Calculus

Running head: CALCULUS 1
Calculus
Name
Institution

CALCULUS 2
Calculus
Question 1
Part i
θ= 3
4 =0.75 rad
Area ABCD=30 cm2
Area sector= θ
360 π r2= θ
2 π π r2 ( ¿ radians ) = θ
2 r 2(¿ 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