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Calculus

The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT/dt = -k(T - A), where T is the temperature of the tea, A is the room temperature, and k is a positive constant. If the water cools from 100°C to 80°C in 1 minute at a room temperature of 30°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

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  1. dT/dt = -k(T-A)
    dT/(T-A) = -k dt
    ln(T-A) = -kt + c
    T-A = c e^(-kt)
    T = A + c e^(-kt)
    Now just plug in you data point, and you have
    30 + c e^-k = 80
    ce^-k = 50
    Now, what you left out is that c = 100-30 = 70
    e^-k = 5/7
    k = ln(7/5) = 0.336
    So, now you know that
    T(t) = 30 + 70e^(-1.25t)
    So now you can find T(4)

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    oobleck
  2. T(t) = 30 + 70e^(-1.25t) isn't correct answer.

    It's like this.

    k = ln ( 7 / 5 )

    T = A + c ∙ e ^ ( - k t )

    T = 30 + 70 ∙ e ^ ( - k t )

    T = 30 + 70 ∙ e ^ [ - ln ( 7 / 5 ) ∙ t ]

    T = 30 + 70 ∙ [ e ^ ln ( 7 / 5 ) ] ^ ( - t )

    T = 30 + 70 ∙ ( 7 / 5 ) ^ ( - t )

    T = 30 + 70 ∙ 1.4 ^ ( - t )

    T(4) = 30 + 70 ∙ 1.4 ^ ( - 4 )

    T(4) = 30 + 70 ∙ 0.26030825

    T(4) = 30 + 18.2215775

    T(4) = 48.2215775 °C

    T(4) = 48°C

    to the nearest °C

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  3. It is better to keep k value in exact form till you get final answer. So use k=ln(5/7) to find T(4).
    Also ln(5/7) can be written as -ln(7/5) using properties of logs
    T(4)=30+70e^(ln(7/5)*4) =48.2

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  4. The temperature, f(t) of a cup of coffee after t minutes can be determined by the equation f(t)=63(0.91)T

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