15- A diabetic patient receives insulin at constant rate from an implanted insulin pump. Insulin has first order elimination kinetics, so the amount of insulin in the blood willobey the given differential equation, where a> 0 is the rate at which insulin is released into their blood by the pump and k, is the fraction of insulin removed from theblood in one unit of time.dM- = a -k, Mdt(a) Assuming M(0) = 0 (i.e., there is no insulin present in the patient's blood at time t= 0), solve the differential equation to find M(t) as a function of t.(b) Find the limit of M(t) as t-00.(c) Assume that a (the rate of release from the pump) is 5 IU/hr and M(t) → 0.2 IU as t+0o. Calculate k1, the rate of insulin elimination.(a) M(t) =(b) lim M(t) =(c) k, =(Type an integer or decimal rounded to three decimal places as needed.)

Name: 15- A diabetic patient receives insulin at constant rate from an impla
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: 15- A diabetic patient receives insulin at constant rate from an implanted insulin pump. Insulin has first order elimination kinetics, so the amount of insulin in the blood willobey the given differential equation, where a> 0 is the rate at which ins

Given : dMdt=a-k1M, M(0) =0 Integrating both sides we get, ∫M(0)M(t)dM(a-k1M)=∫0tdt Let a-k1M=u u1=au2=a-k1M and dM=-duk1 ⇒-1k1∫aa-k1Mduu=∫0tdt⇒-1k1lna-k1Ma=(t-0)⇒lna-k1Ma=-k1t ⇒a-k1M=ae-k1t ⇒a(1-e-k1t)=k1M ⇒M=ak1(1-e-k1t) limt→∞M(t)=limt→∞ak1(1-e-k1t) =ak1(limt→∞1 - limt→∞e-k1t) =ak1(1) =ak1 Hence limt→∞M(t) = ak1 limt→∞M(t)=0.2 IU 0.2=ak1 a=5 IU/hr ,So k1=50.2=25hr-1 k1=25 (hr)-1