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Madison KirbyAP CalculusMrs. Tallman13 April 2015Derivatives, Integrals, and Graphs (Part 2)Natalie Babbitt once wrote, “Things can come together in strange ways. The wood was at the center, the hub of the wheel. All wheels must have a hub. A Ferris wheel has one, as the sun is the hub of the wheeling calendar. Fixed points they are, and best left undisturbed, for without them, nothing holds together.” As with wheels and the sun, the idea of things coming together and having connections applies to the art of calculus as well. The hub of calculus holds different concepts together and shows the connection between everything. These concepts are things such as the derivative, the integral, determining area under a curve and of cross sections, average values, and so much more. Recognizing the connections between these topics is what makes problems, like those to be discussed, solvable. To start, let’s say there is a scientist that measures the depth of the Doe River at Picnic Point. At this point of the river, the width is 24 feet. The velocity of the water at Picnic Point is modeled by vt=16+2sin?(t+10) for 0≤t≤120 minutes, with the units in feet per minute. The scientist’s measurements are taken in a straight line perpendicular to the edge of the river, as shown in Figure 1, and the data for the measurements are shown in table 1, below.-1166849112300-9144017306Figure 1. Diagram of Doe River00Figure 1. Diagram of Doe River-914402036445Table 1 Data Measured00Table 1 Data MeasuredFor table 1, the top row is the distance from one edge of the river to the opposite edge and is measured in feet. The bottom row is the depth measurements the scientist is collecting in feet. These are positive in the table because the scientist is measuring from the river’s bottom to the surface. However, if these points were to be graphed, the depth would be negative since the depth is below the surface. This is displayed in Figure 2 and is only significant for graphing the data. Distance from the river’s edge (ft.)08142224Depth of the water (ft.)07820020320000-10160039531Figure 2. Graph of Data Measured00Figure 2. Graph of Data Measured63502343785(0, 0)(8, -7)(14, -8)(22, -2)(24, 0)123400(0, 0)(8, -7)(14, -8)(22, -2)(24, 0)1234The first problem this scientist faces is approximating the area of the river between the curve and the surface at Picnic Point, in square feet. To do this, the first thing he should do is use the trapezoid rule with the four subintervals indicated in table 1. Figure 3 shows the trapezoids drawn over the graph from Figure 2. The trapezoid rule is used because it estimates the area between the function and the axis. This method will be able to fairly accurately approximate the area because the shape of the trapezoid fits the curve of the data closely. Figure 3. Trapezoid Rule on GraphFigure 3 displays the trapezoid’s drawn onto the function from Figure 2. There are four trapezoids because there are four intervals within the interval from x = 0 to x = 24. However, the data does not have consistent intervals, so solving this problem requires a bit of creativity. The basic formula for trapezoid rule is Tn=?x12?fa+fx1+fx2+fx3+…+fxn-1+12?fb. Tn is the solution, f(x) is the function, a and b are the upper and lower limits of the interval, and ?x is the width of each trapezoid, where ?x=b-an. The general formula is designed for equal interval graphs, but this problem does not have this quality so a different method must be used. To properly calculate this area, each individual trapezoid area must be found separately using the area formula of area=height? (base1+base2)2. The vertical legs are substituted in for base1 and base2, and the horizontal width of each region is substituted for height. For example, the area of the first trapezoid is area1 = 8?(0+7)2 = 28 un2. The second trapezoid is area2 = 45 un2. And the third and fourth are area3 = 40 un2 and area4 = 2 un2. This means that by the trapezoid rule, the area under the surface of the water at Picnic Point is 115 un2.Next, the scientist has 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 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. The water flow is the measurement of how much water is travelling through that area at Picnic Point based on how much fluid the river can hold at this point (the cross-sectional area) and the speed of the river’s water (velocity). This means that the scientist can use the area he computed above and multiply it with the velocity of the water, which was given to be vt=16+2sin?(t+10) for 0≤t≤120 minutes. To set up a problem like this, the formula used is 1b-aabvt?dt?area. The letters a and b are the limits of the function; v(t) is the function, which is f(x) in general terms; and dt is showing that the problem is dealing with values in terms of t which is dx in general. The definite integral is used to make sure that the calculation is evaluating the problem at every point between a and b. Essentially, by slicing the area into small amounts, the height of these slices is what dt represents. After substituting the values in, the formula now looks like 1120-00120(16+2sin?(t+10) ?dt)?115. The result comes out to be 1807.169 ft3/min. This means that the volumetric flow of water at Picnic Point is 1807.169 cubic feet per minute. -7620114617500Now that the scientist has discovered this, he can explore more calculus connections by trying to find the area of the cross-section of the river at Picnic Point based on the function f(x) = 8 sinπ?x24, shown in Figure 4. -11239559527Figure 4. Model of Data00Figure 4. Model of DataFigure 4 is a mathematical model for the graph in Figure 2. This model places the graph above the x-axis, unlike Figure 2. This is because the data entered to find the equation were the positive ones from the table. Essentially, it does not matter whether it is above or below the x-axis because there is still the same amount of area between the curve and the axis. Figure 2 was more for displaying the drawing of the river. To calculate the cross-sectional area using this model, the formula used is area=abfx?dx. Unlike the volumetric flow problem where a and b were the time intervals, a and b in this problem are the distances from one edge of the river to the opposite edge, so a = 0 and b = 24. After substituting the values, the formula becomes area=0248 sin?(π?x24) ?dx. Through the use of the model f(x) = 8 sinπ?x24, the cross-sectional area of the Doe River at Picnic Point is 122.230 ft2. This problem is basically the same as the trapezoid rule above, except a different method was used to figure out the cross-sectional area. The two solutions are fairly similar, however the model may have overestimated the shape of the river’s bottom, which would lead to the larger area. Now that this function has been used to model the data and has been used to compute the cross-sectional area, it can also calculate the volumetric flow through Picnic Point in order to determine whether or not there is a need to divert the water and prevent flooding. If the volumetric flow through Picnic Point exceeds 2100 ft3/min for a 20-minute period, then the water must be diverted. To solve this, the scientist needs to set up the volumetric flow integral, which was explained above. For this problem, the integral would be 160-404060(16+2sin?(t+10) ?dt)?122.230, which calculates the volumetric flow of the water to be 2181.913 ft3/min. This value is greater than the predetermined 2100 ft3/min, which means that yes, the water will need to be. Another situation that requires different stems and features of calculus is the following: 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. Figure 5 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. -825501855970Figure 5. Passenger Flow Graph00Figure 5. Passenger Flow GraphSay the park’s mathematicians need to crunch some numbers and figure out how many people arrive at the ride between t = 0 and t = 3. The first thing they’re going to do is set up a definite integral because this allows for the accumulation of data between two points, which would be from the time the ride opened until three hours later. The integral for this problem is 03(r(t)) ?dt), which become 2?1000+12002+(1200+800)2 after the values from Figure 5 at t = 0 and t = 3 are substituted into the function’s integral. Another possible way to solve this problem is to use the counting squares method. Looking at Figure 6, if the number of squares below the function, and bounded by the axis, is counted between x = 0 and x = 3, and then multiplied by the area of one square, the answer can be found. -850901804535Figure 6. Counting Squares Method00Figure 6. Counting Squares MethodThere are 16 squares in this boundary and one square’s area is 200, so 16?200=3200 people. The result of the calculation for the number of people that arrive at the ride between t = 0 and t = 3 is 3200 people. These mathematicians now wonder whether the number of people waiting in line from t = 2 to t = 3 is increasing or decreasing. Looking at Figure 5, the number of people per hour arriving at the ride (between 1200 and 800 people/hour) is greater than the rate at which people get on the ride (800 people/hour), which means that the number of people waiting in line is increasing in this time frame. Furthering their understanding of this particular ride, the mathematicians wish to know at what time in the day the line for this ride the longest, and how many people are in the line at this time. Using the information previously discovered they know that the line is building up from t ≥ 0 to t < 3, where r(t) is greater than the rate at which people move onto the ride. Also from t > 3 to t ≤ 8, r(t) is less than the rate at which people move onto the ride. This means the line is shorter during these times. The only time that r(t) equals this rate is when the line will be longest, at t = 3. The number of people in line at t = 3 is found by adding the original number of people in line, 700, to the total number of people in line from start to three hours later, found earlier to be 3200, and subtracting the rate time the time, or 800(3). So, the time at which the most number of people are in line for the ride is at 3 hours after it opens and amount of people in line at this time is 1500 people. The next thing the mathematicians decide to do is to write 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. In order to properly create this integral, the initial step is to set it equal to zero, and since the starting line was 700 people this will be added to the integral. So far, the only thing they know is the following: 0 = 700 + some integral. Finding out what ‘some integral’ is where the trickiness begins. This integral is composed of two parts that range from the opening of the ride, 0, to the time of there being no line, t. The first part is the integral of the original equation, so 0tr(x)?dx. The second part is the rate at which people get on the ride, 800t. When the second part is subtracted from the first and set to zero, the time where there is no line can be found. The mathematicians must write 0 = 700 + 0tr(x)?dx-800t. Overall, these two examples were able to illustrate how calculus is like a wheel. Every concept is connected to one another through a hub and can be used to find solutions. In these problems, the definite integral was combined with volumetric flow, and other ideas like counting squares were used to find solutions too. There are endless combinations to math, and the two examples above only show a speck of what’s out there. Works CitedLuberoff, Eli. "Desmos Graphing Calculator."?Desmos Graphing Calculator. Desmos, Inc., 6 Mar. 2015. Web. 6 Mar. 2015. <;. ................
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