University:
High School
Course:
High School Mathematics
Academic year:

2019
Views:

343

Pages:

6
Author:

Milagros W

A GRAPHING CALCULATOR IS REQUIRED FOR THESE QUESTIONS. Fish enter a lake at a rate modeled by the function E given by E(t) = 20 + 15sin(πt/6). Fish leave the lake at a rate modeled by the function L given by L(t) = 4 + 2^(0.1t^4). 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. ∫(0 to 5) E(t) dt = 153.458 ≈ 153 (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)? (∫(0 to 5) L(t) dt) / 5 = 30.295 / 5 = 6.059 ≈ 6 (c) At what time t, for 0 ≤ t ≤ 5, is the greatest number of fish in the lake? Justify your answer. E(t) - L(t) = 0 ⇒ t ≈ 6.204 Amount = ∫(0 to 6.204) (E(t) - L(t)) dt ≈ 80.920 Used Extreme Value Theorem (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. d/dt(E - L) at t=5 ≈ -10.773 So rate is decreasing 2. Let S be the region bounded by the graph of the polar curve r=3 sin(3θ) for 0≤θ≤ 3 π , as shown in the figure above. a) Find the area of S. Area = 1/2 ∫(0 to π/3) r^2 dθ = 1/2 ∫(0 to π/3) (3√sin(3θ))^2 dθ = 3/2 ∫(0 to π/3) sin(3θ) dθ = 3.534 b) What is the average distance from the origin to a point on the polar curve r = 3√sin(3θ) for 0 ≤ θ ≤ π/3? Average distance = (1/π/3) ∫(0 to π/3) 3√sin(3θ) dθ = 2.800 ≈ 1.580 c) There is a line through the origin with positive slope that divides the region S into two regions with equal areas. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of m. slope = m = y/x tan(φ) = y/x = m so φ = tan⁻¹(m) 1/2 ∫(0 to tan⁻¹(m)) r^2 dθ = 1/2 ∫(tan⁻¹(m) to π/3) r^2 dθ d) For k > 0, let A(k) be the area of the portion of region S that is also inside the circle of radius k centered at the origin. Find lim(A(k)) as k approaches infinity. When k gets bigger, the part of r in the first quadrant. x = 0 (r cosθ) when θ = π/2 lim(k→∞) 1/2 ∫(0 to π/2) r^2 dθ = 3.324 The continuous function f is defined on the closed interval [-6, 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 to 5) f(x) dx = 7, find the value of ∫(-6 to -2) f'(x) dx. Show the work that leads to your answer. ∫(-6 to 5) f(x) dx = ∫(-6 to -2) f(x) dx + ∫(-2 to 5) f(x) dx = 7 ∫(-6 to -2) f(x) dx = 7 - ∫(-2 to 5) f(x) dx = 7 - (9 - 9π/4) (b) Evaluate ∫(2 to 5) (2f'(x) + 4) dx 2∫(2 to 5) f'(x) dx + ∫(2 to 5) 4 dx 2[f(5) - f(2)] + 4(3) = 2(0 - (3 - √5)) + 12 = 2 + 2√5 (c) The function g is given by g(x) = ∫(-2 to x) f(t) dt. Find the absolute maximum value of g on the interval [-2, 5]. Justify your answer. g'(x) = f(x) g'(x) = 0 ⇒ x = -1, 5 (d) Find lim(x→0) (10^x - 1) / (arctan x) lim(x→0) (10^x - 1) / (arctan x) = 10^0 - 3(2) / 1 = -5