Differentiation(Maxima)

2009-01-01 6:08 pm

The concentration C(t) units of a drug in a patient's bloodstream t hours after injection is given by
C(t) = 20 t^2 e^(-3t) , for t > 0 or t = 0.

(a) Find the time interval on which C(t) is increasing.
(b) Find the maximun concentration and the number of hours after injection at which it occurs.

回答 (1)

老爺子

2009-01-01 7:06 pm

✔ 最佳答案

(a)
C(t) = 20t2e-3t , for t ≥ 0

C'(t)
= d(20t2e-3t)/dt
= 20t2(de-3t/dt) + 20e-3t(dt2/dt)
= 20t2(-3e-3t) + 20e-3t(2t)
= 40te-3t - 60t2e-3t
= 20e-3tt(2 - 3t)

When C'(t) > 0
20e-3tt(2 - 3t) > 0
0 < t < 2/3

Between 0 h and 2/3 h, C(t) is increasing.

(b)
C(t) = 20t2e-3t , for t ≥ 0
C'(t) = 20e-3tt(2 - 3t)

C"(t)
= d(40te-3t - 60t2e-3t)/dt
= 40t(de-3t/dt) + 40e-3t(dt/dt) -3d(20t2e-3t)/dt
= 40t(-3e-3t) + 40e-3t - 3(40te-3t - 60t2e-3t)
= -120t3-3t + 40e-3t - 120te-3t + 180t2e-3t
= 180t2e-3t - 240te-3t + 40e-3t
= 20e-3t(9t2 - 12t + 2)

When t = 2/3:
C(t) = 20(2/3)2e-3(2/3) = 80/9e2
C't = 20e-3(2/3)(2/3)[2 - 3(2/3)] = 0
C't = 20e-3(2/3)[9(2/3)2 - 12(2/3) + 2] = -40/e2 < 0

The maximum concentration is 80/9e2 units (or 1.203 units) at 2/3 hours after injection.
=

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