Ferris Wheel Rate

Question

The height above the ground in feet of a certain seat on a ferris wheel is given by the following equation:

 

h(t)= 25+20cos(π/4)*(t-2) (that is a pi symbol over the 4) where t represents the number of seconds since the ferris wheel started. The greatest rate of chage in heigh above the ground happens at 4 seconds. What is the rate? 

Answer 1

I am assuming your equation is h(t) = 25 + 20 cos(π/4*(t-2)). If it is h(t) = 25 + 20 cos(π/4)*(t-2), then the rate of change in height is a constant. For the first equation, the rate of change is dh/dt = -5πsin(π/4*(t-2)) .  You can take another derivative, set it to zero and find that t=4, 8,12, ... seconds are all extrema of the function (these actually happen to be minima, but the question may be referring to the magnitude of the rate of change of height). To answer your question, plug t=4 in the last equation to get dh/dt(t=4) = -5πsin(π/2) = -5π. The magnitude of this change in the rate of height is 5π.