Summer 2018 MATH 115 Quiz 5: Comprehensive Pre-Calculus Solutions

This document presents detailed solutions to a Pre-Calculus quiz (MATH 115 Quiz 5, Summer 2018), covering a range of problems including vector analysis, trigonometry, and wave equations. The solutions demonstrate step-by-step approaches to finding bearings, distances using the law of cosines, vector magnitudes and dot products, vector operations, angles between vectors, and characteristics of wave equations such as amplitude, period, and phase shift. Specifically, the solutions address problems involving airspeed and wind direction, distances between airplanes and airports, vector calculations using i and j notation, and the analysis of cosine and sine functions. This resource is valuable for students seeking to understand and master pre-calculus concepts and problem-solving techniques.

Running Head: PRE-CALCULUS
Pre-calculus
Name of the student:
Name of the university:
Course ID:

1PRE-CALCULUS
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................2
Answer 3..........................................................................................................................................3
Answer 4..........................................................................................................................................3
Answer 5..........................................................................................................................................3
Answer 6..........................................................................................................................................4
Answer 7..........................................................................................................................................4
Answer 8..........................................................................................................................................5
Answer 9..........................................................................................................................................5
Answer 10........................................................................................................................................5

2PRE-CALCULUS
Answer 1.
With an airspeed of 288 mph, a plane is heading towards south.
A wind is blowing at 20 mph from a direction of 580 (ϕ) .
Therefore, tanθ= 288
20 =14.4
Or, θ=tan−1 (14.4)=860
Hence, the bearing of the plane = (θ−ϕ )= ( 860−580 )=280
Answer 2.
An observer watches aero-plane 35 miles away from Airport A. The distance between Airport A
and Airport B is 50 miles. The plane is flying at an angle 2850 from the north from Airport A and
Airport B is at angle of 650 from the north.
Angle between vector 35 miles of direction 2850 of areo-plane and vector 50 miles of direction
650 of airport B = 650 + (900 + (2700 - 2850)) = (650 + 900 - 150) = 1400.
The ‘law of cosines’ indicates that-
d2=352+502−2∗35∗50∗cos 1400
Or, d2=352+502−2∗35∗50∗cos 1400
Or, d2=1225+2500−2∗35∗50∗cos 1400
Or, d2=3750−3500∗(−0.766 )
Or, d2=3750−2681
Or, d2=1069
Or, d=32.7
Hence, plane is 32.7 miles far from Airport B.

3PRE-CALCULUS
Answer 3.⃗
RS = ⟨ 12 ,−8 ⟩
The magnitude of the vector = |RS| = √ 122+(−8)2 = √ 144+64 = √208 = 14.22.
Answer 4.
u =⟨ 12 ,−3 ⟩
v = ⟨ −5 ,7 ⟩
The dot product for the given vectors = u.v = u1v1 + u2v2
Here, u1 = 12, u2 = (-3), v1 = (-5), v2 = 7.
Therefore, u.v = 12(-5) + (-3)7 = -60 -21= (-81).
Answer 5.
u = 8i + j
v = -2i – 6j
w = i – 3j
v – (w – u)
= (-2i – 6j) – ((i -3j) – (8i + j))
= -2i – 6j – (i – 3j -8i – j)
= -2i – 6j – i +3j + 8i + j
= (8i – 2i – i) – (6j -3j – j)
= (5i – 2j)

4PRE-CALCULUS
Answer 6.
u = 5i + j
v = -10i – 4j
3v – 2u
= 3(-10i – 4j) – 2(5i + j)
= -30i -12j -10i – 2j
= -(30i +10i) – (12j +2j)
= -40i – 14j
= -(40i + 14j).
Answer 7.
t =⟨ −2 ,−5 ⟩
s = ⟨ 3 ,6 ⟩
Here, t = (t1, t2) and s = (s1, s2).
The angle between two vectors =
cos θ=¿ (⃗ t ⋅⃗ s )
(‖⃗t ‖⋅‖⃗s‖) ¿
(⃗ t ⋅⃗ s ) = t1s1 + t2s2 = (-2)3 + 5.6 = -6 + 30 = 24.
‖⃗t ‖= √ ( −2 ) 2+ ( −5 ) 2 = √4 +25 = √ 29
‖⃗s‖ = √32 +62 = √ 9+36 = √45 = 3 √5
Hence,

5PRE-CALCULUS
cos θ=¿ 24
( √29 ⋅3 √5 ) ¿ = 8
( √145 ) = 0.66436383883
Or, θ = 48.370=0.844 radian.
Answer 8.
The equation of the wave is-
y=−3 cos (x −Π)
Or, y = -3cos (- ( Π−x ¿)
= -3 cos ( Π−x ¿
= -3 cos ( Π
2 +( Π
2 −x )¿
= -3 (-sin (( Π
2 −x)¿
= 3sin ( Π
2 −x ¿
For the equation, y = A sin (ωt + ϕ), A = amplitude of the sine wave.
The amplitude of the wave is 3.
Answer 9.
y = 2cos (2x + Π/2) = -2sin (2x)
While, y = sin (ωt), then period is given by, T = 2 Π
ω .
The time period (T) = 2 Π
2 =Π (as, ω = 2).
Answer 10.
y = -4 + 3sin (4x – π/6)

6PRE-CALCULUS
For the equation, y = A sin (ωt + ϕ), ϕ = phase shift of the sine wave.
The phase shift of a wave equation = π/6.