Modeling trigonometric functions

Question

A small windmill has its center 6 m above the ground and blades 2 m in length. In a steady wind, the tip of one blade makes a complete rotation in 12 seconds. During one minute of rotation, how long is the tip of one blade 7 m or more above the ground?

Answer 1

6 meters above ground=minimum height

2 meter long blades; half max -min height=A= (10-6)/2 = 2

one revolution every 12 seconds= period=2pi/B

general formula is

h(t) = Asin(Bt+C) + D

where D = midline =6+2 = 8=half way between max and min height, half way between 6 and 10

C= phase shift if any

B=2pi(frequency)t = 2pit/period=2pi/t12=pit/6,

A=Amplitude= half (max-min height)= (10-8)/2 =4/2=2 meters = blade length

plug in the values for A,B, D and t=0, to solve for C

at time t=0, h(0) = minimum height = 6 meters

h(0) = -2 + 8 =6= Asin(2pit/period +C) +8

=2sin(pit/6+C)+8 = -2

sin(pit/6 +C)=-1

pit/6+C = arcsin-1= 3pi/2

C= 3pi(0)/2-3pi/2 = 3pi/2= phase shift

h(t)= 2sin(pit/6+3pi/2)+8

check the answer. see if h(0) = minimum height = 6 meters

h(0) =2 =2sin(pi(0)/6+3pi/2) = 2sin(3pi/2)+8 = 2(-1)+8 =6

how long is it 7 meters or more above ground

it starts at 6 m, goes to 8 m in 3 seconds, 10 meters in 6 seconds, back to 8 meters in 9 seconds, back to 6 meters in 12 seconds, from 8 to 8 meters and above 8 is 6 seconds. from 7 to 7 and above is about 7 or 10 seconds longer than 8 to 8, but shorter than 6 to 6, longer than 6 sec, but shorter than 12 seconds

h(t) = 7 = 2sin(pit/6+3pi/2) +8

2sin(pit/6 +3pi/2) =7-8 =-1

sin(pit/6+3pi/2) = -1/2

pit/6+3pi/2= arcsin-.5 = 7pi/6 or -pi/6

pit/6 = 7pi/6-3pi/2 =-pi/3 or -pi/6-3pi/2 =-5pi/3

t = (6/pi)(-pi/3) or (6/pi)(-5pi/3) =-2 or -10

from -2 to -10 = 8 seconds

or you could have done the problem with cosines

only difference would be C, the phase shift would = pi instead of 3pi/2

h(t)= 2cos(pit/6 +pi) + 8