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Calculus/calvulus
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Question
The average speed of a vehicle on a stretch of Route 134 between 6 a.m. and 12 p.m. on a typical weekday is approximated by the following function, f (t), which is measured in miles per hour, and t is measured in hours, with t = 0 corresponding to 6 a.m.
F(t) = 10t - 20√(t) + 30 (0 ≤ t ≤ 6)
a) At what time of the morning commute is the traffic moving at the slowest rate? (Note: Use standard hour: minute clock format when writing your final answer.)
________; ________a.m.
b) What is the average speed of a vehicle at that time?
In most questions I have seen, the capital letter function is the integral of the lower case letter function. Since this makes F(t) to be the integral of f(t), and f(t) is speed, that means that F(t) is distance. The derivative of F(t) is speed, so f(t) = 10 - 10/√t.
Setting this to 0 gives 10 - 10/√t = 0 => 10 = 10/√t, so √t = 1, so t = ±1.
Since the time is positive, t = 1. Since the time is t+6, that means time is 7am.
That is the same as 7:00am.
At t = 0, F(0) = 30.
At t = 6, F(6) = 10*6 - 20√6 + 30 = 60 - 20√6 + 30, which rounds to 41.
If digits are kept to more places, it is actually about 37 seconds after 7am,
but 7am is close enough.
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Scotto
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