/sci/ - Science & Math

>>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.

>>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}<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]