When at rest an average adult breathes in and out every 4s. the average minimum amount of air the lungs is 0.08L and the average maximum amount of air in the lungs is 0.82L. Assume the lungs have a minimum amount of air at t=0 where t is time in seconds.

Question

a) write a function to model the amount of air in the lungs.
b) determine the amount of air in the lungs at 5.5s

Answer 1

Are you assuming the relation follows a trig function, either a sine or cosine ?

Answer 2

cosine

Answer 3

also how do i state all the possible values for angle theta, given that (theta)=-3/8and 0<theta<3pi

Answer 4

one question at a time.

period of your cosine is 4 s
so 2π/k = 4
4k = 2π
k = π/2

so let's start with
A = cos (πt/2)

the min is .08, and the max is .82,
thus a = (.82-.08)/2 = .37

so far: A = .37 cos (πt/2)
the min of that is .37, but we want the min to be .08
so we have to raise our curve by .45

so we have A = .37cos(πt/2) + .45
we want that min of .08 to occur when t = 0
So we have to shift our cosine curve horizontally

and it will look like :
A = .37 cos ( (π/2)(t + d) ) + .45

solving for d when t=0 , A = .08
.37 cos ( (π/2)(0 + d) ) + .45 = .08
cos ( (π/2)(0 + d) ) = -1
Now we know that cos π = 1
thus (π/2)(0 + d) = π
(1/2)d = 1
d = 2

final equation:

A = .37 cos ( (π/2)(t + 2) ) + .45

check:
when t = 0 , we should get .08
A = .37 cos(π) + .45 = -.37-.45 = .08 , good!
when t = 2, we should get our max
A = .37 cos((π/2)(4) + .45 = .82 , good!

looks like equation is good.

so when t = 5.5
A = .37 cos(π/2)(7.5) + .45
= .7116 L

Answer 5

" ...all the possible values for angle Ø, given that Ø =-3/8and 0<theta<3pi "

this makes no sense, did you skip the trig function in front of Ø, such as

sinØ = -3/8 ??

Answer 6

sorry it tan(theta) infront