Calculus Free Response

CALCULUS AB FREE-RESPONSE QUESTIONS CALCULUS AB SECTION II, Part A Time—30 minutes Number of problems—2 A graphing calculator is required for these problems.   t (hours) 0 1 3 6 8 R(t) (liters/hour) 1340 1190 950 740 700 Експортувати в Таблиці 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 above. At time (t = 0), there are 50,000 liters of water in the tank.   (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.   a) (R'(2) = \frac{950 - 1190}{3 - 1} = -120) L/hr² ✓ b) (1(1340) + 2(1190) + 3(950) + 2(740) = 8050) Liters. ✓ [Small graph showing R(t) decreasing] Since R(t) is decreasing, LRAM gives an overestimate of the total amt of H₂O removed. ✓ c) I.C. + IN - OUT (50000 + \int_{0}^{8} 2000e^{-t^{2}/20} dt - 8050 \approx 49786) L 7836.195 ✓ d) (W(0) = 2000) (W(8) = 81.524) Since W(t) and R(t) are continuous and differentiable, by IVT at some time btw t=0 & t=8, W(t) takes on all values btw 2000 & 81.524, so (W(t) = R(t)) since R(t) ≥ 700. ✓ GO ON TO THE NEXT PAGE. -2- Key Observations: Derivative Estimation: The student correctly estimates the derivative using a difference quotient. Left Riemann Sum: The student correctly calculates the left Riemann sum. Overestimate Explanation: The student correctly explains why the left Riemann sum is an overestimate due to the decreasing nature of R(t). Total Water Calculation: The student correctly sets up and calculates the total water in the tank using the initial condition, the integral of W(t), and the Riemann sum. Intermediate Value Theorem: The student correctly applies the Intermediate Value Theorem to explain why the pumping and removal rates must be equal at some time. Checkmarks: The checkmarks indicate that the student's answers were likely marked as correct. CALCULUS AB FREE-RESPONSE QUESTIONS For (t \ge 0), a particle moves along the x-axis. The velocity of the particle at time (t) is given by [v(t) = 1 + 2\sin\left(\frac{t^2}{2}\right).] The particle is at position1 2 at time (t = 4). I.C.   1. secure-media.collegeboard.org secure-media.collegeboard.org (a) At time (t = 4), is the particle speeding up or slowing down? (v(t) = 0) ✓ (b) Find all times (t) in the interval (0 < t < 3) when the particle changes direction. Justify your answer. ✓ (c) Find the position of the particle at time (t = 0). ✓ (d) Find the total distance the particle travels from time (t = 0) to time (t = 3). Split ∫ ✓ a) (v(4) = 1 + 2\sin(8) \approx 2.979) (a(4) = -1.164) Since (v(4)) and (a(4)) have opposite signs, the particle slows down at (t = 4). b) (0 = 1 + 2\sin\left(\frac{t^2}{2}\right)) [Small graph showing sign change] (t \approx 2.707), the (v(t)) changes from (> 0) to (< 0), so the particle changes from moving right to left. END OF PART A OF SECTION II c) (\int_0^4 v(t) dt \approx 5.815) I.P. + 5.815 = 2 I.P. = -3.815 d) (\int_0^3 |v(t)| dt \approx 5.301) or (\int_0^{2.707} v(t) dt + \left| \int_{2.707}^3 v(t) dt \right|) 5.13695 .164247