Math 1B Quiz #11Thursday, 15 November 2007GSI: Theo Johnson-Freydhttp://math.berkeley.edu/∼theojfName:1. (3 pts) Solve the following initial-value problem, to write f as a function of t:dfdt= t ·f2+ 1, f(2) = 0This equation is separable, so we multiply and divide, and then plug in our initialvalue:dfdt= t ·f2+ 1dff2+ 1= t dtarctan f =t22+ Cf = tant22+ Carctan 0 =222+ C0 = 2 + CC = −2f(t) = tant22− 212. (3 pts) Solve the following differential equation, to find a one-parameter family ofsolutions for y as a function of x:2xy0+ y = 6xWe use the method of finding a multiplier, in order to recognize the equation as aseparable one:y0+12xy = 3Zdx2x=12ln xe12ln x=√x√xy0+12xy=√x y0= 3√x√x y = 2x3/2+ Cy = 2x + C/√x23. (4 pts) Carbon has two stable isotopes — carbon-12 (12C) and carbon-13 (13C) —and one relatively common radioactive isotope carbon-14 (14C), produced in the up-per atmosphere by bombardment with cosmic radiation. Plants absorb atmosphericcarbon, and hence the concentration of14C in plants is equal to the atmosphericconcentration. When plants die, they do not absorb any new carbon. The amountof14C in archeological samples is used to date archeological sites.(a) Like all radioactive materials,14C decays at a constant relative rate: the amountthat decays in any given period of time is proportional to the amount present.Write a differential equation modeling the amount of14C in a given amount oftime.Let f(t) be the amount of14C after time t. Thendfdt= −kfWe write −k for the coefficient, since we know that the amount of14C is decreas-ing, and this lets us use a positive k. Remark: the solution to this differentialequation is f(t) = f(0)e−kt.(b) The half-life of14C is 5730 years, and the atmospheric concentration of14C is 600billion atoms per mole (roughly one part per trillion). What is the solution toyour differential equation (relating how much time has elapsed with the amountof14C left)? You do not need to simplify, but you do need to use units.We have f(t) = f (0)e−kt. Before any14C has decayed, i.e. at time t = 0, it isat f(0) = 600 billion atoms per mole. In 5730 years, the total14C has halved:1/2 = e−k 5730years, so k = ln(2)/5730. Thusf(t) =600 × 109rmatoms/molee−(ln(2)/5730 yrs) t(c) A sample from Fell’s Cave, in southern Chile, has a14C concentration of 150billion atoms per mole. Roughly what is the date of the archeological site?By solving the equation above for f(t) = 150 billion atoms per mole, we get theright t. Faster: for the14C to decay from 600 billion atoms per mole to 150billion atoms per mole requires two halvings: thus, two half lifes = 11460 yearshave passed, so the sample is from roughly 9000 B.C.E.3(d) What is the rate of radioactive decay of14C →12C in the sample if the currentconcentration of14C is 150 billion atoms per mole? You do not need to simplify,but you do need to report units.We recall that f0(t) = −kf(t). Then if f(t) = 150 billion atoms per mole, wemust have f0(t) = −(ln(2)/5730 yrs) ×150 ×
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