AP Calculus BC 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts - FRQs - Exam Style Questions
Calc-Ok Question
(B) Find the time \(t\) when the instantaneous rate of change of \(C\) equals the average rate of change of \(C\) on \(0\le t\le 4\). Show the setup.
(C) Assuming the model continues for all \(t>0\), write a limit expression that describes the end behavior of the rate of change in acres, and evaluate it.
(D) At \(t=4\) weeks, measures are taken to counter the spread. Let \(A(t)=C(t)-\displaystyle\int_{4}^{t}0.1\ln(x)\,dx\) model the acres affected for \(4\le t\le 36\). For what \(t\) does \(A\) attain its maximum? Justify.
▶️ Answer/Explanation
(A) Average value
\[ \text{Average}=\frac{1}{4}\int_{0}^{4} C(t)\,dt =\frac{1}{4}\int_{0}^{4} 7.6\arctan(0.2t)\,dt \approx \boxed{2.778\ \text{acres}}. \] (Calculator.)
(B) Instantaneous rate = average rate
Average rate on \([0,4]\): \(\dfrac{C(4)-C(0)}{4}=\dfrac{C(4)}{4}\) with \(C(4)=7.6\arctan(0.8)\). Solve \[ C'(t)=\frac{C(4)}{4}\quad\Rightarrow\quad \frac{38}{25+t^2}=\frac{7.6\arctan(0.8)}{4}. \] Numerical solution on \([0,4]\): \(\boxed{t\approx 2.154\ \text{weeks}}\).
(C) End behavior
\[ \lim_{t\to\infty} C'(t)=\lim_{t\to\infty}\frac{38}{25+t^2}=\boxed{0}. \] The growth rate tends to zero.
(D) Maximizing \(A\) on \([4,36]\)
\(A'(t)=C'(t)-0.1\ln t\). Critical points solve \(\dfrac{38}{25+t^2}=0.1\ln t\). Numerical root: \(\boxed{t\approx 11.442}\). Since \(A'(t)\) changes from \(+\) to \(−\) there (and endpoints give smaller values), \(A\) attains its maximum at \(t\approx 11.442\) weeks.