AP Calculus BC 2.1 Defining Average and Instantaneous Rates of Change at a Point Exam Style Questions - FRQ
Question
(a) Topic-2.1-Defining Average and Instantaneous Rates of Change at a Point
(b) Topic-6.2-Approximating Areas with Riemann Sums
(c) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(d) Topic-5.3-Determining Intervals on Which a Function Is Increasing or Decreasing
1. The temperature of coffee in a cup at time t minutes is modeled by a decreasing differentiable function C, where C ( t ) is measured in degrees Celsius. For \(0\leq t\leq 12\), selected values of C ( t) are given in the table shown.
(a) Approximate C'( 5) using the average rate of change of C over the interval \(3\leq t\leq 7\). Show the work that leads to your answer and include units of measure.
(b) Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the value of \(\int_{0}^{12}C(t)dt\) . Interpret the meaning of \(\frac{1}{12}\int_{0}^{12}C(t)dt\) in the context of the problem.
(c) For \(12\leq t\leq 20\) , the rate of change of the temperature of the coffee is modeled by C'(t)=\(\frac{-24.55e^{0.01t}}{t}\) where C ‘( t) is measured in degrees Celsius per minute. Find the temperature of the coffee at time t = 20.Show the setup for your calculations.
(d) For the model defined in part (c), it can be shown that C “( t) =\(\frac{0.2455e^{0.01t}(100-t)}{t^{2}}\) . For \(12\leq t\leq 20\),determine whether the temperature of the coffee is changing at a decreasing rate or at an increasing rate.Give a reason for your answer.
▶️Answer/Explanation
1(a) Approximate C'(5) using the average rate of change of C over the interval 3 < ¢ < 7. Show the work that leads to your answer and include units of measure.
\(C'(5)= \frac{c(7)- c(3)}{7-3} = \frac{69-85}{4} = -4 \) degrees Celsius per minute
1(b) Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the value of \(\int_{0}^{12} C(t)dt.\) Interpret the meaning of \(\frac{1}{12}\int_{0}^{12} C(t)\\\ dt\) in the context of the problem.
\(\int_{0}^{12} C(t)dt \approx (3-0).C(0) +(7+3).C(3)+ (12-7).C(7)\)
\(=3.100 +4.85+5.69 = 985\)
\(\frac{1}{12}\int_{0}^{12} C(t)\\\ dt\) is the average temperature of the coffee (in degrees Celsius) over the interval from t = 0 to t = 12.
1(c) For 12 < t < 20, the rate of change of the temperature of the coffee is modeled by \(C'(t)=\frac{-2455e^{0.01t}}{t}\), red in degrees Celsius per minute. Find the temperature of the coffee at time t = 20. Show the setup for your calculations.
\(C(20) = C(12) +\int_{12}^{20} C'(t) dt\)
\(=55-14.670812 = 40.329188\)
The temperature of the coffee at time t = 20 is 40.329 degrees Celsius.
1(d) For the model defined in part (c), it can be shown that C”(t) = \(\frac{0.2455e^{0.01t}(100-t)}{t^{2}}\). For 12 < t < 20, determine whether the temperature of the coffee is changing at a decreasing rate or at an
increasing rate. Give a reason for your answer.
Because C”(t) > 0 on the interval 12 < t < 20, the rate of change in the temperature of the coffee, C’(t), is increasing on this interval.
That is, on the interval 12 < t < 20, the temperature of the coffee is changing at an increasing rate.