If the half life of the anti biotic in the

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. If the half-life of the anti-biotic in the bloodstream is 2 hours, what rate of infusion should he prescribe to maintain a long-term amount of 290 mgs in Diana’s bloodstream? 1. Infusion rate = 2 . 137 mgs/min 2. Infusion rate = 1 . 79 mgs/min 3. Infusion rate = 1 . 906 mgs/min 4. Infusion rate = 2 . 021 mgs/min 5. Infusion rate = 1 . 675 mgs/min correct Explanation: Let a be the infusion rate (in mgs per minute). Then the rate at which the amount of anti-biotic in the bloodstream is changing will be a combination of this infusion and the loss through elimination. Thus y ( t ) satisfies the differential equation dy dt = a ky where k is a constant. This is a separable variables equation which after integration be- comes integraldisplay 1 a ky dy = integraldisplay dt. Consequently, 1 k ln( a ky ) = t + C, i.e. , y ( t ) = 1 k parenleftBig a Ae kt parenrightBig , with A an arbitrary constant. Initially, there will be no anti-biotic in the bloodstream, so y (0) = 0 = y ( t ) = a k parenleftBig 1 e kt parenrightBig . The value of k will be determined by the half- life of the anti-biotic in the bloodstream since t = 120 = e 120 k = 1 2 , i.e. , k = (1 / 120) ln 2 = 0 . 00577625 . To main- tain a long-term level of c mgs in the blood- stream, therefore, lim t → ∞ y ( t ) = a k = 290 . Hence Infusion rate 1 . 675 mgs/min. 032 10.0points In a West Texas school district the school year began on August 1 and lasted until May 31. On August 1 a Soft Drink company in- stalled soda machines in the school cafeteria. It found that after t months the machines generated income at a rate of f ( t ) = 400 t 5 t 2 + 8 dollars per month. Find the total income produced during the first semester ending on December 31. 1. income = $92 . 43 2. income = $152 . 43 3. income = $112 . 43 correct 4. income = $132 . 43 5. income = $172 . 43 Explanation: If the first semester lasted from August 1 until December 31, then this period covered the time interval from t = 0 until t = 5 (re- member, the month of August corresponds to
pacheco (jnp926) – Homework 5 – staron – (52840) 17 the time interval (0 , 1]). Thus the total in- come generated during the first semester is given by the integral T first = integraldisplay 5 0 f ( t ) dt = integraldisplay 5 0 400 t 5 t 2 + 8 dt. But integraldisplay 5 0 400 t 5 t 2 + 8 dt = bracketleftBig 400 10 ln(5 t 2 + 8) bracketrightBig 5 0 = 400 10 ln parenleftBig 125 + 8 8 parenrightBig . Hence income = $112 . 43 . 033 10.0points Find the amount A in an account after 9 years when In turn, after exponentiation this becomes A ( t ) = e C e 0 . 05 t with C an arbitrary constant. The constant C is determined by the initial condition A (0) = 100 for A (0) = 100 = e C = 100 . Consequently, the amount in the account after t years is given by A ( t ) = 100 e 0 . 05 t .
Explanation: After integration the differential equation dA dt = 0 . 05 A becomes integraldisplay 1 A dA = integraldisplay 0 . 05 dt. Thus ln A = 0 . 05 t + C.

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