Solve the solution.

Question

                                                                                                         kt

The concentration of bacteria in a sample can be modeled by B(t) = B  e    , where t is in hours and B is the

                                                                                                    0 

concentration in billions of bacteria per liter.

 

 

(a) If the concentration increases by 14% in 5 hours, find K?

(b) if B   = 1.4, find B after 8.5 hours. 

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(c) By what percentage does the concentration increase in each hour?

 

(a) K = ?     (Type an integer or a decimal rounded to four decimal places as needed.)

Answer 1

B(t) = B_0 e^(kt)

 

1.14 B_0 = B_0 e^(5k)

 

1.14 = e^(5k)

 

5k = ln(1.14)

 

k = (ln(1.14))/5 ≈ 0.02620565248128 ≈ 2.6206 % increase / hour

 

B(8.5) = 1.4 e^(8.5(ln(1.14))/5)

 

= 1.4 e^(1.7 ln(1.14))

 

B(8.5) ≈ 1.74930800271248 billions of bacteria per liter

Answer 2

B( t) = B e     , which t ?

Answer 3

(a)   1.14(B) = Be^(kt)

   solve for  K= ln(1.14)/t

                   = ln(1.14)/5

                    =  0.0262/hr   => 2.62% per hr

(b)   we know k = 0.0262/hr 

         B = 1.4e^(0.0262*8.5)

            =1.75 billion of bacteria/liter

(c)   convert answer from part a to percent 2.62%/hour

 

 

Answer 4

Keith

            k= Ln(B/B₀)/T (take Ln of both side of the model equation and solve for k)

 

a. B/B₀ =1.14 (14% increase) T=5 hrs so k=.760/5 = .152 hr^-1 

b. B(8.5) = 1.4e^.152x8.5 = 5.103 bill/L

c. dB/dt=kB so (1/B)dBd/dt) =k is the fractional change in B per unit time so 100k is the percentage increase per hour.

 

Jim