匿名使用者
匿名使用者 發問時間: 科學數學 · 4 年前

exponential growth?

If the grown rate of the number of bacteria at any time t is proportional to the number present at t and doubles in 1 week, how many bacteria can be expected after 2 weeks?

1 個解答

評分
  • Lopez
    Lv 5
    4 年前
    最佳解答

    Let dP / dt = kP , where P denotes the population of bacteria

    Then dP / P = k dt

    ∫ (1/P) dp = ∫ k dt + c1

    ln P = kt + c1

    P = e^( c1 + kt ) = c * e^(kt)

    P0 = P(0) = c , where P0 denotes the initial number, that is , P( t = 0 )

    Hence P = P0 * e^(kt)

    Since the number doubles in 1 week,

    P(1) = P0 * e^(k*1) = 2 * P0

    e^k = 2 , so we have

    P = P0 * e^(kt) = P0 * (e^k)^t = P0 * 2^t

    P(2) = P0 * 2^2 = 4 * P0

    Ans: After 2 weeks, the number will be 4 times of the initial number.

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