Differentiating the Rate of DecayDate: 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 dm ----- = -km where k is a positive constant dt 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 Hello! 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 |
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