6. A drug is injected into a patient\'s blood stream. The concentration of the d

Question

6. A drug is injected into a patient's blood stream. The concentration of the drug in the blood stream t hours after the drug is injected is modeled by the formula C(t) 0.12t t2+t+1 where C is measured in milligrams per cubic centimeter (a) Compute the sensitivity of C at t = 0.5 (corresponding to 30 minutes after injec- tion) (b) Use your answer to part (a) to estimate the change in concentration over the time period from 30 to 33 minutes after injection. (c) Determine the maximum concentration and the time when it occurs.

Explanation / Answer

a)

c(t) = 0.12t/(t2 + t + 1)

t = 0.5

c(t = 5) = 0.12(5)/[(0.5)2 + 0.5 + 1]

= 0.342

b)

at t = 30 minutes

= 1/2 hours

= 0.5

c( t = 0.5) =   0.12(5)/[(0.5)2 + 0.5 + 1]

= 0.342

t = 33 minutes

= 0.55 h

c(t= 0.55) = 0.12(0.55)/[(0.55)2 + 0.55 + 1]

= 0.035

change in C = 0.035 - 0.342

= - 0.307

means when t increases from 30 minutes to 33 minutes then C decreases by 0.307

c)

c(t) = 0.12t/(t2 + t + 1)

in order to find maximum time we use concept of calculus

c(t) = 0.12t/(t2 + t + 1)

= 1/[(t2 + t + 1)/0.12t]

= 1/g(t)

where g(t) =  (t2 + t + 1)/0.12t

= t/0.12 + 1/0.12 + 1/0.12t

so c(t) will be maximum when g(t) is minimum

g(t) = t/0.12 + 1/0.12 + 1/0.12t

g'(t) = 1/0.12 + 0 - 1/0.12t2

put   g'(t) = 0

1/0.12 - 1/0.12t2 = 0

so 1 - 1/t2 = 0

t2 = 1

t = 1

2nd order condition

g''(t) = 2/0.12t3

at t = 1

g''(t) = 2/0.12 > 0

g(t) is minimum at t = 1

hence c(t) is maximum at t = 1

so mamimum c = 0.12(1)/[(1)2 + 1 + 1]

= 012/3

= 0.04

  

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