Question:
| t (minutes) | 0 | 2 | 5 | 9 | 10 |
H(t) | 66 | 60 | 52 | 44 | 43 |
As a pot of tea cools, the temperature of the tea is modeled by a differentiable function H for 0 ≤ t ≤ 10, where time t is measured in minutes and temperature H(t) is measured in degrees Celsius. Values of H(t) at selected values of time t are shown in the table above.
(a) Use the data in the table to approximate the rate at which the temperature of the tea is changing at time t = 3.5. Show the computations that lead to your answer.
(b) Using correct units, explain the meaning of \(\frac{1}{10}\int_{0}^{10}H(t)dt\) in the context of this problem. Use a trapezoidal sum with the four subintervals indicated by the table to estimate \(\frac{1}{10}\int_{0}^{10}H(t)dt\).
(c) Evaluate \(\frac{1}{10}\int_{0}^{10}H'(t)dt\). Using correct units, explain the meaning of the expression in the context of this problem.
(d) At time t = 0, biscuits with temperature 100 C∞ were removed from an oven. The temperature of the biscuits at time t is modeled by a differentiable function B for which it is known that \(B'(t)=-13.84e^{-0.173t}.\) Using the given models, at time t = 10, how much cooler are the biscuits than the tea?
▶️Answer/Explanation
Ans:
(a) \(\frac{H(5)-H(2)}{5-2}=\frac{-8}{3}\frac{0_{C}}{min}\)
(b) \(\frac{1}{10}\int_{0}^{10}H(t)dt \approx \frac{\left [ 2\left ( \frac{66+60}{2} \right )+3\left ( \frac{52+60}{2} \right ) +4\left ( \frac{44+52}{2} \right )+1\left ( \frac{43+44}{2} \right )\right ]}{10}=52.95\)
This represents the average temperature in degrees celsius of the Tea over the interval 0 ≤ t ≤ 10
(c)
\(\int_{0}^{10}H'(t)dt = H(10)-H(0)=43-66 = -23^{0}C\)
This expression shows the total change in temperature in degree Celsius from t = 0 to t = 10.
(d)
\( B'(t)= -13.84e^{-1.73t}\)
\(B(10)= \int_{0}^{10}-13.84e^{-1.73t}+100 = 100-65.817 = 34.1827\)
43 – 34.1827 = 8.817
= 8.817 0C