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Another doctor has a patient whose baseline hormone level is too low. Rather than having to give a series of single shots, she decides to give the patient a continuous intravenous infusion of the hormone supplement. If there were no metabolism, the infusion would cause the concentration o hormone to increase at a constant rate of A mg/L/hr. The velocity constant of elimination for this patient is equal to k hour1. At the initial time, t the patient's hormone concentration is at their baseline. (d) Write down the right side of the differential equation satisfied by y. (Your answer should be in terms of y.) dy dt -ky + A (e) This differential equation can be rewritten in the more familiar form Y = k(M - y) for a particular choice of M. Write down an dt expression for M in terms of A and k. A k (f) What is the patient's initial excess hormone concentration? y(0) = 0 %3D (g) Using the solution of this differential equation, the excess hormone concentration is given by y(t) = M(1 - e-kt). In terms of A and k, what value does the excess hormone concentration approach in the long run? A k mg/L (h) After doing some tests, the patient's velocity constant of elimination is found to be 0.03 hour1. The doctor plans to administer the infusion for 48 hours and wants to raise the excess hormone concentration to 3.8 mg/L at that time. What infusion rate must the doctor choose to reach this target? 0.15 mg/L/hr At this infusion rate, what will the excess hormone concentration be at the half-way time? 2.55 mg/L At this infusion rate, how long will it take for the excess hormone concentration to reach half of the target level? 12.67 hours

Another doctor has a patient whose baseline hormone level is too low. Rather than having to give a series of single shots, she decides to give the patient a continuous intravenous infusion of the hormone supplement. If there were no metabolism, the infusion would cause the concentration o hormone to increase at a constant rate of A mg/L/hr. The velocity constant of elimination for this patient is equal to k hour1. At the initial time, t the patient's hormone concentration is at their baseline. (d) Write down the right side of the differential equation satisfied by y. (Your answer should be in terms of y.) dy dt -ky + A (e) This differential equation can be rewritten in the more familiar form Y = k(M - y) for a particular choice of M. Write down an dt expression for M in terms of A and k. A k (f) What is the patient's initial excess hormone concentration? y(0) = 0 %3D (g) Using the solution of this differential equation, the excess hormone concentration is given by y(t) = M(1 - e-kt). In terms of A and k, what value does the excess hormone concentration approach in the long run? A k mg/L (h) After doing some tests, the patient's velocity constant of elimination is found to be 0.03 hour1. The doctor plans to administer the infusion for 48 hours and wants to raise the excess hormone concentration to 3.8 mg/L at that time. What infusion rate must the doctor choose to reach this target? 0.15 mg/L/hr At this infusion rate, what will the excess hormone concentration be at the half-way time? 2.55 mg/L At this infusion rate, how long will it take for the excess hormone concentration to reach half of the target level? 12.67 hours

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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*only need help with section H 3rd question. The rest are correct
Another doctor has a patient whose baseline hormone level is too low. Rather than having to give a series of single shots, she decides to give the patient a continuous intravenous infusion of the hormone supplement. If there were no metabolism, the infusion would cause the concentration of hormone to increase at a constant rate of A mg/L/hr. The velocity constant of elimination for this patient is equal to k hour1. At the initial time, t = 0, the patient's hormone concentration is at their baseline. (d) Write down the right side of the differential equation satisfied by y. (Your answer should be in terms of y.) dy - -ky + A (e) This differential equation can be rewritten in the more familiar form Y = k(M - y) for a particular choice of M. Write down an dt expression for M in terms of A and k. A M = (f) What is the patient's initial excess hormone concentration? y(0) = 0 (g) Using the solution of this differential equation, the excess hormone concentration is given by y(t) = M(1 - e-kt). In terms of A and k, what value does the excess hormone concentration approach in the long run? A k mg/L (h) After doing some tests, the patient's velocity constant of elimination is found to be 0.03 hour-1. The doctor plans to administer the infusion for 48 hours and wants to raise the excess hormone concentration to 3.8 mg/L at that time. What infusion rate must the doctor choose to reach this target? 0.15 mg/L/hr At this infusion rate, what will the excess hormone concentration be at the half-way time? 2.55 mg/L At this infusion rate, how long will it take for the excess hormone concentration to reach half of the target level? 12.67 hours
Transcribed Image Text:Another doctor has a patient whose baseline hormone level is too low. Rather than having to give a series of single shots, she decides to give the patient a continuous intravenous infusion of the hormone supplement. If there were no metabolism, the infusion would cause the concentration of hormone to increase at a constant rate of A mg/L/hr. The velocity constant of elimination for this patient is equal to k hour1. At the initial time, t = 0, the patient's hormone concentration is at their baseline. (d) Write down the right side of the differential equation satisfied by y. (Your answer should be in terms of y.) dy - -ky + A (e) This differential equation can be rewritten in the more familiar form Y = k(M - y) for a particular choice of M. Write down an dt expression for M in terms of A and k. A M = (f) What is the patient's initial excess hormone concentration? y(0) = 0 (g) Using the solution of this differential equation, the excess hormone concentration is given by y(t) = M(1 - e-kt). In terms of A and k, what value does the excess hormone concentration approach in the long run? A k mg/L (h) After doing some tests, the patient's velocity constant of elimination is found to be 0.03 hour-1. The doctor plans to administer the infusion for 48 hours and wants to raise the excess hormone concentration to 3.8 mg/L at that time. What infusion rate must the doctor choose to reach this target? 0.15 mg/L/hr At this infusion rate, what will the excess hormone concentration be at the half-way time? 2.55 mg/L At this infusion rate, how long will it take for the excess hormone concentration to reach half of the target level? 12.67 hours
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