118_pdfsam_math 54 differential equation solutions odd

# 118_pdfsam_math 54 differential equation solutions odd - t...

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Chapter 3 25. Let M ( t ) denote the mass of carbon-14 present in the burnt wood of the campFre. Then since carbon-14 decays at a rate proportional to its mass, we have dM dt = αM, where α is the proportionality constant. This equation is linear and separable. Using the initial condition, M (0) = M 0 we obtain M ( t )= M 0 e αt . Given the half-life of carbon-14 to be 5600 years, we solve for α since we have 1 2 M 0 = M 0 e α (5600) 1 2 = e α (5600) , which yields α = ln(0 . 5) 5600 0 . 000123776 . Thus, M ( t )= M 0 e 0 . 000123776 t . Now we are told that after t years 2% of the original amount of carbon-14 remains in the campFre and we are asked to determine t .Thu s 0 . 02 M 0 = M 0 e 0 . 000123776 t 0 . 02 = e 0 .
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Unformatted text preview: t t = ln 0 . 02 . 000123776 31 , 606 (years) . 27. The element Hh decays according to the general law of a radioactive decay, which is described by (3.2) (this time, with t measured in years). Since the initial mass of Hh is m = 1 kg and the decay constant k = k Hh = 2 / yr, we get Hh( t ) = e k Hh t = e 2 t . (3.3) or It, the process is more complicated: it has an incoming mass from the decay of Hh and, at the same, looses its mass decaying to Bu. (This process is very similar to brine solution 114...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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