Pre-calculus: Solved Assignments and Essays

This page contains solved assignments and essays related to pre-calculus. It includes answers to problems related to vectors, dot product, wave equations, and more. The content is suitable for students studying pre-calculus in college or university.

Running Head: PRE-CALCULUS
Pre-calculus
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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.