1. A car travels 20 mi/hr for 2 hours and then 30 mi/hr for 3 hours.
a) For 0<t<2 let c(t) be the distance the car has traveled in miles from time 0 to time t, where t is measured in hours. Find c(t).
b) Extend your function c(t) from part (a) to the time period 2 <= t <= 5. You will have a piecewise continuous function.
c) For the function c(t) in the previous problem, find c'(1) and c'(3).
a) c = 20 t , 0<t<2
b) c = 40 + 30 (t-2) , 2<t<5
c) Use the first equation to get c(1) and the second equation to get c(3)
They call it c', but it's the same thing as c at the value of t selected.
a) To find the function c(t) for 0<t<2, we can break down the problem into two parts: the distance covered during the first 2 hours traveling at 20 mi/hr, and the distance covered during the remaining time.
The distance covered during the first 2 hours is given by:
d1 = 20 * 2 = 40 miles
To find the distance covered during the remaining time, we need to calculate the time traveled after the first 2 hours. Let's call this time t_r:
t_r = t - 2
We know that the car travels at 30 mi/hr during this time period, so the distance covered during t_r is:
d2 = 30 * t_r = 30 * (t - 2) = 30t - 60 miles
Therefore, the total distance c(t) covered from time 0 to time t is the sum of d1 and d2:
c(t) = d1 + d2 = 40 + (30t - 60) = 30t - 20 miles
b) To extend the function c(t) from 2 <= t <= 5, we need to consider the distance covered during this time period. Since the car is traveling at 30 mi/hr, we can use the same formula as in part (a) for t_r:
d3 = 30 * t_r = 30 * (t - 2) = 30t - 60 miles
Therefore, the function c(t) for 2 <= t <= 5 is given by:
c(t) = d3 = 30t - 60 miles
Combining the results from parts (a) and (b), we have the following piecewise continuous function for c(t):
c(t) = 30t - 20 for 0 < t < 2
c(t) = 30t - 60 for 2 <= t <= 5
c) To find c'(1) and c'(3), we need to differentiate the function c(t) with respect to t.
For the function c(t) = 30t - 20, when 0 < t < 2, the derivative c'(t) is simply the derivative of 30t - 20:
c'(t) = 30
So, c'(1) = 30.
For the function c(t) = 30t - 60, when 2 <= t <= 5, the derivative c'(t) is again the derivative of 30t - 60:
c'(t) = 30
Therefore, c'(3) = 30.
To summarize:
c'(1) = 30
c'(3) = 30