Spring Paqcket Answers

A GRAPHING CALCULATOR IS REQUIRED FOR THESE QUESTIONS. Fish enter a lake at a rate modeled by the function E(t) = 20 + 15 sin(πt/6). Fish leave the lake at a rate modeled by the function L(t) = 4 + 2^(0.02t). Both E(t) and L(t) are measured in fish per hour, and t is measured in hours since midnight (t = 0). (a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 AM (t = 5)? Give your answer to the nearest whole number.   ∫₀⁵ E(t) dt = 153.458 (b) What is the average number of fish that leave the lake per hour over the 5-hour period from midnight (t = 0) to 5 AM (t = 5)? (1/5) ∫₀⁵ L(t) dt = 30.295 / 5 = 6.059 (c) At what time, 0 ≤ t ≤ 5, is the greatest number of fish in the lake? Justify your answer. Amt = ∫₀ᵗ (E(t) - L(t)) dt = 6.204 t_max = 6.204 (d) Is the rate of change in the number of fish in the lake increasing or decreasing at 5 AM (t = 5)? Explain your reasoning. Rate of change is E - L. (E - L)' at t = 5 is -10.723 So rate is decreasing 2. The velocity of a particle, P, moving along the x-axis is given by the differentiable function v_p(t), where v_p(t) is measured in meters per hour and t is measured in hours. Selected values of v_p(t) are shown in the table above. Particle P is at the origin at time t = 0.   (a) Justify why there must be at least one time t, for 0.3 ≤ t ≤ 2.8, at which v_p'(t), the acceleration of particle P, equals 0 meters per hour per hour.   Solution: From the table, we can see that v_p(t) increases from 0 to 55, then decreases from 55 to -29, and then increases again from -29 to 55. This means that there must be at least one point in between these changes where the velocity is neither increasing nor decreasing, which implies that the acceleration (v_p'(t)) is 0. (b) Use a trapezoidal sum with the three subintervals (0, 0.3), (0.3, 1.7), and (1.7, 2.8) to approximate the value of ∫₀⁴ v_p(t) dt. (1/2)((55+0)(0.3) + (29+55)(1.4) + 2(55-29)(1.1)) = 40.75 (c) A second particle, Q, also moves along the x-axis so that its velocity for 0 ≤ t ≤ 4 is given by v_Q(t) = 45√(t)cos(0.063t²). Find the time interval during which the velocity of particle Q is at least 60 meters per hour. Find the distance traveled by particle Q during the interval when the velocity of particle Q is at least 60 meters per hour.   v_Q(t) ≥ 60 3.519 ≤ t ≤ 1.866 ∫₁.⁸⁶⁶³⁵¹⁹ v_Q(t) dt = 106.109 (d) At time t = 0, particle Q is at position x = -90. Using the result from part (b) and the function v_Q(t) from part (c), approximate the distance between particles P and Q at time t = 2.8. -90 + ∫₀²⁸ v_Q(t) dt = 45.938 45.938 - 40.75 = 5.188 NO CALCULATOR IS ALLOWED FOR THESE QUESTIONS. Slope = 2 [Graph of a function f consisting of two line segments and a quarter-circle centered at (5,3)] The continuous function f is defined on the closed interval -6 ≤ x ≤ 5. The figure above shows a portion of the graph of f, consisting of two line segments and a quarter of a circle centered at the point (5,3). It is known that the point (3, 3 - √5) is on the graph of f. (a) If ∫_{-6}^5 f(x) dx = 7, find the value of ∫_{-6}^{-2} f(x) dx. Show the work that leads to your answer. ∫_{-6}^5 f(x) dx = ∫_{-6}^{-2} f(x) dx + ∫_{-2}^5 f(x) dx = 7 ∫_{-6}^{-2} f(x) dx = 7 - ∫_{-2}^5 f(x) dx = 7 - (11 - (9π/4)) = 7 - 11 + (9π/4) = -4 + (9π/4) (b) Evaluate ∫_{-2}^5 (2f'(x) + 4) dx. ∫_{-2}^5 (2f'(x) + 4) dx = 2∫_{-2}^5 f'(x) dx + ∫_{-2}^5 4 dx = 2(f(5) - f(-2)) + 4(5 - (-2)) = 2(0 - (3 - √5)) + 4(7) = 2(-3 + √5) + 28 = 2√5 - 6 + 28 = 2√5 + 22 (c) The function g is given by g(x) = ∫_{-2}^x f'(t) dt. Find the absolute maximum value of g on the interval -2 ≤ x ≤ 5. Justify your answer. g'(x) = f'(x) g'(x) = 0 when x = -1, 3, 5 g(-2) = 0 g(3) = 3 - √5 g(5) = 0 The absolute maximum value of g occurs at x = 3, and the maximum value is 3 - √5. (d) Find lim_{x→3} (10 - 3(x-2))/(f(x) - f(2)). lim_{x→3} (10 - 3(x-2))/(f(x) - f(2)) = 10 - 3(0) / (f(3) - f(2)) = 10 / (3 - √5 - 0) = 10 / (3 - √5)