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Question 1

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A cup of coffee is cooling. Its temperature after ttt minutes is C(t)C(t)C(t) degrees Celsius, a decreasing, differentiable function. For 0≤t≤120 \le t \le 120≤t≤12, some of its values are in the table.

ttt (minutes)03712
C(t)C(t)C(t) (°C)100856955
  1. (a)\text{(a)}(a)

    Use the average rate of change of CCC over 3≤t≤73 \le t \le 73≤t≤7 to approximate C′(5)C'(5)C′(5). Show the working and give units.

  2. (b)\text{(b)}(b)

    Use a left Riemann sum with the three subintervals in the table to approximate ∫012C(t) dt\displaystyle\int_0^{12} C(t)\,dt∫012​C(t)dt. Then say what 112∫012C(t) dt\dfrac{1}{12}\displaystyle\int_0^{12} C(t)\,dt121​∫012​C(t)dt means for the coffee.

  3. (c)\text{(c)}(c)

    For 12≤t≤2012 \le t \le 2012≤t≤20, the cooling is modelled by C′(t)=−24.55 e0.01ttC'(t) = -\dfrac{24.55\,e^{0.01t}}{t}C′(t)=−t24.55e0.01t​ degrees Celsius per minute. Find the temperature at t=20t = 20t=20. Show the setup.

  4. (d)\text{(d)}(d)

    For that model, C′′(t)=0.2455 e0.01t (100−t)t2C''(t) = \dfrac{0.2455\,e^{0.01t}\,(100 - t)}{t^2}C′′(t)=t20.2455e0.01t(100−t)​. For 12<t<2012 < t < 2012<t<20, is the temperature changing at an increasing rate or a decreasing rate? Give a reason.

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