Calculus Assignment: Solving Equations, Arcs, and Graph Analysis

This Calculus assignment provides detailed solutions to several problems involving trigonometric functions and equations. The first question focuses on calculating the area of a sector and the difference in arc lengths. The second question involves solving trigonometric equations, including finding solutions for θ within specified ranges. The assignment also includes graphical analysis, where the student plots y=cosx and y=sin2x on the same axes and determines solutions for the equation sin2x - cosx = 1 by identifying points where the graphed function equals 1. The solutions for x are provided in degrees. Desklib offers a wide range of similar solved assignments and study resources for students.

Running head: CALCULUS 1
Calculus
Name
Institution

CALCULUS 2
Calculus
Question 1
Part i
θ= 3
4 =0.75rad
Area ABCD=30 cm2
Area sector= θ
360 π r2 = θ
2 π π r2 ( ¿ radians ) =θ
2 r2 (¿ radians)
Area ABCD= 0.75
2 ( ( 3 k ) 2− ( 2 k ) 2 ) =30 cm2
( ( 3 k ) 2− ( 2 k ) 2 )=30× 2
0.75 =8 0
9 k 2−4 k2=8 0
5 k2=80 , k2=1 6
k = √16=4 cm
Part ii
Length of arc= θ
2 D=θ
2 ×2 r=θr
Length of arc AB=0.75 ( 2 K )=0.75 ( 2 × 4 )=6 cm
Length of arc CD=0.75 ( 3 K ) =0.75 ( 3× 4 )=9 cm
Difference between the length of the two arcs¿ ( 9−6 ) cm=3 cm
Question 2
Part a

CALCULUS 3
(1−cosθ)(1+cosθ)
tanθ =( 1−cos2 θ)
tanθ
But, tanθ=sinθ /cos θ
(1−cos2 θ)
tanθ = (1−cos2 θ)
(sinθ /cos θ)= cos θ(1−cos2 θ)
(sinθ)
But, cos2 θ+sin2 θ=1 ,1−cos2 θ=sin2 θ
cos θ(1−cos2 θ)
(sinθ) =cos θ(sin2 θ)
(sinθ ) =sinθ cos θ
Part b(i)
3 sinθ =2+ 1
sinθ
We multiply the equation by sinθ to obtain
3 sin2 θ=2 sinθ+1
3 sin2 θ−2 sinθ−1=0 … .. equation 1
Let sinθ=m
Then equation 1 becomes
3 m2−2 m−1=0
3 m2−3 m+ m−1=0
3 m ( m−1 )+1 ( m−1 )=0
( 3 m+1 ) ( m−1 )=0

CALCULUS 4
when ( 3 m+1 )=0 , m=−1
3
when , ( m−1 )=0 ,m=1
Therefore, sinθ=1∨sinθ=−1
3
when sinθ=1 ,θ=sin−1 1=90 °
When sinθ=−1
3 , θ=sin−1
(−1
3 )=199.47 ° ( 3 rd quadrant ) 0 r θ=340.53° (4 th quadrant)
Hence, θ=90 ° ,199.47 ° ,340.53 °
Part b(ii)
sin 2 θ=0.5
2 θ=sin−1 0.5=30 ° (1 st quadrant )∨ ( 180−30 )=150 ° ( second quadrant)
2 θ=30 ° , 150°
θ=30 °
2 , 150°
2
θ=15° , 75 °
Question 3
Part a
x 0 30 60 90 120 150 18
0
210 240 270 300 330 360
cosx 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1
Sin2x 0 0.87 0.87 0 -0.87 -0.87 0 0.87 0.87 0 -0.87 -0.87 0
A graph of y=cosx∧ y =sin 2 x on the same axes is shown in figure 1

CALCULUS 5
Figure 1: A graph of y=cosx∧ y =sin 2 x
Part b
Solutions for sin 2 x−cosx=1
To get the solutions, we draw a graph of ( sin 2 x−cosx ) .The solutions are the points where
( sin 2 x−cosx )=1as shown in figure 2.
x 0 30 60 90 120 150 18
0
210 240 270 300 330 360
cosx 1 0.8
7
0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1
Sin2x 0 0.8
7
0.87 0 -0.87 -0.87 0 0.87 0.87 0 -0.87 -0.87 0
Sin2x-cosx -1 0 0.37 0 -0.37 0 1 1.74 1.37 0 -1.37 -1.74 -1

CALCULUS 6
Figure 2: A graph of( sin 2 x−cosx )
As we can see from figure 2, there are two points where ( sin 2 x−cosx )=1 hence the equation has two
solutions for 0 ≤ x ≤ 360
Part c
The solutions of ( sin 2 x−cosx )=1 in the interval 0 ≤ x ≤ 360 are x=180 °∧x=249. 63 °