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Differentiating the Rate of Decay

Date: 3/14/96 at 11:53:36
From: John Duncan Burns
Subject: Differentiation

Dear Dr Math,

Hello. I am a 16-year-old studying A level math and further math 
in England.

Our class has been trying to work out this problem without much 
luck. We need to do variable separables but are getting stuck 
after this point.

A lump of radioactive subtance is disintegrating at time 't' days 
after it was first observed to have mass 10 grams and

      -----   = -km          where k is a positive constant

Find the time, in days, for the substance to reduce to 1 gram in 
mass, given that its half life is 8 days. (the half-life is the 
time in which half of any mass of the substance will decay)

Please help. Thank you, 

Andrew Burns.

Date: 3/26/96 at 15:39:41
From: Doctor Sebastien
Subject: Re: Differentiation


dm/dt = -km
We now use the method of variable separables:

(1/m)dm/dt = -k
Both sides are divided by m.

Integrating both sides w.r.t. t - from now on, I will write 
'Integrate' instead of the integrate sign.

Integrate (1/m dm/dt) dt = Integrate (-k) dt
Integrate (1/m) dm = Integrate (-k) dt
ln(m) = -kt + c
m = e^(-kt + c)
m = (e^(-kt))*(e^c)
m = Ae^(-kt), where A = e^c
At time t = 0, we know that the mass is 10g. 
At time t = half-life, mass = 5g

When t = 0, that is, at the start of the experiment,
m = Ae^(-k*0) = A
Therefore, A = 10
Therefore, m = 10e^(-kt)

When t = half-life = 8 days,
5 = 10*e^(-k*8)
1/2 = e^(-8k)
e^(-8k) = 1/2
-8k = ln (1/2)
-k = (1/8)ln (1/2)
Therefore, m = 10 e^{(1/8)ln (1/2) * t} = 10 e^{(t/8)ln(1/ 2)}

When m = 1g, 1 = 10 e^{(t/8)ln(1/2)}
e^{(t/8)ln(1/2)} = 0.1
(t/8)ln(1/2) = ln (0.1)
t/8 = ln(0.1) / ln (1/2)
t = (8 ln(0.1)) / ln (1/2)
t = 26.6

Therefore, after 26.6 days, the mass of the radioactive substance 
is 1 gram.

-Doctor Sebastien,  The Math Forum

Associated Topics:
High School Calculus

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