Answered

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(c). Under a set of controlled laboratory conditions, the size of the population P of a certain bacteria culture at time t (in seconds) is given by the function P(t)= 3t² +3e' +10, t≥0.

(1) What is the size of the population after 1 minute?

(ii) Find the average rate of change of P at t = 2 and t = 6?

(iii) How fast is the size of the population changing after 1 minute?

[Verify your answer by MATHEMATICA and attach the printout of the commands and output

C Under A Set Of Controlled Laboratory Conditions The Size Of The Population P Of A Certain Bacteria Culture At Time T In Seconds Is Given By The Function Pt 3t class=

Sagot :

LammettHash

(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is

P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155

In Mathematica, it's convenient to define a function:

P[t_] := 3t^2 + 3E^t + 10

(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)

Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.

(ii) The average rate of change of P(t) over an interval [a, b} is

(P(b) - P(a))/(b - a)

Then the ARoC between t = 2 and t = 6 is

(P(6) - P(2))/(6 - 2) ≈ 321.030

In M,

(P[6] - P[2])/(6 - 2)

and you can also include N just as before.

(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :

P'(t) = 6t + 3e^t

Evaluate the derivative at t = 60 :

P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰

The approximate value is quite large, so I'll just leave its exact value.

In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.

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