
-1.9310 = -0.000124t
t = 15573</div>

Hence the cave dwellers lived approximately 15,500 years ago.

>>5493567

Equation (1.22) expresses the relationship between the variable quantity 'x' and the variable time 't'. We are therefore at last in a position to answer the original question: How long ago did the cave dwellers live? By a chemical analysis of the charcoal, the chemist was able to determine the ratio of the amounts of <span class="math">C^{14}[/spoiler] to <span class="math">C^{12}[/spoiler] present at the time of the discovery of the cave. A comparison of this ratio with the fixed ratio of these two carbons in living trees disclosed that 85.5 percent of the amount of <span class="math">C^{14}[/spoiler] present at death had decomposed. Hence 0.145'A' units of <span class="math">C^{14}[/spoiler] remained. Substituting this value for 'x' in (1.22), we obtain
<span class="math">0.145 = Ae^{-0.000124t}[/spoiler]
<span class="math">0.145 = e^{-0.000124t}[/spoiler]
<span class="math">log0.145 = -0.000124t[/spoiler]
<span class="math">-1.9310 = -0.000124t[/spoiler]
<span class="math">t = 15573[/spoiler]

Hence the cave dwellers lived approximately 15,500 years ago.

>>5493536
By writing <span class="math">\frac{dx}{dt}\cdot[/spoiler] you imply that the amount of C14 atoms is a differentiable function in t.
There is always an integer amount of atoms and every purely integer-valued differentiable function is constant.

>>5493599
True, but when you're accelerating, you can still have a discrete velocity at any point in time. However, when you bring time back into consideration, velocity is still changing with respect to time. Hence acceleration is dv/dt.
The same thing applies in your case.

>>5493390
Blacks out produce whites in america at a 6:1 ratio

>>5493385
>hanging in facebook
>see stupid post
>get mad abou stupid post

u w0t m8?

>>5493599
>implying this matters when N_A = 6.02214129×10^23

>>5493613
>thinks velocity is not continuous

inb4 planck velocity

>>5493385
>ive seen fossilized cooked hams from like 15 years ago and hats from the 70's
>hats from the 70's