Stage 2 Specialist Mathematics: Mosquito Vector Analysis

This assignment uses vector laws to represent the position and velocity of two mosquitoes in space. Part 1 establishes parametric equations for the mosquitoes' positions over time and calculates their speeds and the time required to reach specific points. Part 2 focuses on determining the minimum distance between the mosquitoes and the time at which this occurs, using vector subtraction and calculus to find the time at which the derivative of the distance is zero. Part 3 explores how to hit the two mosquitos with a single spray. The assignment demonstrates the application of vector calculus in modeling and analyzing the movement of objects in a three-dimensional space, including calculations of distance, velocity, and optimization of relative positions.

Running head: STAGE 2 SPECIALIST MATHEMATICS
STAGE 2 SPECIALIST MATHEMATICS
Name of the Student
Name of the University
Author Note

1STAGE 2 SPECIALIST MATHEMATICS
Introduction:
In this particular assignment the vector laws are used to represent the position and velocity of
two mosquitoes in space.
Part 1:
Let the initial position of the first mosquito is (a,b,c) = (1,2,3) and the velocity vector = [d,e,f]
= [2,4,3].
Also, let the initial position of the second mosquito is (g,h,l) = (4,5,6) and the velocity vector
= [m,n,p] = [1,3,4].
Hence, the position of the first mosquito at time t is
[1 + 2t, 2 + 4t, 3 + 3t]
And the position of second mosquito at time t is
[4 + t, 5 + 3t, 6 + 4t]
These are the parametric equations of position of mosquitos at time t where unit of position is
in cm.
The velocity of 1st mosquito is v1 = 2i + 4j + 3k
Hence, speed = ||v|| = √22+42+ 32 = 5.385.
Now, let the two positions along the straight line for first and second mosquito are [7, 14, 12]
and [8, 17, 22].
Hence, the time required to reach [7, 14, 12] for first mosquito is
t1 = (final x position – initial x position)/x-component of velocity vector = (7 – 1)/2 = 3
Now, the time required for 2nd mosquito to reach [8, 17, 22] is

2STAGE 2 SPECIALIST MATHEMATICS
t2 = (final x position – initial x position)/x-component of velocity vector = (8-4)/1 = 4
seconds.
Part 2:
Now, at time t the position of first mosquito is P = [1 + 2t, 2 + 4t, 3 + 3t]
Also, at time t the position of second mosquito is Q = [4 + t, 5 + 3t, 6 + 4t]
Hence, the vector ⃗ PQ = (4+t – (1+2t))i + ((5+3t)-(2+4t))j + ((6+4t) – (3+3t))k
= (3 - t)i + (3-t)j + (3+t)k
Now, the distance between the two mosquitos at time t is
||⃗PQ||= √ ( 3−t )2 + ( 3−t )2 + ( 3+t )2
Now, when the two mosquitos are closest or distance between the mosquitos are minimum
when the time derivative is zero for its square term.
Hence,
d
dt (||⃗PQ||2
)=0
 d
dt ( ( 3−t )¿ ¿2+ (3−t )2+ ( 3+t )2 )¿ = 0
 −4 ( 3−t ) +2 ( 3+t )=0
 −¿12 + 4t + 6 + 2t = 0
 6 t=6 => t = 1
Hence, the two mosquitos are closest at time instance t = 1 sec.
The distance between them at t = 1 sec is
D ( t ) ∨¿t =1 = √ ( 3−1 ) 2 + ( 3−1 )
2 + ( 3+1 )
2 ¿ = √4 +4 +16 = √24 = 4.899 cm.

3STAGE 2 SPECIALIST MATHEMATICS
Part 3:
Now, in order to hit the two mosquitos with just one spray the insect repellent position must
be such that it is in the line connected by the positions of two mosquitos at any time instant t.
Hence, the vector equation of the insect repellent is ⃗
PQ =¿ (3 - t)i + (3-t)j + (3+t)k
Or,⃗
Q P=( ( 1+2 t )−( 4+ t))i + ((2+4t) – (5+3t))j + ((3+3t) - (6+4t))k
= (−3+t )i + (-3+t)j + (-3 -t)k