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AP Calculus BC 5.1 Using the Mean Value Theorem - FRQs - Exam Style Questions

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Calc-Ok Question

A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function \(f\), where \(f(t)\) is measured in gallons per second and \(t\) is measured in seconds since pumping began. Selected values of \(f(t)\) are given in the table.
\(t\) (seconds)06090120135150
\(f(t)\) (gal/s)00.10.150.10.050

(a) Using correct units, interpret the meaning of \(\displaystyle \int_{60}^{135} f(t)\,dt\) in the context of the problem. Use a right Riemann sum with the three subintervals \([60,90],\,[90,120],\,[120,135]\) to approximate \(\displaystyle \int_{60}^{135} f(t)\,dt\).

(b) Must there exist a value of \(c\), for \(60<c<120\), such that \(f'(c)=0\)? Justify your answer.

(c) The rate of flow of gasoline (gal/s) can also be modeled by \[ g(t)=\frac{t}{500}\cos\!\left(\left(\frac{t}{120}\right)^{\!2}\right), \quad 0\le t\le 150. \] Using this model, find the average rate of flow of gasoline over \(0\le t\le 150\). Show the setup for your calculations.

(d) Using the model \(g\) from part (c), find the value of \(g'(140)\). Interpret the meaning of your answer in the context of the problem.

Most-appropriate topic codes:

TOPIC 6.2: Riemann Sums (Right endpoint) — part (a)
TOPIC 6.4: FTC & Accumulation (interpretation of \(\int f\)) — part (a)
TOPIC 5.1: Mean Value Theorem — part (b)
TOPIC 8.1: Average Value of a Function — part (c)
TOPIC 4.1: Interpreting the Derivative in Context — part (d)
▶️ Answer/Explanation
(a) Interpretation and right Riemann sum
Interpretation with units: \[ \int_{60}^{135} f(t)\,dt \] represents the total number of gallons of gasoline pumped into the tank from \(t=60\) s to \(t=135\) s. (gal/s integrated over seconds \(\Rightarrow\) gallons).
Right Riemann sum (widths \(30, 30, 15\); right endpoints \(90,120,135\)): \[ \int_{60}^{135} f(t)\,dt \approx f(90)(90-60)+f(120)(120-90)+f(135)(135-120). \] \[ = (0.15)(30)+(0.10)(30)+(0.05)(15) = 4.5+3.0+0.75 = \boxed{8.25\ \text{gallons}}. \]
(b) Existence of \(c\) with \(f'(c)=0\) on \((60,120)\)
\(f\) is differentiable \(\Rightarrow\) continuous on \([60,120]\).
Compute the average rate of change: \[ \frac{f(120)-f(60)}{120-60}=\frac{0.10-0.10}{60}=0. \] By the Mean Value Theorem, there exists \(c\in(60,120)\) such that \[ f'(c)=0. \] Yes, such a \(c\) must exist.
(c) Average rate of flow using \(g(t)\)
Average value over \([0,150]\): \[ \frac{1}{150}\int_{0}^{150} g(t)\,dt = \frac{1}{150}\int_{0}^{150}\frac{t}{500}\cos\!\left(\Big(\tfrac{t}{120}\Big)^{2}\right) dt. \] Substitute \(u=\left(\tfrac{t}{120}\right)^{2}\) so \(du=\tfrac{t}{7200}\,dt\) and \(t\,dt=7200\,du\).
Then \[ \int_{0}^{150}\frac{t}{500}\cos(u)\,dt =\frac{7200}{500}\int_{u=0}^{u=(150/120)^{2}}\cos u\,du =14.4\,\sin u\Big|_{0}^{(150/120)^{2}}. \] \[ =14.4\,\sin\!\left(\frac{25}{16}\right). \] Average value: \[ \boxed{\frac{1}{150}\int_{0}^{150} g(t)\,dt = \frac{14.4}{150}\sin\!\left(\frac{25}{16}\right) = \frac{12}{125}\sin\!\left(\frac{25}{16}\right) \approx 0.096\ \text{gal/s}.} \]
(d) \(g'(140)\) and interpretation
\(g(t)=\dfrac{t}{500}\cos\!\left(\left(\tfrac{t}{120}\right)^{2}\right)\).
Let \(u(t)=\left(\tfrac{t}{120}\right)^{2}\) so \(u'(t)=\tfrac{t}{7200}\).
Product/chain rule: \[ g'(t)=\frac{1}{500}\cos u(t)\;-\;\frac{t}{500}\sin u(t)\cdot \frac{t}{7200} =\frac{1}{500}\cos u(t)\;-\;\frac{t^{2}}{3{,}600{,}000}\sin u(t). \] At \(t=140\): \(u(140)=\left(\tfrac{140}{120}\right)^{2}=\tfrac{49}{36}\). \[ g'(140)=\frac{1}{500}\cos\!\left(\frac{49}{36}\right) -\frac{49}{9000}\sin\!\left(\frac{49}{36}\right) \approx \boxed{-0.0049\ \text{(gal/s)}^{2}}. \] Interpretation: At \(t=140\) s, the rate at which gasoline is flowing into the tank is decreasing at about \(0.005\) gallon per second per second.
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