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. (a) Set up a differential equation for the quantity, Q, in milligrams, of the drug in the body at time t hours. (Your answer will contain the unknown constants r and k.)
(b) Solve this differential equation, assuming there is no drug in the body initially. (Your answer will contain r and k.)
(c) What is the limiting long-run value of Q?

a.)Q' = -kQ + r

b.)Q =

c.) Limiting long run value of Q =

Step-by-step explanation:

a.)

The rate of change is directly proportional to the quantity, Q

⇒ ∝ Q

⇒ = -kQ ( because it is decreasing )

Also given, quantity is increasing with a constant rate r

⇒ = -kQ + r

⇒Q' = -kQ + r

b.)

As we have

= -kQ + r

⇒ = dt

⇒∫ = ∫dt

⇒-

⇒log(-kQ + r) = -kt -kC

Taking exponential both side, we get

⇒-kQ + r =

⇒-kQ = -r +

⇒Q = -

⇒Q = -

⇒Q = -      .......(1)

Now,

At t = 0, Q = 0

0 = - B

⇒

⇒B = r

∴ equation (1) becomes

Q = -

⇒Q =

c.)

for limiting long run value of Q

               =  

               =

⇒Limiting long run value of Q =

-13

step-by-step explanation:

f(x)=2x-5 for f(-4)

f(-4)   means let x=-4

f(-4) = 2(-4) -5

f(-4) =-8-5

      = -13

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09