A drug is administered intravenously at a constant rate of r mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality k>0. (Set up and) Solve a differential equation for the quantity, Q, in milligrams, of the drug in the body at time t hours. Assume there is no drug in the body initially. Your answer will contain r and k.
Q =
Graph Q against t. What is Q?, the limiting long-run value of Q?
Q?=
If r is doubled (to 2r), by what multiplicative factor is Q? increased?
Q? (for 2r) = Q? (for r)
Similarly, if r is doubled (to 2r), by what multiplicative factor is the time it takes to reach half the limiting value, 12Q?, changed?
t (to 12Q?), for 2r) = t (to 12Q?), for r)
If k is doubled (that is, we use 2k instead of k), by what multiplicative factor is Q? increased?
Q? (for 2k) = Q? (for k)
On the time to reach 12Q??
t (to 12Q?), for 2k) = t (to 12Q?), for k)

See explanation

Step-by-step explanation:

Solution:-

- We are told that a drug is administered to a patient at a rate of ( r ). The drug present in the patient at time t is Q. The drug also leaves the patient's body at a rate proportional to the amount of drug present at time t.

- We will set up the first order ODE for the rate of change of drug ( Q ) in the patient's body.

- We know that the rate of inflow is the rate at which the drug is administered that is ( r ) and the flow out is proportional to the amount currently present in the patient's body ( k*Q ). Where ( k ) is the constant of proportionality:

- Express the ODE in the standard form:

- The integrating factor ( u ) for the above ODE would be:

- Use the standard solution of ( Q ) using the integrating factor ( u ):

Where, c: the constant of integration.

- The initial value problem is such that there is no drug in the patient body initially. Hence, Q ( 0 ) = 0:

- The solution to the ODE is:

                 .... Answer

- We can use any graphing calculator to plot the amount of drug ( Q ) in the patient body. The limiting value of the drug in the long-run ( t -> ∞ ) can be determined as follows:

              Lim  ( t -> ∞ ) [ Q ( t ) ] = Lim  ( t -> ∞ ) [  

              Lim  ( t -> ∞ ) [ Q ( t ) ]  =  

- The long-run limiting value of drug in the body would be ( r / k ).

- If the rate of drug administrative rate is doubled then the amount of ( Q ) at any time t would be:

- The multiplicative factor is 2.

- To reach half the limiting value ( 0.5* r / k ) the amount of time taken for the double rate ( 2r ) of administration of drug would be:

- Similarly for the normal administration rate ( r ):

- The multiplicative factor ( M ) of time taken to reach half the limiting value is as follows:

- Similarly repeat the above calculation when the proportionality constant ( k ) is doubled.

x-3y=4

step-by-step e

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5