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STAGE 1- INVESTIGATON FOLIO

Devanshee Parekh SACE NO: 986543F Ms Mouffat

The Two Yachts Problem Introduction The focus of this investigation is to determine the pathways of 2 yachts (A and B), both will have different vectors and initial coordinates. Parametric equations will be used to assist in this investigation. Parametric equations are equation that utilise another independent variable, time (t). Through parametric equation it is possible to find the intersection points, but theoretical time will always follow. In the investigation parametric equations will be used to determine positions of the yachts. Parametric calculations ensure that no collisions occur between yachts. Throughout the investigation, the calculator TI-84 Plus CE will be used aid in the calculations. The equations of Yacht A and Yacht B that will be used for the calculations are: Yacht A has the initial position (-10,4) and velocity vector [2,-1] Yacht B has the initial position (4, -13) and the velocity vector [-1,3]

Exercise 1 1) Explain why the position of each yacht at the time t can be written as the vector equations. [ X A , Y A ] =[ −10,4 ]+t [ 2,−1] [ X B ,Y B ]= [ 4 ,−13 ]+t [−1,3 ] With each time increment the distance of the yacht increases by the length of the velocity of the vector [2,-1] 2) a) Write the above equations in parametric form, and on squared paper plot the paths of the yachts when t= 0, 1, 2, 3, 4… Yacht A The equation of Yacht A in terms of x ( t ) and y (t) is: ¿> x ( t ) =−10+2 t ¿> y ( t )=4 −t Combining both equation in the terms of y (t) while also eliminating t from the equation: ¿>2 t=x+ 10 x ¿>t= +5 2 x ¿> y ( t )=4−( )+5 2 −x −1 2 Table 1: Displays the coordinates of Yacht A in relation to the time x-coordinate y- coordinate Time (hours) -10 4 0 -8 3 1 -6 2 2 -4 1 3 -2 0 4 0 -1 5 ¿> y ( t )=

Yacht B The equation of Yacht B in terms of x ( t ) and y (t) is: ¿> x ( t ) =4 −t ¿> y ( t )=−13+3 t Combining both equation in the terms of y (t) while also eliminating t from the equation: ¿>t= 4 − x ¿> y ( t )=−13+3(4−x ) ¿> y ( t )=−13+12−3 x ¿ ¿> y ( t )=−1+3 x ¿ ¿> y ( t )=−3 X −1 Table 2: Displays the coordinates of Yacht A in relation to the time x-coordinate y- coordinate Time (hours) 4 -13 0 3 -7 1 2 -4 2 1 -1 3 0 2 4 -1 5 5

Graph 1: Depicts the pathways of Yacht A and B over time. The purple points display the intersection points of Yacht A and B. b) For the yachts to collide that have to be at the same destination at the same time. But in this case the yachts do not collide, they only cross pathways. They do not

collide as Yacht A reaches intersection point at 5 hours and Yacht B reaches the intersection point at 4 hours. c) The distance formula will be used to determine the distance and the speed will be calculated after that using the speed, distance and time triangle. 2 2 Distance formula: ( x2 − x1 ) +( y2 − y 1) Triangle:

√

Speed of Yacht A: 2 2 −8−( −10) +( 3− 4 ) ¿ d YA = √ ¿ d YA = √ ( 2 ) +( −1) d YA = √ 5 2

2

Therefore, d s= t 5 √ s= 1 s= √5 km/h

Speed of Yacht B 2 2 d YB =√ ( 3−4 ) + (−10+13 ) 2 2 d YB =√ ( −1 ) +( 3 ) d YB =√ 10 km/h Therefore, d s= t 10 √ s= 1 s=√10 km/h

3) Find the position vector of B relative to A The formula used to find the position vector is: AB=(

a2−a 1 ) b2−b 1

Position vector: Yacht A= (−10,4 ) Yacht B= ( 4 ,−13) 4− (−10 ) AB=( ) − 13− 4 AB=( 14 ) −17 4) Use the result of Question 3 to show that if d is the distance between the yachts at any time t then d 2=25 t 2−220 t + 485 The formula will be used to determine the distance between each Yacht at different times and will be used to prove that the formula is correct: Time Yacht A Yacht B D2 (distance) (km) 0 (-10,4) (4,-13) 485 1 (-8,3) (3,-10) 290 2 (-6,2) (2,-7) 145 3 (-4,1) (1,-4) 50 4 (-2,0) (0,-1) 5 5 (0,-1) (-1,2) 10

To show that the above equation is true the distance formula ( 2 2 d 2=( x 2− x 1 ) +( y 2− y 1 ) ) will be used. 2 2 d 2=( x 2− x 1 ) +( y 2− y 1 ) 2 2 2 d =(−13+ 3 t−4 + t ) +( 4−t +10−2t ) 2 2 2 d =(−17+4 t ) + ( 14−3 t ) 2 2 2 d =289−136 t+16 t + 196−84 t + 9 t d 2=25 t 2−220 t + 485 as required

5) Find the minimum value of d 2 and the time t when this occurs. Make sure that you justify your answer. d 2=25 t 2−220 t + 485 is a concave up parabola, this is concluded because the t is squared and the coefficient of it is positive. To find the minimum distance the following formula will be used: −b Minimum= 2a −(−220) 2 (25 ) Minimum=4.4 Therefore, at 4.4 hours the ships were the closest distance from each other. To find the distance at 4.4 hours, 4.4 needs to be substituted in the place of t. The distance at 4.4 hours is: 2 2 d =25 t −220 t + 485 d 2 =25(4.4)2 −220 (4.4 )+ 485 2 d =1 So, the minimum value of d2 is 1 Minimum=

6) Hence find the minimum distance between the yachts and the time this occurs: The minimum distance between the yachts is: d 2=1 d=1 So, the minimum value of d is 1 and it occurs at 4.4 hours. 7) Suppose Yacht B wishes to rendezvous with Yacht A at 3.5 hours. If A and B start from their original positions and A maintains its original path, find the velocity vector [a,b] for Yacht B Position of Yacht A at 3.5 hours: To find this value the 3.5 will be substituted into the original equations of t in terms of x and y. x ( t )=−10+ 2t y ( t) =4−t x ( t )=−10+2(3.5) y ( t) =4−( 3.5) x ( t )=−3 y ( t) =0.5 Therefore, after 3.5 hours Yacht A will be at (-3,0.5)

Now, to find the velocity vector, first the position vector will be found with and then both value will be divided by 3.5 to find the velocity vector. Original (4,-13), destination (-3,0.5) AB=( −3−4 ) 0.5 + 13 AB=(

−7 ) 13.5

−7 AB=( 3.5 ) 13.5 3.5 −2 AB=( 27 ) 7

Exercise 2 Create a situation of your own with different initial positions and velocity vectors for two yachts. −1 Yacht A= Initial position (-1,4) Vector ( 5 ) 1 Yacht B= Initial position (2,-3) Vector ( ) −2

1) a) Yacht A The equation of Yacht A in terms of x ( t ) and y (t) is: ¿> x ( t ) =−1−t ¿> y ( t )=4 +5 t Combining both equation in the terms of y (t) while also eliminating t from the equation: ¿>t=−1−x ¿> y ( t )=4 +5(−1−x ) ¿> y ( t )=4 −5 −5 x ¿> y ( t )=−1−5 x Table 1: Displays the coordinates of Yacht A in relation to the time x-coordinate y- coordinate Time (hours) -1 4 0 -2 9 1 -3 14 2 -4 19 3 -5 24 4

-6

29

5

Yacht B The equation of Yacht B in terms of x ( t ) and y (t) is: ¿> x ( t ) =2+ t ¿> y ( t )=−3−2t Combining both equation in the terms of y (t) while also eliminating t from the equation: ¿>t=x−2 ¿> y ( t )=−3−2(x−2 ) ¿> y ( t )=−3−2 x +4 ¿> y ( t )=1−2 x Table 2: Displays the coordinates of Yacht A in relation to the time x-coordinate y- coordinate Time (hours) 2 -3 0 3 -5 1 4 -7 2 5 -9 3 6 -11 4 7 -13 5 2)

Graph 2: Depicts the pathways of Yacht A and B over time.

b) For the yachts to collide that have to be at the same destination at the same time. But in this case the yachts do not collide, they only cross pathways. c) The distance formula will be used to determine the distance and the speed will be calculated after that using the speed, distance and time triangle. 2 2 Distance formula: ( x2 − x1 ) +( y2 − y 1) Triangle:

√

Speed of Yacht A: 2 2 −2−( −1) +(9−4 ) ¿ d YA = √¿ d YA = √ (−1) +( 5 ) d YA = √ 26 2

2

Therefore, d s= t 26 s= √ 1 s= √26 km/h

Speed of Yacht B −3 −5−(¿ ) ¿ ¿ ( 3−2 )2 +¿ d YB = √¿

2 2 d YA =√ ( 1 ) + ( −2 ) d YA =√ 5 km/h

Therefore, d s= t 5 √ s= 1 s=√5 km/h

3) Find the position vector of B relative to A The formula used to find the position vector is: AB=(

a2−a 1 ) b2−b 1

Position vector: Yacht A= (−1,4 ) Yacht B= ( 2 ,−3) 2− (−1) AB=( ) −3−4 AB=(

3 ) −7

4) Use the result of Question 3 to show that if d is the distance between the yachts at any time t then d 2=25 t 2−220 t + 485 The formula will be used to determine the distance between each Yacht at different times and will be used to prove that the formula is correct: Time Yacht A Yacht B D2 (distance) (km) 0 (-1,4) (2,-3) 277 1 (-2,9) (3,--5) 221 2 (-3,14) (4,-7) 490

3 4

(-4,19) (-5,24)

(5,-9) (6,-12)

865 1417

To find the distance formula the following equation will be used: 2 2 2 ( d =( x 2− x 1 ) +( y 2− y 1 ) ) d 2=( x 2− x 1 ) +( y 2− y 1 ) − 1− t 2+t−( ¿) ¿ ¿ ¿ 2 d =¿ 2 2 2 d =( 2+ t + 1+t ) +(−3 −2 t−4 −5 t ) 2 2 2 d =( 3+2 t ) + (−7−7 t ) d 2=49+98 t + 49 t 2 + 9 + 12 t + 4 t 2 2 2 d =53 t +110 t+58 2

2

5) Find the minimum value of d 2 and the time t when this occurs. Make sure that you justify your answer. d 2=53 t 2+110 t+ 58 is a concave up parabola, this is concluded because the t is squared and the coefficient of it is positive. To find the minimum distance the following formula will be used: −b Minimum= 2a −110 2( 53 ) Minimum=−1.037735849 Hours ∈Theretical Time Minimum=

For the initial position and the position vectors chose for the yachts, they will not cross in real time. This is concluded because the minimum of the equation found is negative meaning that the yachts collided before time started. However, if the paths were extended geographically into the previous time period, before hours started, the paths of the yachts would have collided. Therefore, the minimum value of d2 is at 0 hours as the boats travel further apart in real time. 2 2 d 2=( x 2− x 1 ) +( y 2− y 1 ) 2 2 d 2=( 2−(−1 )) +( −3 −4 ) 2 2 d 2=( 3 ) +( −7) 2 d =9+ 49 d 2=58 km So, the minimum value of d2 is 58 6) Hence find the minimum distance between the yachts and the time this occurs: The minimum distance between the yachts is:

d 2=58 d= √58 So, the minimum value of d is 58 and occurs at 0 hours. 7) Suppose Yacht B wishes to rendezvous with Yacht A at 3.5 hours. If A and B start from their original positions and A maintains its original path, find the velocity vector [a,b] for Yacht B Position of Yacht A at 3.5 hours: To find this value the 3.5 will be substituted into the original equations of t in terms of x and y. x ( t )=−1−t y ( t) =4+ 5t x ( t )=−1−3.5 y ( t) =4+ 5(3.5) x ( t )=−4.5 y ( t) =21.5 Therefore, after 3.5 hours Yacht A will be at (-4.5, 21.5) Now, to find the velocity vector, first the position vector will be found with and then both value will be divided by 3.5 to find the velocity vector. Original (2,-3), destination (-4.5,21.5) AB=(−4.5−2) 21.5+3 AB=(−6.5 ) 24.5

Exercise 3 Create a third scenario where the two yachts collide. Position of Yacht A at 2 hours: To find this value the 2 will be substituted into the original equations of t in terms of x and y. x ( t ) =−1−t y ( t) =4+ 5t x ( t ) =−1−2 y ( t) =4+ 5(2) x ( t ) =−3 y ( t )=14 Therefore, after 2 hours Yacht A will be at (-3,14) Original (2,-3), destination (-3,14) AB=(−3−2 ) 14 +3 AB=(−5) 17 So, the two yachts will collide at 2 hours, the starting position of Yacht A will be (2, -3) and 1 the vector will be ( ) . The initial position of Yacht B will be (-1,4) and the vector will be −2 (−5 ) . Both yachts will collide at the coordinates (-3,14). 17

Exercise 4 Yacht A now starts form an initial position (-4,13) and moves with velocity vector

(−21 )

Yacht B starts at (3,-3), but his rudder jams and the yacht moves in a circle with the parametric equation: π π x ( t ) =3+ cos t , y (t )=−3+sin ( t ) 2 2

( )

Graphs of the equation:

Calculate to determine if the jammed rudder causes a collision. 1) Finding the equation of Yacht A without t:

.

Yacht A= (-4,13) and x ( t )= (− 4+t ) t=x+ 4 y ( t )= (13−2 t ) y ( t )=13 −2 ( x+ 4) y ( t )=5 −2 x

(−21 ) y ( t )=13 −2 x−8

2) Turing the parametric equation into non-parametric equation to find the point of intersection: π x ( t )=3+cos ( t ) 2 π y ( t )=−3+sin( t ) 2 π x−3 =cos t 2 π y +3=sin t 2 2 2 cos θ+ sin θ=1 2 2 ( x−3) + ( y +3) =1 Therefore, the centre of the circle is (3,-3) and the radius is 1. The centre of the circle is also the starting point of Yacht B.

( ) ( )

3) Finding the x coordinates of the points of intersection of Yacht B. The linear equation of Yacht A was substituted: ( x−3 )2 + ( (5−2 x)+3 ) 2=1 2 2 x +6 x +9+ ( 8−2 x ) =1 x 2+6 x +9+64−32 x +4 x 2=1 2 5 x −38 x +73 =1 5 x2 −38 x +72= 0 2 5 x −18 x−20 x +72= 0 5 x(x −4)−18(x−4)=0 (5 x−18)−( x −4 )=0 18 x= ∧4 5 x=3.6∧4 4) Finding the y- coordinates of the points of intersection of Yacht B y ( t )=5 −2 ( 3.6 ) y ( t )=−2.2 Therefore, the first intersection point is (3.6, -2.2) y ( t )=5 −2 ( 4 ) y ( t )=−3

Therefore, the second intersection point is (4,-3) 5) Finding the time at which each intersection occurred: t=x+ 4 t=3.6+4 t=7.6 hours time for first intersection t=x+ 4 t=4 +4 t=8 hours time for second intersection

6) Finding the coordinates of Yacht A after 7.6 hours and 8 hours to determine if the yachts collided: π (7.6) 2 x ( t )=3+cos¿ x ( t )=3.809016992 Therefore, the yachts did not collide as 7.6 hours as the x coordinates do not match for Yacht A and B. π (8) 2 x ( t )=3+cos¿ x ( t )=4 The x coordinates match

π (8) 2 y ( t) =−3+sin ¿ x ( t ) =−3 The y coordinates match

Therefore, the coordinates of Yacht A and B match at 8 hours meaning they collided.

Discussion From the investigation it was found that in Exercise 1, the yachts do not collide while their pathways do collide at 4.4 hours. In Exercise 2 it was found that with the pathways of the yachts collide in theoretical time meaning they collide before time started. Leading to the conclusion that from starting they will move away from each other. The use of parametric equations allowed for the investigator to calculate the time as well as the coordinates of the yachts. Due to parametric equation it is now possible to determine the time in a regular Cartesian plane. Vector were also used and substituted to make parametric equations. Vectors can be used to find parametric equations because they determine the velocity which is a function of time and parametric equations consist of the variable time. Moreover, it was found that with more study of calculus the minimum can now be found using differentiation rather than parametric equations. However, there were a few assumptions and limitations present during the investigation. The main assumption was that the ocean was flat and that the speed and the magnitude of the Yachts were constant. Obviously, this is not true as the ocean is constantly moving, therefore the magnitude would also never be constant. The speed of the yachts are also not

constant as different forces of waves require different speeds to sail. The main limitation of this investigation was that only 2 motions were considered (linear and parametric), while there are many more such as sin and cos. The TI_84 PLUS CE calculator was used, the assumption was made that the values provided by the calculator were correct, this is also a limitation as it could have provided wrong values as well.

Conclusion In conclusion, it was found that the Yachts will not collide with the initial vectors given. What else can I write?...