Suppose that the number of cars, C, on 1st Avenue in a city over a period of tim

Question

Suppose that the number of cars, C, on 1st Avenue in a city over a period of time t, in days, is graphed on a rectangular coordinate syatem where time is on the horizontal axis. Further suppose that the number of cars driven on 1st Avenue can be modeled by an exponental function,c=p*at(c=p*a^t) where p is the number of cars on the road on the first day recorded and it is the number of days. Decide how you would prefer to commute to work each day. First choose a value for "p" between 100 and 200; this is the initial number of cars on the road. Second, chose the value for "a"; this is the growth factor - you can choose"a" to be any number between 0 and 1 "or" choose "a" to be any number greater than 1. Insert the chosen values for "p" and "a" into the formula listed above. Use the formula to find the number of cars, C on 1st Avenue, at any three values of time,t, in days that you want. Show your calculations and put units on ypur final answer.

Explanation / Answer

For both problems, let's assume P = 150. On day 1, t will equal 0, so either way there will be 150 cars on the road because a0 = 1.

First, let's look at what happens if we choose a value of a that is between 0 and 1. I will choose 0.5

We can express this equation as C(t) = 150(0.5)t

On day two, the number of cars will be halved. Remember that day two is one day since day 1, so t = 1

C(1) = 150(0.5)1 = 150(0.5) = 75 cars

Each day, the number of cars will be half of the previous day. So, on day 4 (t=3)

C(t) = 150(0.5)3 = 18.75 cars, which we will round to 19 cars.

On day 10 (t=9), we probably won't have any cars on the road:

C(9) = 150(0.5)9 = 0.29 cars

You can't have 0.29 of a car, so in practice C would equal 0. The graph of the function would be a curve from 150 and would approach the x-axis curving downward without ever touching it. It would be concave up ( like a slide). This is called exponential decay. I would provide a diagram, but Cramster is giving me an error, sorry.

Now, if we used a number greater than 1 for p, such as 2, then every day the number of cars would be twice as much as the previous and the equation would be:

C(t) = 150(2)t

So, after 3 days (t=2),

C(2) = 150(2)2 = 150(4) = 600

This graph would also be concave up, but it would curve upward. This is called exponential growth.

To answer the question:

If p were between 1 and 0, I would definitely drive. However, if P were greater than 1, it would probably be best to just walk after a few days because 1st Avenue would be crammed very quickly! :)

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