newell_lec11-15_100713

# 24 assuming the same values of k and k and a

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Unformatted text preview: nute. The concentration of the new solution is 1/4 lb. per gallon. The water/salt solution is kept well mixed and drained from the tank at a rate of 1 gallon per minute. (i) Write down a diﬀerential equation for x(t), the amount of salt in lbs. in the tank at time t. What is x(0)? (ii) Solve the diﬀerential equation with this initial value. Write an explicit expression for x(t). (iii) What are the values x(0), x(100), x(400), i.e., the amount of salt in the tank initially, after 100 minutes, after 400 minutes. (iv) Let c(t) be the concentration of salt in the brine solution at time t. Express c(t) as a function of x(t) and t. To what value does c(t) ﬁnd as t → ∞? Explain this result. Question 4: Radiocarbon dating is based on the fact that living material, (wood, plants, bones), accumulates a small amount of Carbon 14, a radioactive isotope of carbon. At the plant’s death, the isotope begins to decay at a rate proportional to the amount present. Let Q(t) be the amount of the isotope in grams at time t, measured in years. Write an equation for Q(t) relating the rate of change (i.e. dQ(t) ), with the dt amount present, (i.e. Q(t)). Show that the time z for Q(t) to halve in value, i.e. Q(z ) = 1 Q(0) is independent of Q(0), the starting amount. Given the half life, z , 2 of C 14 is 5568 years, how long has some previously living matter been dead if the remains contain only 20% of the original amount contained when the “plant” was alive? Question 5: Given that the population doubling time for a population that grows under the Malthus Law dy = ky is 100 ln 2 years, ﬁnd k . How long does it take the population dt to increase by a factor of 5? Using the same value of k , and assuming the population saturation level K is 106 , ﬁnd the population y (t) after 300 ln 2 years given the initial population was 100,000. 24 Assuming the same values of k and K and a harvesting rate of h = year, ﬁnd lim y (t) where y (t) satisﬁes the ode t→∞ ￿ dy y￿ = ky 1 − −h dt K for case (a) y (0) = 5000, 000 (b) y (0) = 200, 000. 3 16 · 104 per Question 6: Given that y1 (t) = et is a solution of d2 y dy (t + 1) + y = 0, 2 dt dt ﬁnd a second, linearly independent solution y2 (t). Solve the initial value problem given y (1) = 2, y ￿ (1) = 1. t Quest...
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