Bradford T. answered 01/26/21
Retired Engineer / Upper level math instructor
h(t) = Asin(B(t+C)) + D Assuming no dampening
D is the midpoint = (50+10)/2 = 30
The range of sin() is ±1, so
h(3) = A(1)+30 = 50 --> A = 20
h(7) = 20(-1)+30 = 10 checks out
B = 2π/Period and Period = 2(7-3)=8 --> B = 2π/8 = π/4 (because half cycle is from max to min)
For minimum and maximum
3π/4+Cπ/4 = π/2 --> C=-1
7π/4+Cπ/4 = 3π/2 --> C=-1
h(t) = 20sin(π/4(t-1)) + 30
b)
h(6) = 20sin(5π/4) + 30 = 20(-√2/2) + 30 = 15.86 inches
c)
20sin(π/4(t-1)) + 30 = 23
Solve for t
20sin(π/4(t-1)) = -7
π/4(t-1) = sin-1(-7/20) = -0.1138π
t = 4(-0.1138) +1 = 0.5447 seconds First time
Second time:
20sin(π/4( 0.5447-1-x)) + 30 = 23
x = 5 seconds later, so t = 5.447 seconds