1. The Great Pyramid of Khufu<br />11430027876500<br />Faraad Armwood<br />Math 2432<br />Mrs. Feller<br />Faraad Armwood<br />Math 2432<br />Mrs. Feller<br />The Great Pyramid of Khufu<br />Among the many applications of integral calculus to physics and engineering, we consider three determinate factors: work, force due to water pressure, and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths), our strategy is to break up a physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and evaluate the resulting integral.<br />Work<br />The term work can be described as the amount of effort required to perform a task. However, in physics work is referred to as, a scalar quantity that can be roughly described as the product of a force times the distance through which it acts, and it is called the work of the force. More precisely, only the component of a force in the direction of the movement of its point of application generates work. In general, if a object moves along a straight line with position function s(t), the force F on the object is defined by Newton’s Second Law of Motion. IN the case of constant acceleration, the force F is also constant, and the work done can be defined by the equation,<br />1371600609600<br />1 W= Fd work = force x distance<br />If the force (F) is measured in newtons and the distance (d) is measure in meters, then the unit for work (W) is newton-meters, or joules. If F is measure in pounds and d is measured in feet, then the unit for W is foot-pound (ft-lb).<br />Suppose an object is moving along the x-axis in the positive direction, from a to b, and at each point x between and b a force f (x) acts on the object, where f is a continuous function. We can divide the interval [a,b] into n subintervals and each point having a distance equal to delta x. Now assume the force a this point is f (xi), where the point xi is the ith subinterval. The work that is done given all these factors can be approximated by,<br />1943100742950<br />2 Wi≈fxi ∆x<br />Thus the total work can be approximated by,<br />1943100996950<br />3 W≈ i=1fxi ∆x<br />It seems as though our approximation will become better if n is larger. Therefore we can define the work done in moving the object from a to b as the limit of this quantity as n ∞. <br />148590015240000<br />4 W = limn->∞i=1fxi Δx = abfx dx<br />Faraad Armwood<br />Math 2432<br />Mrs. Feller<br />The Problem: Suppose that it took 20 years to construct the great pyramid of Khufu at Gizeh in Egypt. This pyramid is 500 feet high and has a square base with edge length 750 feet. Suppose also that the pyramid is made of rock with density ρ = 120 lb/cubic foot. Finally suppose that each laborer did 160 ft-lb/hour of work in lifting rocks from ground level to their final position in the pyramid and worked 12 hours daily for 330 day/year. How many laborers would have been required to construct the pyramid?<br />To better understand this problem, let’s draw an image.<br />342900031527750342900383857533147003609975017145004067175750ft750ft3429003609975038862001438275500ft500ft3429000409575365760040957517145007524752400300246697500457200280987501257300109537519431001323975102870016668750217170018954758001002238375262890028562302171700455930148590045593034290068453014859006845303429003199130<br />Being as though the pyramid has a square base we can split the pyramid in half and use similar triangles.<br />2857500271462537537524003001457325ss24003004286251714500145732524003004286259144002600325914400428625-1143003971925By using similar triangles we can find the area of each slab (slice) of the pyramid. Below is a proportion set up by using similar triangles.By using similar triangles we can find the area of each slab (slice) of the pyramid. Below is a proportion set up by using similar triangles.4114800771525500-y500-y3657600428625036576001457325038862004286254800600260032504800600428625050292004286250052578001343025500ft500ft 500375 = (500-y)s/2<br />Cross multiply to find the solution for this equation and solve for s.<br /> <br /> 500s = 375000 – 750y <br />Divide by 500 to solve for s.<br />S = 375000-750y500<br />Since this fraction can be simplified do so.<br />S = 750 – 1.5y<br />Since we divided our pyramid into to parts using similar triangles we can solve for the area of each slap by squaring our equation for s. The fact that our base is a square allows us to assume that each side is equal.<br />Area = (750-1.5y)2 <br />Therefore our volume with respect to our change in y can be written as,<br />Volume = (750-1.5y)2 dy<br />Now let’s take into account that we are raising each slab up a y distance.<br />(750-1.5y)2 y dy<br />Finally multiply by the weight of each slab, and take the integral from o to 500ft.<br />0500120750-1.5y2y dy<br />Now since we know how much work was needed to lay each slab on the pyramid on an interval of [0,500], we can begin to calculate the amount of workers needed to build the pyramid. Suppose that each laborer did 160 ft-lb/hour of work in lifting rocks from ground level to their final position in the pyramid and worked 12 hours daily for 330 day/year. Using this schedule of labor for each worker, let’s calculate the amount of work one person can do in twenty years, and divide the total amount of work by this number to get the amount of workers.<br />160 ft-lbhourx 12 hoursday= 1920 ft-lbday<br />Notice the hours in this conversion table cancel and the units you are left with are ft-lb/day, so each person can do 1920 ft-lbs a day.<br />Now let’s calculate the amount of work one person can do in a year, assuming that they work 330 days per year.<br />1920 ft-lbdayx330 daysyear= 633600 ft-lbyear<br />As with our previous conversion table, we have units that cancel and we’re left with 633600 ft-lbs per year. This number represents the amount of work one individual can do in a year. Also, since it took 20 years to build the pyramid, multiply the work that an individual can do in one year by 20.<br />633600 ft-lbyear x 20 years1 = 12672000 ft-lbs <br />According to our solution, an individual given this particular work schedule can do 1.2672 x 107ft-lbs of work in 20 years. Since we know how much work one person can do in twenty years, divide the total amount of work needed by how much each person could do and this will give you how many workers it took to build the pyramid.<br />Workers = Total Amount of Work NeededAmount of work one indvidual could do <br /> = 1.40625x 10121.2672 x 107<br /> <br /> = 110,973 workers <br />The amount of workers required to build the pyramid is approximately 110,973 workers.<br />
![The Great Pyramid of Khufu<br />11430027876500<br />Faraad Armwood<br />Math 2432<br />Mrs. Feller<br />Faraad Armwood<br />Math 2432<br />Mrs. Feller<br />The Great Pyramid of Khufu<br />Among the many applications of integral calculus to physics and engineering, we consider three determinate factors: work, force due to water pressure, and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths), our strategy is to break up a physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and evaluate the resulting integral.<br />Work<br />The term work can be described as the amount of effort required to perform a task. However, in physics work is referred to as, a scalar quantity that can be roughly described as the product of a force times the distance through which it acts, and it is called the work of the force. More precisely, only the component of a force in the direction of the movement of its point of application generates work. In general, if a object moves along a straight line with position function s(t), the force F on the object is defined by Newton’s Second Law of Motion. IN the case of constant acceleration, the force F is also constant, and the work done can be defined by the equation,<br />1371600609600<br />1 W= Fd work = force x distance<br />If the force (F) is measured in newtons and the distance (d) is measure in meters, then the unit for work (W) is newton-meters, or joules. If F is measure in pounds and d is measured in feet, then the unit for W is foot-pound (ft-lb).<br />Suppose an object is moving along the x-axis in the positive direction, from a to b, and at each point x between and b a force f (x) acts on the object, where f is a continuous function. We can divide the interval [a,b] into n subintervals and each point having a distance equal to delta x. Now assume the force a this point is f (xi), where the point xi is the ith subinterval. The work that is done given all these factors can be approximated by,<br />1943100742950<br />2 Wi≈fxi ∆x<br />Thus the total work can be approximated by,<br />1943100996950<br />3 W≈ i=1fxi ∆x<br />It seems as though our approximation will become better if n is larger. Therefore we can define the work done in moving the object from a to b as the limit of this quantity as n ∞. <br />148590015240000<br />4 W = limn->∞i=1fxi Δx = abfx dx<br />Faraad Armwood<br />Math 2432<br />Mrs. Feller<br />The Problem: Suppose that it took 20 years to construct the great pyramid of Khufu at Gizeh in Egypt. This pyramid is 500 feet high and has a square base with edge length 750 feet. Suppose also that the pyramid is made of rock with density ρ = 120 lb/cubic foot. Finally suppose that each laborer did 160 ft-lb/hour of work in lifting rocks from ground level to their final position in the pyramid and worked 12 hours daily for 330 day/year. How many laborers would have been required to construct the pyramid?<br />To better understand this problem, let’s draw an image.<br />342900031527750342900383857533147003609975017145004067175750ft750ft3429003609975038862001438275500ft500ft3429000409575365760040957517145007524752400300246697500457200280987501257300109537519431001323975102870016668750217170018954758001002238375262890028562302171700455930148590045593034290068453014859006845303429003199130<br />Being as though the pyramid has a square base we can split the pyramid in half and use similar triangles.<br />2857500271462537537524003001457325ss24003004286251714500145732524003004286259144002600325914400428625-1143003971925By using similar triangles we can find the area of each slab (slice) of the pyramid. Below is a proportion set up by using similar triangles.By using similar triangles we can find the area of each slab (slice) of the pyramid. Below is a proportion set up by using similar triangles.4114800771525500-y500-y3657600428625036576001457325038862004286254800600260032504800600428625050292004286250052578001343025500ft500ft 500375 = (500-y)s/2<br />Cross multiply to find the solution for this equation and solve for s.<br /> <br /> 500s = 375000 – 750y <br />Divide by 500 to solve for s.<br />S = 375000-750y500<br />Since this fraction can be simplified do so.<br />S = 750 – 1.5y<br />Since we divided our pyramid into to parts using similar triangles we can solve for the area of each slap by squaring our equation for s. The fact that our base is a square allows us to assume that each side is equal.<br />Area = (750-1.5y)2 <br />Therefore our volume with respect to our change in y can be written as,<br />Volume = (750-1.5y)2 dy<br />Now let’s take into account that we are raising each slab up a y distance.<br />(750-1.5y)2 y dy<br />Finally multiply by the weight of each slab, and take the integral from o to 500ft.<br />0500120750-1.5y2y dy<br />Now since we know how much work was needed to lay each slab on the pyramid on an interval of [0,500], we can begin to calculate the amount of workers needed to build the pyramid. Suppose that each laborer did 160 ft-lb/hour of work in lifting rocks from ground level to their final position in the pyramid and worked 12 hours daily for 330 day/year. Using this schedule of labor for each worker, let’s calculate the amount of work one person can do in twenty years, and divide the total amount of work by this number to get the amount of workers.<br />160 ft-lbhourx 12 hoursday= 1920 ft-lbday<br />Notice the hours in this conversion table cancel and the units you are left with are ft-lb/day, so each person can do 1920 ft-lbs a day.<br />Now let’s calculate the amount of work one person can do in a year, assuming that they work 330 days per year.<br />1920 ft-lbdayx330 daysyear= 633600 ft-lbyear<br />As with our previous conversion table, we have units that cancel and we’re left with 633600 ft-lbs per year. This number represents the amount of work one individual can do in a year. Also, since it took 20 years to build the pyramid, multiply the work that an individual can do in one year by 20.<br />633600 ft-lbyear x 20 years1 = 12672000 ft-lbs <br />According to our solution, an individual given this particular work schedule can do 1.2672 x 107ft-lbs of work in 20 years. Since we know how much work one person can do in twenty years, divide the total amount of work needed by how much each person could do and this will give you how many workers it took to build the pyramid.<br />Workers = Total Amount of Work NeededAmount of work one indvidual could do <br /> = 1.40625x 10121.2672 x 107<br /> <br /> = 110,973 workers <br />The amount of workers required to build the pyramid is approximately 110,973 workers.<br />](https://image.slidesharecdn.com/completeprojectdraft-111010202227-phpapp02/85/Mathe-2432-Project-1-320.jpg)
