Order id 945826 Calculus 1.

Order id 945826 Calculus
1. As from the figure:
s= 20
cos θ
I = sin ( θ )
20
cos (θ )
¿ sin ( θ ) cos ( θ )
20 sin ( θ ) cos ( θ ) =1
2 sin ( 2θ )∴ I =sin ( 2 θ )
40
For maximum intensity dI
dθ =0 ⟹ dI
dθ = cos ( 2θ )
20 =0cos ( 2 θ ) =0
θ= π
4 +πn∨θ=3 π
4 + πn for n=0 ,1 , 2 ,3 …for n=0 ,θ= π
4 ∨ 3 π
4 Thus θ= π
4 rad ∨45 °
tan ( θ ) = Height
20 Height=20 tan ( 45° ) ¿ 20 ft
2. A cylindrical soda can is being designed to hold a volume of approximately 21.7 in3.
a) Minimize surface area
Since we know the volume we can solve for h in terms of r
V =π r2 h∴ h= 21.7
π r2 S=2 π r2 +2 πrhS=2 (π r 2+πr ( 21.7
π r2 ))=2 π r2 +43.4 r−1
For minimum r, dS
dr =0 dS
dr =4 πr− 43.4
r 2 =0 4 π r3 =43.4r =3
√ 43.4
4 π =1.511564285∈¿
∴ h= 21.7
π (1.511564)2 =3.023128569∈¿
b) The assumption that my read to that error are:
i) The cylindrical soda is a closed cylinder
ii) The surface area function is differentiable for all values of r