It takes the skytrain two minutes to travel from Brentwood Station to Holdom Station. Let t be the time (in minutes) after 12:00 noon. Suppose the train leaves Brentwood Station at 11:59 am (so t=-1), and arrives at 12:01 pm (so t=1). The speed of the train at time t is exactly f(t) kilometers per hour, where
f(t)=(120/1+t^2)-60, for -1 ≤ t ≤ 1.
a) What is the maximum speed of the train when traveling between the stations?
b) Assuming that the track is straight, what is the distance between the two stations? ( Hint: The speed of the train at time t in kilometers per minutes is the function g(t)=f(t)/60).
c) What is the average speed, vave, of the train when traveling between the stations? Express your answer in kilometers per hour.
To find the maximum speed of the train when traveling between the stations, we need to determine the highest point on the graph of f(t).
a) To find the maximum, we need to find the critical points. These occur when the derivative of f(t) is equal to zero or does not exist. Let's find the derivative and solve for when it equals zero:
f'(t) = (-240t)/(1 + t^2)^2
Setting f'(t) = 0:
(-240t)/(1 + t^2)^2 = 0
This expression is equal to zero when the numerator is equal to zero, so we must have t = 0.
To confirm if this is a maximum value, we can check the second derivative.
f''(t) = (240(3t^2 - 1))/(1 + t^2)^3
Plugging in t = 0:
f''(0) = (240(3(0)^2 - 1))/(1 + (0)^2)^3 = -240 < 0
Since f''(0) is negative, we can conclude that t = 0 corresponds to a maximum value.
Now, we plug t = 0 into f(t) to find the maximum speed:
f(0) = (120/1 + (0)^2) - 60 = 60 km/h
Therefore, the maximum speed of the train when traveling between the stations is 60 km/h.
b) To find the distance between the two stations, we need to calculate the area under the graph of f(t) over the interval [-1, 1]. Since the speed is given in km/h and time is given in minutes, we need to convert the speed function f(t) into g(t) in km/minutes:
g(t) = f(t) / 60 = (1/60)(120/1 + t^2) - 1
To find the distance, we integrate g(t) from -1 to 1:
distance = ∫[from -1 to 1] (1/60)(120/1 + t^2) - 1 dt
= ∫[from -1 to 1] (120/60)(1/(1 + t^2)) - (1/60) dt
= (∫[from -1 to 1] 2/(1 + t^2) dt) - (1/60) ∫[from -1 to 1] dt
Evaluating the integrals:
distance = [2arctan(t)] [from -1 to 1] - (1/60)[t] [from -1 to 1]
= 2(arctan(1) - arctan(-1)) - (1/60)(1 + 1)
= 2(arctan(1) + arctan(1)) - (1/30)
= 2(π/4 + π/4) - (1/30)
= (π/2) - (1/30)
Therefore, the distance between the two stations is (π/2) - (1/30) kilometers.
c) The average speed, vave, is given by the total distance divided by the total time taken:
vave = distance / total time taken
The total time taken from 11:59 am to 12:01 pm is 2 minutes, so the average speed is:
vave = [(π/2) - (1/30)] / 2
Simplifying:
vave = π/4 - 1/60
Therefore, the average speed of the train when traveling between the stations is (π/4) - (1/60) kilometers per hour.
(a) find where df/dt = 0
or, just note that f(t) is an even function.
(b) since f(t) is the velocity, the distance is
∫[-1/60,1/60] f(t) dt
(c) avg speed is distance/time