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# The Integral and the Derivative Calculus Ideas

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It is often said that one gets lost in the story, or the world, or even own thoughts, but what kind of lost is being referred to? The lost where you get so mixed up in the story, so overwhelmed that you can’t find which way is up, which way is down, or even have the slightest idea of where to turn next? Or is the lost where you’re so emerged in the storyline, in the character’s lives and personal feelings, that you hope you’ll never have to surface again, that you can just continue to live in the story book world. That’s how I am in math: I want to be sucked into the world, the language of mathematics, the thousands of theorems, and formulas, and laws, and I want to never come out again.

This paper will combine several calculus ideas that all stem off 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 is the graph changing as x increases or decreases. The integral, or antiderivative as it is commonly called, is the opposite because it can be thought of as undoing the differentiation.

A scientist measures the depth of the Doe River at Picnic Point. The river is 24 feet wide at this location. The measurements are taken in a straight line perpendicular to the edge of the river. The data are shown in the table below. The velocity of the water at Picnic Point, in feet per minute, is modeled by v(t) = 16 + 2sin( ) for 0 ≤ t ≤ 120 minutes.

Distance from the river’s edge(feet) 0 8 14 22 24

Depth of the water (feet) 0 7 8 2 0

a) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the area of the cross section of the river at Picnic Point, in square feet. Show the computations that lead to your answer.

The first step is to know what the trapezoidal rule is and what it does. The trapezoid rule divides the area under the curve of a definite integral into trapezoids with varying widths and heights based on the data given. The area of these trapezoids are then added together 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 )w

where h1 and h2 represent the heights of the trapezoids, and w represents the width of the trapezoids. The area of the first trapezoid, with heights 0 and 7 and a width of 8, is 28 ft2, the area of the second trapezoid, with heights 7 and 8 and a width of 6, is 45 ft2, the area of the third trapezoid with heights 8 and 2 and a width of 8, is 40 ft2, and the area of the fourth and final trapezoid, with heights 2 and 0 and a width of 2, is 2 ft2. The area under the curve, using trapezoidal rule, can be found using the equation

〖Area〗_Total=A_1+A_2+A_3+A_4

so the total area of the cross section of the river is 115 ft2.

b) The volumetric flow at a location along the river is the product of the cross-sectional area and the velocity of the water at that location. Use your approximation from part (a) to estimate the average value of the volumetric flow at Picnic Point, in cubic feet per minute, from t = 0 to t =120 minutes.

The average value of volumetric flow can be calculated by the equation

1/(b-a) ∫_a^b▒〖c*f(x)□(24&dx)〗

where a is the starting point, b is the ending point, c is a constant, and f(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 letter a, 115 ft2, and the equation is the equation given for the velocity of the river water, v(t) = 16 + 2sin( ). The calculation of the average value is then

1/(120-0) ∫_0^120▒〖115*v(t)□(24&dt)〗

and is evaluated to equal 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 depth of the water, in feet, at Picnic Point x feet from the river’s edge. Find the area of the cross-section of the river at Picnic Point based on this model.

The area of the cross section can be calculated by integrating the given equation from the starting value to the ending 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 the f(x) is the given equation. Within the context of the problem, the starting point would be 0 and the ending point would be 24 as that is the total width of the river, and equation would be the above given equation. Using the equation and the given data, the cross sectional area can be found to be 122.23 ft2.

d) Recall that the volumetric flow is the product of the cross sectional area and the velocity of the water at a location. To prevent flooding, water must be diverted if the average value of the volumetric flow at Picnic Point exceeds 2100 cubic feet per minute for a 20-minute period. Using your 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 the water must be diverted?

The average value of the volumetric flow can be calculated using the same formula as above

1/(b-a) ∫_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 the constant, and the same velocity equation, v(t) = 16+2sin(√(x+10)), the average value of the volumetric flow is 2181.89 ft3/min. Therefore, the flow would have to be diverted away from that part of the river.

2. There are 700 people in line for a popular amusement-park ride when the ride begins operation in the morning. Once it begins operation, the ride accepts passengers until the park closes 8 hours later. While there is a line, people move onto the ride at a rate of 800 people per hour. The graph below shows the rate, r(t), at which people arrive at the ride throughout the day. Time t is measured in hours from the time the ride begins operation.

How many people arrive at the ride between t = 0 and t = 3? Show the computations.

The number of riders who arrived at the ride between t=0 and t=3 can be calculated using the trapezoidal rule and the graph below. This can be done because the area under the curve on the given interval corresponds to the number of riders who arrived in that time frame. Two trapezoids can be made on the given interval, one from t=0 to t=2 and the second on the interval 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 resulting total area under the curve and number of passengers who arrived at the ride between t=0 and t=3 was 3200 passengers.

Is the number of people waiting in line to get on the ride increasing or decreasing between t = 2 and t = 3? Justify.

The number of people waiting to get on the ride is increasing between t=2 and t=3 because the rate at which passengers get onto 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, it is still larger than 800 passengers an hour.

At what time (t) is the line for the ride the longest? How many people are in line at that time? Justify.

It is given that passengers ride the ride then exit at a rate of 800 passengers an hour and based on the graph, on the time interval t=0 to t=3, people are arriving at the line faster than they are leaving the line, so the line is therefore growing. At t=3, however, people start getting in line at a much slower rate so the line will grow more slowly. At time t=3 the line is the longest it would ever be. The number of passengers can be found by adding the number of people in line at t=0, 700, to the number of passengers who got into line between t=0 and t=3, 3200 and subtracting the number of passengers who rode the ride in the same time interval, 800*3 or 2400. The number of passengers in line at t=3 is 1500 passengers.

Write, but do not solve, an equation involving an integral expression of r whose solution gives the earliest time t at which there is no longer a line for the ride.

An equation to find the earliest time at which there will be no one in line is

0=700+∫_0^t▒r(x)dx-800t

where t is the time. This is the general form of the equation used in letter c. The number of people in line at given time t can be found by adding the number of people in line at t=0, 700, to the number of passengers who get in line from the start, t=0, to that point, t, which is represented by the integral from 0 to t of the rate at which people arrive in line, and then the number of people who already rode and exited the ride is subtracted from this sum. This equation is set equal to 0 to find when there will be no one in line.

Mathematics is a beautiful language that only some can understand. It takes a special person to be able to wrap their head around pieces of information, decide which of thousands of formulas, theorems, and principals apply, and make something out of almost nothing. It’s almost like magic! It’s strange, it’s complex, it’s intriguing, it’s amazing: it’s the wonderful world of mathematics.

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