AP Calculus BC 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points - FRQs - Exam Style Questions
Calc-OkQuestion
Fish enter a lake at a rate modeled by \(E(t)=20+15\sin\!\big(\tfrac{\pi t}{6}\big)\). Fish leave at a rate modeled by \(L(t)=4+2^{\,0.1t^{2}}\). Both \(E(t)\) and \(L(t)\) are in fish/hour, and \(t\) is hours since midnight \((t=0)\).
(a) How many fish enter the lake during the 5 hours from midnight \((t=0)\) to \(5\) A.M. \((t=5)\)? Give the nearest whole number.
(b) What is the average number of fish per hour that leave the lake from \(t=0\) to \(t=5\)?
(c) For \(0\le t\le 8\), at what time is the greatest number of fish in the lake? Justify your answer.
(d) At \(t=5\) A.M., is the rate of change of the number of fish in the lake increasing or decreasing? Explain.
Most-appropriate topic codes (CED):
• TOPIC 8.3: Using Accumulation Functions & Definite Integrals in Applied Contexts — part (a)
• TOPIC 8.1: Average Value of a Function — part (b)
• TOPIC 5.2 / 5.5: EVT, Critical Points & Candidates Test — part (c)
• TOPIC 4.1: Interpreting the Meaning of the Derivative in Context — part (d)
• TOPIC 8.1: Average Value of a Function — part (b)
• TOPIC 5.2 / 5.5: EVT, Critical Points & Candidates Test — part (c)
• TOPIC 4.1: Interpreting the Meaning of the Derivative in Context — part (d)
▶️ Answer/Explanation
(a) Number entering on \([0,5]\)
\[ \int_{0}^{5}\!\!E(t)\,dt=\int_{0}^{5}\!\!\Big(20+15\sin\!\tfrac{\pi t}{6}\Big)dt =\underbrace{\int_{0}^{5}\!20\,dt}_{=100}+\underbrace{15\int_{0}^{5}\!\sin\!\tfrac{\pi t}{6}\,dt}_{(*)}. \]
For \((*)\): \(\displaystyle \int \sin(kt)\,dt=-\frac{1}{k}\cos(kt)\), with \(k=\tfrac{\pi}{6}\).
\[ 15\left[-\frac{1}{k}\cos(kt)\right]_{0}^{5} =15\left[-\frac{6}{\pi}\cos\!\left(\tfrac{\pi t}{6}\right)\right]_{0}^{5} =-\frac{90}{\pi}\Big(\cos\tfrac{5\pi}{6}-\cos 0\Big). \]
\(\cos\tfrac{5\pi}{6}=-\tfrac{\sqrt3}{2}\), \(\cos 0=1\) \(\Rightarrow\) term \(=\dfrac{90}{\pi}\!\left(1+\tfrac{\sqrt3}{2}\right)\).
Total: \[ 100+\frac{90}{\pi}\!\left(1+\frac{\sqrt3}{2}\right)\approx 153.457690\ \text{fish}. \]
Nearest whole number: 153 fish. (Applied accumulation context: Topic 8.3.)
(b) Average number leaving on \([0,5]\)
Average value on \([a,b]\): \(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)\,dx\). (Topic 8.1.)
\[ \text{Average}=\frac{1}{5}\int_{0}^{5}\!\!L(t)\,dt =\frac{1}{5}\!\left(\int_{0}^{5}\!4\,dt+\int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\right) =\frac{1}{5}\!\left(20+\int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\right). \]
The integral \(\displaystyle \int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\) has no elementary antiderivative; using numerical integration (e.g., refined Riemann/Simpson) gives \(\displaystyle \int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\approx 10.29519\).
Hence \[ \text{Average}\approx \frac{20+10.29519}{5}=6.059038\ \text{fish/hour}. \]
(c) Time when the number of fish is greatest on \([0,8]\)
Let \(N(t)\) be the number of fish in the lake since midnight. Then \(N'(t)=E(t)-L(t)\).
A maximum of \(N(t)\) occurs where \(N'(t)=0\) and changes from \(+\) to \(-\) (EVT/critical points + candidates test). (Topics 5.2 & 5.5.)
Solve \(E(t)-L(t)=0\): \(20+15\sin\!\big(\tfrac{\pi t}{6}\big)=4+2^{\,0.1t^{2}}\).
Using Newton’s method starting near \(t_0=6.2\):
• \(f(t)=E(t)-L(t)\), \(f'(t)=E'(t)-L'(t)\), with \(E'(t)=\tfrac{15\pi}{6}\cos\!\big(\tfrac{\pi t}{6}\big)\), \(L'(t)=2^{\,0.1t^{2}}\ln 2\cdot(0.2t)\).
• \(t_{1}=t_{0}-\dfrac{f(t_{0})}{f'(t_{0})}\approx 6.2037\).
Refined value: \(t\approx \mathbf{6.20356}\) hours. Sign check: for \(t<6.20356\), \(E-L>0\); for \(t>6.20356\), \(E-L<0\).
Therefore the greatest number of fish occurs at \(t\approx 6.204\) h.
(d) Is the rate \(N'(t)\) increasing or decreasing at \(t=5\)?
\(N”(t)=E'(t)-L'(t)\). Compute at \(t=5\):
\(E'(t)=\dfrac{15\pi}{6}\cos\!\big(\tfrac{\pi t}{6}\big)=\dfrac{5\pi}{2}\cos\!\big(\tfrac{\pi t}{6}\big)\).
\(\cos\!\big(\tfrac{5\pi}{6}\big)=-\tfrac{\sqrt3}{2}\Rightarrow E'(5)=-\dfrac{5\pi\sqrt3}{4}\approx -6.8017.\)
\(L'(t)=2^{\,0.1t^{2}}\ln 2\cdot(0.2t)\Rightarrow L'(5)=2^{2.5}\ln 2\cdot 1= (4\sqrt2)(\ln 2)\approx 5.656854\times 0.693147\approx 3.9209.\)
\(N”(5)=E'(5)-L'(5)\approx -6.8017-3.9209=-10.7226<0\).
Therefore the rate \(N'(t)\) is decreasing at \(t=5\) A.M.
\[ \int_{0}^{5}\!\!E(t)\,dt=\int_{0}^{5}\!\!\Big(20+15\sin\!\tfrac{\pi t}{6}\Big)dt =\underbrace{\int_{0}^{5}\!20\,dt}_{=100}+\underbrace{15\int_{0}^{5}\!\sin\!\tfrac{\pi t}{6}\,dt}_{(*)}. \]
For \((*)\): \(\displaystyle \int \sin(kt)\,dt=-\frac{1}{k}\cos(kt)\), with \(k=\tfrac{\pi}{6}\).
\[ 15\left[-\frac{1}{k}\cos(kt)\right]_{0}^{5} =15\left[-\frac{6}{\pi}\cos\!\left(\tfrac{\pi t}{6}\right)\right]_{0}^{5} =-\frac{90}{\pi}\Big(\cos\tfrac{5\pi}{6}-\cos 0\Big). \]
\(\cos\tfrac{5\pi}{6}=-\tfrac{\sqrt3}{2}\), \(\cos 0=1\) \(\Rightarrow\) term \(=\dfrac{90}{\pi}\!\left(1+\tfrac{\sqrt3}{2}\right)\).
Total: \[ 100+\frac{90}{\pi}\!\left(1+\frac{\sqrt3}{2}\right)\approx 153.457690\ \text{fish}. \]
Nearest whole number: 153 fish. (Applied accumulation context: Topic 8.3.)
(b) Average number leaving on \([0,5]\)
Average value on \([a,b]\): \(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)\,dx\). (Topic 8.1.)
\[ \text{Average}=\frac{1}{5}\int_{0}^{5}\!\!L(t)\,dt =\frac{1}{5}\!\left(\int_{0}^{5}\!4\,dt+\int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\right) =\frac{1}{5}\!\left(20+\int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\right). \]
The integral \(\displaystyle \int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\) has no elementary antiderivative; using numerical integration (e.g., refined Riemann/Simpson) gives \(\displaystyle \int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\approx 10.29519\).
Hence \[ \text{Average}\approx \frac{20+10.29519}{5}=6.059038\ \text{fish/hour}. \]
(c) Time when the number of fish is greatest on \([0,8]\)
Let \(N(t)\) be the number of fish in the lake since midnight. Then \(N'(t)=E(t)-L(t)\).
A maximum of \(N(t)\) occurs where \(N'(t)=0\) and changes from \(+\) to \(-\) (EVT/critical points + candidates test). (Topics 5.2 & 5.5.)
Solve \(E(t)-L(t)=0\): \(20+15\sin\!\big(\tfrac{\pi t}{6}\big)=4+2^{\,0.1t^{2}}\).
Using Newton’s method starting near \(t_0=6.2\):
• \(f(t)=E(t)-L(t)\), \(f'(t)=E'(t)-L'(t)\), with \(E'(t)=\tfrac{15\pi}{6}\cos\!\big(\tfrac{\pi t}{6}\big)\), \(L'(t)=2^{\,0.1t^{2}}\ln 2\cdot(0.2t)\).
• \(t_{1}=t_{0}-\dfrac{f(t_{0})}{f'(t_{0})}\approx 6.2037\).
Refined value: \(t\approx \mathbf{6.20356}\) hours. Sign check: for \(t<6.20356\), \(E-L>0\); for \(t>6.20356\), \(E-L<0\).
Therefore the greatest number of fish occurs at \(t\approx 6.204\) h.
(d) Is the rate \(N'(t)\) increasing or decreasing at \(t=5\)?
\(N”(t)=E'(t)-L'(t)\). Compute at \(t=5\):
\(E'(t)=\dfrac{15\pi}{6}\cos\!\big(\tfrac{\pi t}{6}\big)=\dfrac{5\pi}{2}\cos\!\big(\tfrac{\pi t}{6}\big)\).
\(\cos\!\big(\tfrac{5\pi}{6}\big)=-\tfrac{\sqrt3}{2}\Rightarrow E'(5)=-\dfrac{5\pi\sqrt3}{4}\approx -6.8017.\)
\(L'(t)=2^{\,0.1t^{2}}\ln 2\cdot(0.2t)\Rightarrow L'(5)=2^{2.5}\ln 2\cdot 1= (4\sqrt2)(\ln 2)\approx 5.656854\times 0.693147\approx 3.9209.\)
\(N”(5)=E'(5)-L'(5)\approx -6.8017-3.9209=-10.7226<0\).
Therefore the rate \(N'(t)\) is decreasing at \(t=5\) A.M.