Free Response

(Note: Because I don't want to write the equations over and over again, I'll refer to them as R(t), S(t), and Y(t).)

1. R(t) is the rate at which sand is removed, so how much sand was actually removed is the integral of R(t) from t=0 to t=6. Numerically, fnInt (R(t), t, 0, 6)=31.816 cubic yards (approximately).

2. The amount of sand on the beach at time t is the initial 2500 cubic yards plus the total additional sand that is added. The total amount of sand added is the amount of sand added (the integral of S(t) from t=0 to t=x) minus the amount of sand removed (the integral of R(t) from t=0 to t=x). Mathematically, it looks like Y(t)=2500+fnInt (S(t)-R(t), t, 0, x).

3. The rate at which the total amount of sand on the beach changes is the instantaneous rate of change of Y(t) (its derivative). So, at any time time, the are is nDeriv (Y(t))=nDeriv (2500+fnInt (S(t)-R(t), t, 0, x) )=S(t)-R(t). At time t=4, the rate of change of sand is S(4)-R(4)=-1.909 cubic yards/hours.

4. From above, the rate of change of sand is Y '(t)=S(t)-R(t). As with other functions, the first thing we do is find the critical points. Found on the calculator, the point is t=5.118, as that is when Y '(t)=0, so the critical points are t=0, t=5.118, and t=6. If you look on the calculator, you can see that the sign of Y '(t) is negative until t=5.118, where it becomes positive, and remains positive when it it reaches t=6, so the only possibility for a minimum is t=5.118, as the endpoints, t=0 and t=6 would both be maximums. Therefore, the minimum value of sand is Y(5.118)=2500+fnInt (S(t)-R(t), t, 0, 5.118)=2492.369 cubic yards (approximately).

Hmm....I've looked through a couple of other people's blogs, and I'm seeing a number of answers for the last two parts. If anyone has questions on what I did, or felt I did something incorrectly, please comment and let me know.

(P.S. If anyone is wondering how I input my functions into the calculator, I made Y1=S(t), Y2=R(t), and Y3=Y1-Y2.)