Unformatted text preview: 37. 38. 39. 40. 41. AU) : Ane'o'mﬂ' = 0.0m0 for t=(lnlOD)/O.13153=35.01 and ﬁnd that it will be
about 35 years until the region is again inhabitable. Taking r = 0 when the body was formed and r = T now, the amount QU) of 233U in
the body at time t (in years) is giVen by QU) = Que—h, where k = (1n 2)](4.5]><109).
The given information tells us that _QLJ_ : 0.9
ovmn After substituting Q(T) = Que—H, we solve readily for e” = 19/9, so
T = (l/k)ln(l9/9) == 4.86x109. Thus the body was formed approximately 4.86 billion years ago. Taking t = 0 when the rock contained only potassium and t = T now, the amount
Q0) of potassium in the rock at time t (in years) is given by Q0) = Qoe'k’, where
k = (In 2)/(l.28><109). The given information tells us that the amount AU) of argon at
time t is Al!) == HQWQUH
and also that A(T) = Q(T). Thus osonx9mn. After substituting Q(T) = Q6” we readily solve for T = (lnlO/ln2)(1.28X109) = 4.25X109. Thus the age of the rock is about 1.25 billion years. Because A = 0 the differential equation reduces to 2" = id", so T0) = 252‘“. The
fact that H20) = 15 yields 1.: = (1/20)ln(5/3), and finally we solve 5 = 25.94" for t = (1n 5)”: z 63 min. The amount of sugar remaining undissolved after 3‘ minutes is given by AU) 2 Anew:
we find the value of k by solving the equation an) = Aget = 0.75210 for k = —ln0.75 = 0.28768. To find how long it takes for half the sugar to dissolve, we solve
the equation AU) = A084" :JiAa for t=(lr‘1 2)f0.28768==2.41 minutes. (a) The light intensity at a depth of 1 meters is given by ﬁx) = 1084'“. We solve
the equation Itx) = JOEL“ : é!“ for x = (1n 2)ll.4 x 0.495 meters. Section 1.4 17 ...
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