Question
t (minutes) | 0 | 2 | 5 | 9 | 10 |
H(t) (degrees Celsius) | 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 \(\int_{0}^{10}H'(t)dt.\) Using correct units, explain the meaning of the expression in the context of this problem.
(d) At time t = 0, biscuits with temperature 100 0C 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)
\(\frac{1}{10}\int_{0}^{10}H(t)dt\approx \frac{\left [2(\frac{66+180}{2}) +3(\frac{52+60}{2})+4(\frac{44+52}{2})+1(\frac{43+44}{2}) \right ]}{10}=52.95\)
This represents the average temperature in degree 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)
\(B'(t)=-13.84e^{-.173t}\)
\(B(10)=\int_{0}^{10}-13.84e^{-.173t}+100 = 100-65.817 = 34.1827\)
43-341827 = 8.817
= 88170C Celsius