An invasive-species area is modeled by an arctangent function (with its derivative supplied). Students find the average value over an interval, solve for a time where the instantaneous rate equals the average rate of change, write and evaluate a limit describing the end behavior of the rate, and maximize a modified accumulation-based model over an interval.
An invasive-species area is modeled by an arctangent function (with its derivative supplied). Students find the average value over an interval, solve for a time where the instantaneous rate equals the average rate of change, write and evaluate a limit describing the end behavior of the rate, and maximize a modified accumulation-based model over an interval.
Read the complete question in the official College Board FRQ PDF ›
Original Tian2 solution. Each part identifies which rubric points are earned and why.
The average value of $C$ on $[0,4]$ (calculator active, radian mode) is $$\frac{1}{4-0}\int_0^4 C(t)\,dt = \frac{1}{4}\int_0^4 7.6\arctan(0.2t)\,dt.$$ Evaluating the definite integral numerically gives $\int_0^4 C(t)\,dt = 11.112896$, so $$\frac{1}{4}(11.112896) = 2.778224 \approx 2.778.$$ From $t=0$ to $t=4$ weeks, the average number of acres affected is about $2.778$ acres.
2 points. P1: the average-value setup — the integral of C on [0,4] together with evidence of division by 4 (the correct final answer in the presence of the correct integral counts as that evidence). P2: the numeric answer 2.778 (or 2.778224 truncated/rounded to three decimals). A bare 2.778 with no integral shown forfeits P1; an inappropriately rounded value forfeits P2.
The average rate of change of $C$ on $[0,4]$ is $$\frac{C(4)-C(0)}{4-0} = \frac{5.128031-0}{4} = 1.282008.$$ Set the instantaneous rate equal to this and solve on $0\le t\le 4$: $$C'(t)=\frac{38}{25+t^2}=1.282008 \;\Rightarrow\; t = 2.154298 \approx 2.154.$$ The instantaneous rate of change of $C$ equals its average rate of change over $[0,4]$ at $t\approx 2.154$ weeks.
2 points. P3: presents the average rate of change of C on [0,4] — any of (C(4)-C(0))/4, C(4)/4, 5.128/4, or the value 1.282 earns it (because C(0)=0). Note: simply writing t=1.282 earns neither P3 nor P4. P4: the correct solution t=2.154 supported by the equation C'(t) = average rate of change.
The rate of change of the number of affected acres is $C'(t)=\dfrac{38}{25+t^2}$. Its end behavior is described by $$\lim_{t\to\infty} C'(t) = \lim_{t\to\infty}\frac{38}{25+t^2}.$$ As $t\to\infty$ the denominator grows without bound while the numerator stays fixed at $38$, so $$\lim_{t\to\infty}\frac{38}{25+t^2}=0.$$ The rate at which acres become affected approaches $0$ in the long run.
2 points. P5: a correct limit expression — either lim_{t->inf} C'(t) or lim_{t->inf} C(t) earns it. P6: the value 0. Caution: writing lim C(t) (which would be 7.6*(pi/2) for C itself) makes the response ineligible for P6, and plugging infinity into the formula as arithmetic (38/(25+inf^2)=0) is treated as scratch work, not credit.
Here $A(t)=C(t)-\displaystyle\int_4^t 0.1\ln x\,dx$, so by the Fundamental Theorem of Calculus $$A'(t)=C'(t)-0.1\ln t = \frac{38}{25+t^2}-0.1\ln t.$$ The maximum on $[4,36]$ occurs at a critical point or an endpoint. Solving $A'(t)=0$ on $[4,36]$ gives $$\frac{38}{25+t^2}=0.1\ln t \;\Rightarrow\; t = 11.441700 \approx 11.442,$$ the only critical number in the interval. Comparing values (candidates test): $$A(4)=5.128031,\quad A(11.442)=7.316978,\quad A(36)=1.743056.$$ The largest value occurs at the interior critical point, so $A$ attains its maximum at $t\approx 11.442$ weeks.
3 points. P7: considers A'(t)=0 (equivalently C'(t)-0.1 ln t = 0 or C'(t)=0.1 ln t); discussing a sign change of A' or 'critical points of A' also earns it — but merely writing t=11.4417 does not. P8: a valid global justification; via the candidates test, correctly evaluate A at t=4, t=11.442, and t=36 (correct to the first decimal), or argue A'>0 before and A'<0 after t=11.442. P9: the answer t=11.442 (or 11.441). A purely local First/Second Derivative Test argument forfeits P8 but can still earn P9 with the correct answer.
Our worked solutions and practice questions are original instructional content created by Tian2 AP. They are aligned to the concepts and skills described in College Board’s Course and Exam Description and are not reproductions of, or affiliated with, College Board’s official materials.
