The Re-write of: THE MISSING AIRPLANE ALIAA ISMAIL 900081536

Today I was informed that a plane that left Cairo airport last night was missing somewhere on its route west. The plane had a new engine installed. As good as this might sound, it is actually very risky. Therefore, the mechanic asked the pilot to inform him of the plane’s speed every 10 minutes of the journey, so that he can detect the engine’s behavior. After the first hour had passed, the mechanic noticed that the engine was functioning properly by showing promising results according to which he informed the pilot to report the plane’s speed every 15 minutes. Half an hour later the communication was cut and the last words of the pilot were “I’ve to make an emergency landing right away!” Since the communications were cut off, the exact location of the plane was and still remains unknown. The passengers’ lives are now in danger unless we can get to them as soon as possible. So, the authorities came to me as MACT-132 Expert to seek help in finding the location of the plane. I’ve been given some information to help: I) II) III) IV) V) VI)

The speed every ten minutes for the first hour and every 15 minutes for the next 30 minutes. I have the towns on the route of the flight with the distance in between them. I know the direction of the flight, which is west. I have the maximum speed of the plane, which is 600 km/h and the minimum, which is 100 km/h. I have the maximum acceleration and deceleration which is 1800 km/h2 I have the jerk, double jerk, and triple jerk.

Now, I’m supposed to use the available information to calculate the exact location with the minimum error possible in order to find the passengers before it is too late to send the search party. The search party can only cover a total distance of 100 km and they have to start from one of the following towns:

According to this diagram I assumed that the total distance the plane had covered must not be more than the sum of all the distances between the towns “A, B, C, D, E, F” added to 100km as the search party stated that they can cover a total distance of 100km: 200 + 75 + 125 + 150 + 100 + 125 + 100 = 875 km THE TIME TAKEN FOR THE AIRPLANE’S EMERGENCY LANDING In accordance with the pilot’s statement “ I am going to have to make an emergency landing right away!” I assumed that the pilot used the plan’s maximum deceleration, which is 1800 km/hr2.knowing that his speed after an hour and a half was 320km/hr, I recognized immediately that it was possible to know the exact time that the plane will take to land. Using the equation: v = u + at which has been derived from: Δv v − v0 = Δt Δt v = v0 + aΔt dv a= dt dv = a dt v Δt ⌠ ⌠ dv = a dt ⌡ ⌡ v0 0 v − v0 = aΔt

a=

v = v0 + aΔt v = u + at

(final velocity = initial velocity + (acceleration x time)

Where v = 0, u = 320, a = 1800, and t is the unknown. 0 = 320 + (-1800)t 1800t =320 t = 0.178 hours = 10.67 minutes.

If it takes the plane 10.67 minutes to land then the distance covered by the plane will be: The distance covered by the plane before landing:

Speed (km/hr) = 320 km

Time (hours) = 0.1778 hours Therefore the area of this triangle represents the distance covered by the plane before landing which is = Area = 0.5 x (10.67/60) x 320 Area = 28.45

THE MAXIMUM AND MIMIMUM POSSIBLE DISTANCES Before trying to solve the mystery using any mathematical rules I deiced to draw a graph of the absolute minimum and absolute maximum distance the plan can cover using the time intervals and different speeds from the mechanic’s records, the maximum acceleration and deceleration the plan can accomplish, and the rate of change of the plane’s acceleration.

Can the plane change from its maximum acceleration to its maximum deceleration? (The maximum acceleration and deceleration = 1800 km/hr2) Yes it can In order for the plane to do so it’s rate of change of acceleration (the double jerk) has to be above 3600 km/hr3 and it’s double jerk is 6000-km/ hr3. I calculated the actual distance by drawing the slope of the max acceleration and deceleration. Then, I divided my areas into small triangles, rectangles and trapezoids. Then, add all these areas together I was able to get the max and min areas.

The maximum value is 658.33 km The minimum value is 463.56 km Therefore the distance covered by the airplane has to be between the maximum and the minimum distances.

THE LEFT-HAND POINT AND RIGHT-HAND POINT APPROXIMATIONS Keeping the maximum and minimum values in mind I decided to start by using the left-hand point and right-hand point approximations, so I started by plotting a speed/time graph using my previous calculation of the amount of time taken to land and the table below:

The left hand point approximation is a form of approximation in which we use the left point as our measurement tool for the rectangles. We divide the area under the curve into rectangles which can be calculated easily by knowing the base and the height of the rectangle. The left hand point approximation gave me a value of 541.67 km The right hand end point is the same idea of the left hand end point but instead we use the right point as our measurement tool. The right hand point approximation gave me a value of 536.67 km The average between the two values is 539.17 km

â€œIn other words, the absolute value of the error of the left-hand point and right-hand point approximation is bounded by a constant multiplied by 1/n.â€?(1)

THE MID-POINT RULE Then I tried the Mid-point rule which is supposed to be more accurate than the left and right hand end points rules.

≈ Mn = ∆x [f(x1) + f(x2) +…+ f(xn)] (2) The midpoint rule requires that we take f(x) medium between two points. So, since we don’t have the function. We cannot make up midpoints. While applying the mid-point rule I have to take in consideration that all partitions must be equal, so I have to base all my calculations on the given speed and time readings only. So I chose my mid points to be: x = 1/6 x=½ x = 5/6 x = 5/4 So I will use the first 3 points in the first calculation: M3 = (1/3) [300 + 390 + 280] M3 = 323.33 km M1 = (0.5/1) [400] M1 = 200 km

Mn = M3 + M 1 Mn = 323.33 + 200 = 523.33 km In order to calculate the total distance using the mid- point rule I would have to add the distance for the emergency landing. That distance will equal to the area under the graph from 90 minutes until it hits the x-axis with gradient 1800. Area = 0.5 x (10.67/60) x 320 Area = 28.45 so the distance covered = 28.45 km So the total distance given by the mid-point rule will not be 523.33 km but it will be (523.33 + 28.45 = 551.78 km)

Like any other rule, the mid-point rule portrays a significant amount of error. This error can be calculated by the following formula: ׀ET =׀K (b – a) 3/24n2(3) ׀ET =׀K (b – a) 3/24n2 Where K is ׀f″(x)׀ K = 6000 Since the calculation for Trapezoidal rule was done in 2 parts M3 and M1 then we would have to calculate the error for M3 and M1 then I will add them together in order to find the error in Mn. Error in M3 ׀EM =׀6000 (1 – 0) 3/ 24 (32) ׀EM =׀27.777777778 ׀EM =׀27.78 Error in M1 ׀EM =׀6000 (1.5 – 1) 3/ 24 (12) ׀EM =׀31.25 Total error in the Mid-point rule ׀EM =׀27.78 + 31.25 = 59.03

TRAPEZOIDAL RULE Later I tried working with trapezoidal rule: ≈ Tn = (∆x/2)[f(x0) + 2f(x1) +2f((x2) …+f(xn-1)+f(xn)] (4)

Since n is the number of partitions and according to this method all the partitions must be equal, but the partitions we have on our speed time readings allows the partitions to vary, we have 10 minutes partitions and 15 minutes partitions therefore we have to divide the our calculation in two as follows: Tn = T6 + T 2 ∆x = (b – a)/n ∆x1 = (1 – 0)/6 ∆x1 = 1/6 T6 = ((1/6)/2)[0 +(2)(300)+(2)(420)+(2)(390)+(2)(370)+(2)(280) + 380] T6 = 325 km ∆x2 = (1.5 – 1)/2 ∆x2 = 0.5/2 ∆x2 = 0.25 T2 = (0.25/2) [380 + (2)(400) + 320] T2 = 187.5 km Tn = 325 km + 187.5 km = 512.5 km

The distance given by the trapezoidal rule is the distance, which the airplane has covered over the first 90 minutes of its journey, but it is not the total distance, covered by the airplane before landing. In order to get an estimation of the total distance covered by the airplane before landing we would have to take into consideration that the pilot had to perform an emergency landing on the ninetieth minute of his journey, we would also have to use the calculation done earlier of the landing time, which turned out

to be 10.67 minutes. I would calculate the distance traveled in those 10.67 minutes by calculating the area under that certain part of the graph. That part is a triangle therefore the area will be = 0.5 (base x height) Area = 0.5 x (10.67/60) x 320 Area = 28.45 so the distance covered = 28.45 km So the total distance given by the trapezoidal rule will not be 512.5 km but it will be (512.5 + 28.45 = 540.95 km) The result given by the trapezoidal rule was 540.95 km but as we know that the trapezoidal rule is just an approximation so it must contain an amount of error. I did not know what to do at that point so I visited my instructor of MACT-132 course, Dr. Ryan Derby-Talbot, who advised me to open my textbook () and search for the formula used to calculate the error in the trapezoidal rule. So I found the following formula: ׀ET =׀K(b – a)3/12n2(5) ׀ET =׀K(b – a)3/12n2 Where K is ׀f″(x)׀ K = 6000 Since the calculation for Trapezoidal rule was done in 2 parts T6 and T2 then we would have to calculate the error for T6 and T2 then I will add them together in order to find the error in Tn. Error in T6 ׀ET =׀6000(1– 0)3 / 12(6)2 ׀ET =׀13.8888888889 ׀ET = ׀13.89 Error in T2 ׀ET =׀6000(1.5– 1)3 / 12(2)2 ׀ET = ׀15.625

Total error in Trapezoidal rule ׀ET =׀15.625 + 13.89 = 29.515

SIMPSON’S RULE Then I read about another method of calculating the distance covered by the airplane, which is “Simpson’s Rule”. Simpson’s rule is “Another rule for approximate integration results from using parabolas instead of straight line segments to approximate a curve (3).” So I started by extracting its formula out of the textbook…

≈ Sn = (∆x/3)[f(x0) + 4f(x1) +2f((x2)+4f(x3)+…+2f(xn-2)+4f(xn-1)+f(xn)](6)

Here in Simpson’s Rule “n” (the number of partitions) has to be an even number, and all partitions must be equal, but the partitions we have on our speed time readings are not all equal six of them are 10 minutes partitions and the other two are 15 minutes partitions therefore we have to divide the our calculation in two as follows: Sn = S6 + S2 ∆x = (b – a)/n ∆x1 = (1 – 0)/6 ∆x1 = 1/6 S6 = ((1/6)/3)[0 +(4)(300)+(2)(420)+(4)(390)+(2)(370)+(4)(280) + 380] S6 = 324.44 km ∆x2 = (1.5 – 1)/2

∆x2 = 0.5/2 ∆x2 = 0.25 S2 = (0.25/3) [380 + (4)(400) + 320] S2 = 191.67 km Sn = 324.44 km + 191.67 km = 516.11 km Since this is only the distance covered by the airplane in the first 1.5 hours of the journey then it is not the total distance covered by the airplane before landing, so I will have to add the distance covered during the emergency landing. This distance was calculated while carrying out the trapezoidal rule calculations and it appeared to be 28.45 km. So the total distance covered by the airplane according to Simpson’s rule is: 516.11 km + 28.45 km = 544.56 km The result given by the Simpson’s rule was 544.56 km but as we know that Simpson’s rule is just an approximation so it must contain some error. I did not know what to do at that point so I visited my instructor of MACT-132 course, Dr. Ryan Derby-Talbot, who advised me to open my textbook (Calculus 6E by James Stewart) and search for the formula used to calculate the error in the trapezoidal rule. So I found the following formula: ׀ET =׀K (b – a) 5/180n4(7) ׀ET =׀K (b – a) 5/180n4 Where K is ׀f″(x)׀ K = 1.8 x 106 Therefore we would have to calculate the error for T6 and T2 then I will add them together in order to find the error in Tn.

Error in S6 ׀Es =׀1.8 x 106 (1– 0) 5 / 180(6) 4 ׀Es =׀7.72

Error in S2 ׀Es =׀1.8 x 106 (1.5– 1) 5 / 180(2) 4 ׀Es =׀19.53 Total error in Simpson’s Rule ׀Es =׀15.625 + 13.89 = 27.25 * from my calculations I found out that most accurate one is Simpson’s Rule as it has the smallest error bound, while the midpoint the least accurate one with the biggest error bound since we used less point in that method. Usually, midpoint is more accurate than trapezoidal but in this case the trapezoidal had more points, so it is more accurate.

Conclusion:After doing my calculations using the data given to me by the mechanic, I made good approximations and put in mind the percentage of error that would have happened: Since I found out that Simpsonâ€™s Rule is the most accurate method giving the least error I decided to work out the distance by subtracting the error and adding it to the value, which I have achieved from Simpsonâ€™s rule:

544.56 - 27.25 < The actual distance > 544.56 + 27.25 517.31 < The actual distance > 571.81

Since the actual distance lies between 517.31 and 571.81 then the search will actually have to happen within 54.50 km. As I know town (D) is 550km away from the airport therefore it would be worthwhile to start from town (D). According to this calculation the airplane would be found between 32.69 km east and 21.81 km west of town (D). The search group could begin their search from town (D) but I think it would be advisable for them to search 45km due west and 55km due east of town (D).

REFERENCE (1) “Quotation about the error in the left/right hand point rule” from http://math.furman.edu/~dcs/book/c4pdf/sec42.pdf (2) “The mid-point Rule” form Calculus 6th edition by James Steward page 532 (3) “Error in mid-point Rule” from Calculus 6th edition by James Steward page 535 (4) “Trapezoidal Rule” from Calculus 6th edition by James Steward page 533 (5) “Error in Trapezoidal Rule” from Calculus 6th edition by James Steward page 535 (6) “Simpson’s Rule” from http://mathworld.wolfram.com/SimpsonsRule.html (7) “Error in Simpson’s Rule” from Calculus 6th edition by James Steward page 539 (8) [Proof of “v = u + at”] was achieved from http://hypertextbook.com/physics/mechanics/motion-equations/

THE MISSING AIRPLANE

Published on Sep 12, 2011

Now, I’m supposed to use the available information to calculate the exact location with the minimum error possible in order to find the pass...

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