As a Ferris wheel rotates, the height, h, meters, above the ground of one of its cars, can be modeled by the function h(t)=10sin {pi (t-o.5)} + 15, where t is time, in minutes. At what value of time is the instantaneous rate of change the greatest positive value ?

Question

1. T= 0.5 minutes
2. T= 1 minutes
3. T= 0 minutes
4. T= 1.5 minutes

Answer 1

sinθ has its maximum rate of change when θ=0 (that is, when cosθ = 1)
so, you want π(t-0.5) = 0
t = 0.5

Answer 2

The instantaneous rate of change is represented by the first derivative.
To find where that instantaneous rate of change is a maximum we set the 2nd derivative equal to zero.
h(t)=10sin {pi (t-o.5)} + 15
h(t)= 10sin (π(t - 0.5)) + 15

h ' (t) = 10cos(π(t - 0.5)*π
= 10πcos(π(t - 0.5)

h '' (t) = -10πsin(π(t - 0.5)*π
= -10π^2 sin(π(t - 0.5)
-10π^2 sin(π(t - 0.5) = 0
sin(π(t - 0.5) = 0
π(t - 0.5) = 0 or π(t - 0.5) = π/2
t - 0.5 = 0 or t-.5 = 1/2
t = 0.5 OR t = 1

at t = .5, h'(.5) = 10πcos(π(.5 - 0.5) = 10π
at t = 1, h'(1) = 10πcos(π(1 - 0.5) = 0

which value of t gave us the max and which gave us the minimum velocity?