It is often said that we get lost in history, or in the world, or even in our own thoughts, but what kind of loss are we referring to? The loss where you're so confused in the story, so overwhelmed that you can't find which way is up, which way is down, or even have any idea where to turn next? Or is it the loss in which you have so emerged in the plot, in the lives of the characters and in personal feelings, that you hope to never have to emerge again, that you can simply continue to live in the storybook world. This is how I am in mathematics: I want to be sucked into the world, into the language of mathematics, into the thousand theorems, and formulas, and laws, and I never want to leave it again. Say no to plagiarism. Get a tailor-made essay on "Why Violent Video Games Shouldn't Be Banned"? Get an Original Essay This article will combine several calculus ideas that all stem from two main topics: the integral and the derivative. The derivative is the instantaneous rate of change of a function at a specific point. Another way to think about the derivative is how the graph changes as x increases or decreases. The integral, or the antiderivative as it is commonly called, is the opposite because it can be thought of as canceling out the differentiation. A scientist measures the depth of the Doe River at Picnic Point. The river is 24 feet wide at this location. Measurements are taken in a straight line perpendicular to the edge of the river. The data is shown in the table below. The water velocity at Picnic Point, in feet per minute, is modeled by v(t) = 16 + 2sin( ) for 0 ≤ t ≤ 120 minutes. Distance from river edge (feet) 0 8 14 22 24 Depth of water (feet) 0 7 8 2 0a) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the cross-sectional area of ​​the river at Picnic Point, in square feet. Show the calculations that lead to your answer. The first step is to know what the trapezoid rule is and what it does. The trapezoid rule divides the area under the curve of a defined integral into trapezoides with varying widths and heights based on the data provided. The areas of these trapezoids are then added to find an approximation of the total area under the curve. The known and accepted formula for the area of ​​the trapezoid is〖Area〗_T=1/2 (h_1+h_2 )where h1 and h2 represent the heights of the trapezoides and w represents the width of the trapezoides. The area of ​​the first trapezoid, with heights 0 and 7 and width 8, is 28 ft2, the area of ​​the second trapezoid, with heights 7 and 8 and width 6, is 45 ft2, the area of ​​the third trapezoid with heights 8 and 2 and width 8, is 40 ft2, and the area of ​​the fourth and final trapezoid, with heights 2 and 0 and width 2, is 2 ft2. The area under the curve, using the trapezoidal rule, can be found using the equation〖Area〗_Total=A_1+A_2+A_3+A_4so the total cross-sectional area of ​​the river is 115 ft2.b) The volume flow in a location along the river is the product of the cross-sectional area and the velocity of the water at that location. Use the approximation from part (a) to estimate the average value of volumetric flow at Picnic Point, in cubic feet per minute, from t = 0 to t =120 minutes. The average value of the volumetric flow can be calculated by the equation 1/ (ba) ∫_a^b▒〖c*f(x)□(24&dx)〗where a is the starting point, b is the ending point, c is a constant ef(x) is the equation. This is the mean value theorem and can be used because the function is continuous on the closed interval of [a,b] and differentiable on the open interval (a,b). In the context of this problem, the starting point is 0, the ending point is 120, the constant is the cross section of the river found in the letter a, 115 ft2, and the equation is the equation given for the velocity of the water of the river , v(t) = 16 + 2sen( ). The calculation of valueaverage is therefore 1/(120-0)∫_0^120▒〖115*v(t)□(24&dt)〗and is evaluated equal to 1807.17 and the units are ft3/min based on the units of the constant (ft2 ) multiplied by the units of the equation (ft/ min).c) The scientist proposes the function f, given by f(x) = as a model for the water depth, in feet, at Picnic Point x feet from the shore of the river. Find the cross-sectional area of ​​the river at Picnic Point based on this model. The cross-sectional area can be calculated by integrating the given equation from the initial value to the final value. The general equation for integration is ∫_a^b▒〖f(x)□(24&dx)〗where a is the starting point, b is the ending point, and f(x) is the given equation. In the context of the problem, the starting point would be 0 and the ending point would be 24 since this is the total width of the river, and the equation would be the equation given above. Using the equation and data provided, the cross-sectional area can be equal to 122.23 ft2.d) Recall that volumetric flow is the product of the cross-sectional area and the velocity of the water at a given location. To avoid flooding, water must be diverted if the average volumetric flow rate at Picnic Point exceeds 2,100 cubic feet per minute for a 20-minute period. Using the answer from part c), find the average value of the volumetric flow during the time interval 40 ≤ t ≤ 60 minutes. Does the value indicate that water needs to be diverted? The average value of the volumetric flow rate can be calculated using the same formula as above1/(ba) ∫_a^b▒〖c*f(x)□(24&dx)〗but this time using the values ​​of 40 minutes for a, 60 minutes for b, 122.23 ft2 as a constant and the same velocity equation, v(t) = 16+2sin(√(x+10)), the average value of the volumetric flow rate is 2181.89 cubic feet/min . Therefore the flow should be diverted away from that part of the river.2. There are 700 people lined up for a famous amusement park ride when the ride starts operating in the morning. Once the ride begins, it accepts passengers until the park closes 8 hours later. While there is a line, people move at a rate of 800 people per hour. The graph below shows the rate, r(t), at which people get to the race during the day. Time t is measured in hours from the time the ride starts running. How many people arrive at the race between t = 0 and t = 3? Show calculations. The number of cyclists arriving at the race between t=0 and t=3 can be calculated using the trapezoidal rule and the graph below. This is possible because the area under the curve in the given interval corresponds to the number of cyclists arriving in that time interval. Two trapezoids can be created on the given interval, one from t=0 to t=2 and the second on the interval from t=2 to t=3. For the first trapezoid, the heights are 1000 and 1200, and the width is 2, so the area is 2200, and for the second trapezoid the heights are 1200 and 800, with a width of 1, so the area is 1000 Therefore, the result the total area under the curve and the number of passengers arriving at the ride between t=0 and t=3 was 3200 passengers. Does the number of people queuing to ride the ride increase or decrease between t = 2 and t = 3? Justify.The number of people waiting to board the ride increases between t=2 and t=3 because the rate at which passengers board the ride is 800 passengers per hour and while the rate at which passengers arrive at the ride is decreasing between t=2 and t=3, the number of passengers is still above 800 per hour. At what time (t) is the queue for the ride longest? How many people are in line at that time? Justify. It is assumed that passengers ride the route and then get off at a rate of 800 passengers per hour, and according to the graph, in the time interval from t=0 to t=3, people arrive on the line faster than they leave the race..