9514 1404 393
Answer:
a. h(t) = -20t +400
b. θ(t) = arctan(2 -2/15t); dθ/dt = -30/(1125 -120t +4t^2)
c. 15 seconds; 100 ft
Step-by-step explanation:
a. The initial height of the elevator is 400 ft. The rate of change of height is -20 ft/s, so the height equation can be ...
h(t) = -20t +400
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b. The tangent of the angle above the line of sight is "opposite"/"adjacent":
tan(θ) = (h(t) -100)/(150) = -2/15t +2
θ(t) = arctan(2 -2/15t) . . . . radians
The derivative of the angle function is ...
dθ/dt = 1/(1+(2 -2/15t)^2)(-2/15)
dθ/dt = -30/(1125 -120t +4t^2)
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c. The value of dθ/dt will have a peak where the denominator has a minimum, at t = -(-120)/2(4)) = 15. (The quadratic vertex coordinate is t=-b/(2a).)
The elevator appears to be moving fastest at t=15 seconds.
The height at that time is ...
h(15) = 400 -20(15) = 100
The elevator appears to be moving fastest when it is at eye level, 100 ft above the ground.
