So, ran into a bit of a confusing question on some Calculus homework. The question went as follows (I'm quoting the question here):
"Oh no! John's chickens have all escaped their coop. At time $t=0$ there are no more chickens in the coop. The piece-wise linear graph below show the rate $R(t)$, in number of chickens per minute, at which Mrs. Poland is putting the chickens back in their coop during a $9$ minute period."
There were a couple parts of this question that I was able to figure out, like "How many chickens are put into the coop from time $t=0$ to $t=9$, which I figured out to simply just be the integral of this function from $0$ to $9$ (which was $26$), but the part where I'm getting lost is the second and third parts of the question. The second part is as follows:
I tried to find the anti-derivative, but it doesn't seem to have one? So I found the definite integral of this function from $0$ to $9$ and it gave me $18.12$ but I'm not too sure if that's right/the actual amount of chickens that exited during that amount of time. I'm also confused on how I'd go about finding the amount of chickens in the coop at one moment, taking into account the function $E(t)$ and $R(t)$. Any help would be appreciated on being pointed in the right direction here and sorry for the long question, just needed a bit of explaining to make my problem clear. Oh and one more question to add onto there, how would I go about finding whether or not at one moment the amount of chickens in the coop is increasing/decreasing?
Best Answer
The amount of chickens exited in the time interval is indeed $\lfloor \int_0^9 E(t)dt\rfloor =18$. At any time $t$, the amount of chickens in the coop will be the amount entered minus the amount exited (as we start with $0$ chickens inside the coop), i.e. $$\int_0^t \left(R(t) -E(t) \right) dt$$ Also, note that the 'net' rate at any moment is $R(t)-E(t)$. If this is positive, the amount of chickens is increasing, and decreasing otherwise.