AP Calculus BC 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals - FRQs - Exam Style Questions
Calc-Ok Question
| \(t\) (minutes) | \(C(t)\) (°C) |
|---|---|
| 0 | 100 |
| 3 | 85 |
| 7 | 69 |
| 12 | 55 |
(a) Estimate \(C'(5)\) using the average rate of change of \(C\) over the interval \(3\le t\le 7\). Include units.
(b) Use a left–Riemann sum with subintervals \([0,3],\,[3,7],\,[7,12]\) to approximate \(\displaystyle \int_{0}^{12} C(t)\,dt\). Then interpret \(\displaystyle \frac{1}{12}\int_{0}^{12} C(t)\,dt\) in the context of the problem.
For \(12\le t\le 20\), the rate of change of the coffee’s temperature is given by \(\displaystyle C'(t)=\frac{-24.55\,e^{0.01t}}{t}\), where \(C'(t)\) is measured in °C per minute.
(c) Find \(C(20)\). Show the setup used to obtain your answer.
It is also known that \(\displaystyle C”(t)=\frac{0.2455\,e^{0.01t}(100-t)}{t^{2}}\) for \(12<t<20\).
(d) For \(12<t<20\), determine whether the coffee’s temperature is changing at an increasing rate or a decreasing rate. Justify your answer.
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▶️ Answer/Explanation
Use the secant over \([3,7]\): \[ C'(5)\approx \frac{C(7)-C(3)}{7-3}=\frac{69-85}{4}=-4\ \text{°C/min}. \]
\[ \int_{0}^{12} C(t)\,dt \approx (3-0)\,C(0) + (7-3)\,C(3) + (12-7)\,C(7) =3(100)+4(85)+5(69)=\boxed{985\ \text{°C·min}}. \] Average temperature over the first 12 minutes: \[ \frac{1}{12}\int_{0}^{12} C(t)\,dt \approx \boxed{82.083\ \text{°C}}. \] (This is the coffee’s average temperature from \(t=0\) to \(t=12\) minutes.)
Accumulation with initial value \(C(12)=55\): \[ C(20)=C(12)+\int_{12}^{20} C'(t)\,dt =55+\int_{12}^{20}\frac{-24.55\,e^{0.01t}}{t}\,dt \approx \boxed{40.329\ \text{°C}}. \]
For \(12<t<20\), \(e^{0.01t}>0\), \(t^2>0\), and \(100-t>0\). Thus \(C”(t)>0\). Therefore \(C'(t)\) is increasing on \((12,20)\), so the temperature is changing at an increasing rate (the cooling rate becomes less negative).