1 answer

The concentration of a chemical in the bloodstream t hours after injection into muscle tissues is...

Question:

The concentration of a chemical in the bloodstream t hours after injection into muscle tissues is given by the following. C(t
C$ 1.15 C(D 0.15 0.1 0.05 1 1.5 CD 0.15 G10 1.15 005 1 1.5 2 253 0 0.5 1 1.5 2 253 t = 2.0801 hr (C) Use calculus to determin
The concentration of a chemical in the bloodstream t hours after injection into muscle tissues is given by the following. C(t) 120 (a) Complete the table and use it to approximate the time when the concentration is greatest. (Round your answers to three decimal places.) + 0 0.5 1 1.5 2.5 C() 0.092 0.179 0 .247 0.286 293 0.278 t2.5 hr (b) Use a graphing utility to graph the concentration function and use the graph to approximate the time when the concentration is the greatest. CIB 0.15
C$ 1.15 C(D 0.15 0.1 0.05 1 1.5 CD 0.15 G10 1.15 005 1 1.5 2 253 0 0.5 1 1.5 2 253 t = 2.0801 hr (C) Use calculus to determine analytically the time when the concentration is greatest.

Answers

I am assuming you need solution of part C only as part A and B are correctly solved!

c(t) = 5t 1 27+t3 In order to find when the concentration, we will rese concept q maxima and minima of calcules. first deriva3 d(t) = 5 cm (2876) ckey = 5(221 at the (1274837 end (27+372 3 c(t) = 5[(27+43) (1) -+10+3t?] (274832^ct) = 5/27+63 -3 +] (27+€32 56 (t) = 5 (27 - 2+3) 127+ + 2 for maximum or minimum cl (t) = 0 = 5(27-2€ o ( 27+32 = 27-2ť =t = 3 (2)/ c(t) will be maximum at to Byte if c (és va co So, taking second derivative of c(t) c(t) = d c(t) dt-dr5(27-2+3) 7 Duoef ks 24? (29sctt) = 57(2346) (27-24) 1 -(27-28) d(27++327 - ((27+ +32 2zlity = 5 (27te) (6-2 (3) - 1 l (27-2)(2+)). (27437 (27 +23) 345(t) = 5 1-6++(27+esj?= 2(27-28). (27+€3).(0+3+²) 7 (27++3;4 at t=3 (2) 1/3| (중) - - (2) - 227-19 1태를녀름에 223 (24tN LG x 9. 3.</p><p>14 | 2f 22 el) - FI-12 )-12-3 2. (2 )()7 (27끌다 (6) - 1 2 ) - 2(2747)(27)() (27 )(0) 일) (24 ) ]re(一)--卷(!) 7(高) (4) 一 ) 十 ) = (): 本-) - - 0. 1036 le cul ) < 0 so (lt) és maximum at t = 3 21/3

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