# Thread: (A o r)(t) results too high

1. ## (A o r)(t) results too high

The problem is to find the area of the circular region covered by emitted pollutant that cannot rise vertically.
The pollutant begins emitting at 8:00am or t = 0 represents 8:00 am.
The radius of the circle of pollutants is r(t) = 2t miles.
A(r) = pi or 3.14r^2 that represents the area of a circle of radius r.

Find what the circular region would be by noon. I made t=4
then plugged it into the function.
A[r(t)] so I came up with
r(4) = 2(4) = 8 miles
then plug the 8 into the A function to be 3.14 * 8^2 = 200.96 miles. That can't be right, it is too many miles of coverage.
Where did I go wrong in my thinking?

2. ## Nothing wrong dear

There is nothing wrong with the computation. Observe that you are talking about area here, not length.

Lets perform a small approximation

What would be the area of a square of side 16? It will be 256 square units. Your circle is entirely inside that square (that is why I chose 16). So it should have lesser area. And ya it does!

The moral is to compare the suspicious numbers by approximating it with a reliable way of computation.

3. ## What about a domain for this?

Thank you I will try to do approximation more to test my work.

Can you tell me something about domains or input values.
It not a question on this assignment, but I am also studying domain and range on other problems.

Would the domain be [0,9] assuming the business closes at 5 PM?

4. Would the domain be [0,9] assuming the business closes at 5 PM?
You mean the "pollutant emitting" business?
If so and the function is r(t) = 2t, then you are right. Range would be 2[0,9] = [0,18].