A tabular flow-rate function (gasoline being pumped into a tank) is given. Students estimate an accumulated quantity using a right Riemann sum, interpret a definite integral with correct units, apply an existence theorem to the derivative, and then use a supplied closed-form rate model to find an average rate over an interval and an instantaneous rate of change at a point.
A tabular flow-rate function (gasoline being pumped into a tank) is given. Students estimate an accumulated quantity using a right Riemann sum, interpret a definite integral with correct units, apply an existence theorem to the derivative, and then use a supplied closed-form rate model to find an average rate over an interval and an instantaneous rate of change at a point.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
**Interpretation.** Since $f(t)$ is the rate of flow in gallons per second, $\int_{60}^{135} f(t)\,dt$ accumulates that rate over time, so it represents the **total number of gallons of gasoline pumped into the tank from $t=60$ seconds to $t=135$ seconds.** **Right Riemann sum** with subintervals $[60,90]$, $[90,120]$, $[120,135]$. A right sum uses the value of $f$ at the right endpoint of each subinterval, times the subinterval width: $$\int_{60}^{135} f(t)\,dt \approx f(90)(90-60)+f(120)(120-90)+f(135)(135-120).$$ Reading the table $f(90)=0.15,\ f(120)=0.1,\ f(135)=0.05$: $$= (0.15)(30)+(0.1)(30)+(0.05)(15) = 4.5 + 3.0 + 0.75 = \boxed{8.25 \text{ gallons}}.$$
3 points. P1 (interpretation with units): must state 'gallons of gasoline pumped/added' AND the time interval t=60 to t=135. P2 (form of Riemann sum): show the sum f(90)(30)+f(120)(30)+f(135)(15) with the six factors; at least five of the six factors must be correct. P3 (answer): 8.25. The compact form (0.15)(30)+(0.1)(30)+(0.05)(15)=8.25 earns both P2 and P3. A bare answer of 8.25 with no shown products earns neither P2 nor P3. Any arithmetic error in the sum forfeits P3.
We are asked whether some $c$ in $(60,120)$ must satisfy $f'(c)=0$. Because $f$ is differentiable, $f$ is **continuous on $[60,120]$ and differentiable on $(60,120)$**, so the Mean Value Theorem applies. Compute the average rate of change of $f$ over $[60,120]$: $$\frac{f(120)-f(60)}{120-60}=\frac{0.1-0.1}{60}=\frac{0}{60}=0.$$ By the Mean Value Theorem there exists a $c$ with $60<c<120$ such that $f'(c)$ equals this average rate of change, i.e. $f'(c)=0$. **Yes**, such a value of $c$ must exist.
2 points. P1: present f(120)-f(60)=0, or (0.1-0.1)/60=0, or f(60)=f(120). P2 (answer with justification): requires P1, plus stating f is continuous because it is differentiable (or equivalent), plus answering 'yes'. MVT or Rolle's Theorem is acceptable; an argument citing the Intermediate Value Theorem cannot earn P2.
The **average rate of flow** of $g$ over $[0,150]$ is the average value of $g$: $$\frac{1}{150-0}\int_{0}^{150} g(t)\,dt,\qquad g(t)=\frac{t}{500}\cos\!\left(\left(\frac{t}{120}\right)^{2}\right).$$ Evaluating with a graphing calculator (radian mode): $$\int_{0}^{150} g(t)\,dt = 14.399504,\qquad \frac{1}{150}(14.399504)=0.0959967.$$ The exact value is $\frac{12}{125}\sin\!\left(\frac{25}{16}\right)$. So the average rate of flow of gasoline over $0\le t\le 150$ is approximately $\boxed{0.096}$ gallon per second (a value of $0.095$ is also accepted).
2 points. P1 (average value formula): present (1/150)∫_0^150 g(t)dt (single or multiple steps). P2 (answer): 0.096 or 0.095. A bare ∫_0^150 g dt = 0.0959967 (mislabeling the integral as the average) earns neither point. Calculator in degree mode does not earn P1. Answers should be correct to three decimals.
Differentiate the model with a calculator and evaluate at $t=140$: $$g'(140)\approx -0.004908.$$ (The exact value is $\frac{1}{500}\cos\!\left(\frac{49}{36}\right)-\frac{49}{9000}\sin\!\left(\frac{49}{36}\right)$.) **Interpretation.** $g'$ is the rate of change of the flow rate. Since $g'(140)\approx-0.005<0$, **the rate at which gasoline is flowing into the tank is decreasing at a rate of about $0.005$ gallon per second per second at time $t=140$ seconds.**
2 points. P1: g'(140) ≈ -0.004908 (report as -0.005 or -0.004); the value may appear only inside the interpretation. P2 (interpretation): must present a numerical value for g'(140) AND state that the rate of flow of gasoline is changing at that rate AND 'at t=140'. 'Decreasing at a rate of -0.005' or 'increasing at a rate of 0.005' does NOT earn P2 (sign/direction mismatch). Degree mode gives g'(140)=0.001997 or 0.00187 and does not earn the value point.
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