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AP Calculus BC 2.3 Estimating Derivatives of a Function at a Point - FRQs - Exam Style Questions

AP Calculus BC - FRQs and Answer - All Topics

Calc-Ok Question

Water is pumped into a tank at a rate modeled by \(W(t)=2000e^{-t^{2}/20}\) liters per hour for \(0\le t\le 8\), where \(t\) is measured in hours. Water is removed from the tank at a rate modeled by \(R(t)\) liters per hour, where \(R\) is differentiable and decreasing on \(0\le t\le 8\). Selected values of \(R(t)\) are shown in the table. At time \(t=0\), there are 50,000 liters of water in the tank.
\(t\) (hours)01368
\(R(t)\) (liters/hour)13401190950740700
(a) Estimate \(R'(2)\). Show the work that leads to your answer. Indicate units of measure.
(b) Use a left Riemann sum with the four subintervals indicated by the table to estimate the total amount of water removed from the tank during the 8 hours. Is this an overestimate or an underestimate of the total amount of water removed? Give a reason for your answer.
(c) Use your answer from part (b) to find an estimate of the total amount of water in the tank, to the nearest liter, at the end of 8 hours.
(d) For \(0\le t\le 8\), is there a time \(t\) when the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank? Explain why or why not.

Most-appropriate topic codes (CED):

• TOPIC 2.3: Estimating Derivatives of a Function at a Point (from tables/graphs) — part (a) 
• TOPIC 6.2: Approximating Areas with Riemann Sums — part (b)
• TOPIC 8.3: Using Accumulation Functions & Definite Integrals in Applied Contexts — part (c) 
• TOPIC 1.16: Working with the Intermediate Value Theorem (IVT) — part (d)

▶️ Answer/Explanation

(a) Estimate \(R'(2)\)
Use a symmetric difference quotient with nearby data at \(t=1\) and \(t=3\):
\[ R'(2)\approx\frac{R(3)-R(1)}{3-1}=\frac{950-1190}{2}=-120\ \text{liters/hr}^2. \] Units: liters per hour per hour.

(b) Left Riemann sum on \([0,1],[1,3],[3,6],[6,8]\)
\[ \int_0^{8}R(t)\,dt \approx 1\cdot R(0)+2\cdot R(1)+3\cdot R(3)+2\cdot R(6) =1(1340)+2(1190)+3(950)+2(740). \] \[ =1340+2380+2850+1480 = \boxed{8050\ \text{liters removed}}. \] Because \(R\) is decreasing, a left sum **overestimates** the integral.

(c) Water in the tank at \(t=8\)
Net amount \(=\) initial \(+\displaystyle\int_{0}^{8}W(t)\,dt – \text{(removed)}\). Using a calculator, \[ \int_{0}^{8}2000e^{-t^2/20}\,dt \approx 7836.195325\ \text{L}. \] Therefore, \[ \text{Total} \approx 50{,}000 + 7836.195325 – 8050 \approx \boxed{49{,}786\ \text{liters (nearest liter)}}. \]

(d) Existence of \(t\) with \(W(t)=R(t)\)
Let \(F(t)=W(t)-R(t)\). Both \(W\) and \(R\) are continuous on \([0,8]\), so \(F\) is continuous. \[ F(0)=2000-1340>0,\qquad F(8)=2000e^{-64/20}-700\approx 81.5-700<0. \] Since \(F(0)>0\) and \(F(8)<0\), the **Intermediate Value Theorem** ensures some \(c\in(0,8)\) with \(F(c)=0\), i.e., \(\boxed{W(c)=R(c)}\).

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