2005 FR 5

R(t)= 2+5sin(4piT/25)------removes sand
S(t)= 15t/(1+3t)-----------adds sand

1. The amount of sand that will be removed from Sandy beach can be expressed by the integral of the function R(t) on the interval 0 to 6.
Calculator: fnInt(2+5sin(4piT/25),T,0,6) = 31.81593137 = approx. 31.816
Considering that the units is cubic yards, the answer is 31.816 cubic yards.

2. To find the TOTAL amount of sand at any given time, we must subtract the integral of R(t) from the integral of S(t), since fnInt(R(t)) is the amount of sand being hauled OUT of Sandy Beach and fnInt(S(t)) is the amount of sand being hauled IN to Sandy Beach. However, the integrals of these functions must have a certain limitation, an interval. Since we are looking for the integral from t=0 up to any period of time, we can name this variable x. Therefore, we are finding the integral from 0 to x of (S(t)- R(t)) [calculator:(S(t)- R(t),T,0,x]. Also, since there are 2500 cubic yards of sand at t=0, 2500 must be added to the equation.
Calculator: Y(t)= fnInt((15t/(1+3t)-(2+5sin(4piT/25)),T,0,x)+ 2500
or in simpler terms: Y(t)= fnInt(S(t)-R(t),T,0,x) +2500

3. To find the rate at which the total amount of sand is changing at t=4, we use the derivative of Y(t), Y'(t). Since Y(t)= fnInt(S(t)-R(t),T,0,x) +2500, we use nDeriv( to find its derivative., which is S(t)-R(t). Since we are looking for the instantaneous rate of change at t=4, we plug in 4 into S(t)-R(t).
R(4)-S(4)=4.615384615- 6.524135262=-1.908750647=approx -1.909
Since we are finding a rate, the units is cubic yards per hour. The answer is -1.909 cubic yards per hour.

4. So, now we must find what point in time, in between 0hrs and 6hrs, the amount of sand reaches an all-time low. Since the amount of sand on a beach cannot be negative (that just wouldn't be reasonable), we are looking for the smallest positive output of Y(t) on the interval 0-6. First, we must find the critical points, so we use the equation of the derivative of Y(t), Y'(t), and set it equal to 0. We are looking for when the outputs of Y'(t) switch from negative to positive.
0=S(t)-R(t)=Y'(t)
Using my calculator to graph S(t)-R(t), i use 2nd calc, Zero to find when the graph has a y value equal to 0.
Y'(t)=0 at t=5.1178653=approx. 5.118
When testing critical points, the end points must also be checked. This means the critical points are t=0, t=6, and t=5.118. When looking at the graph of Y'(t), it becomes clear that t=0 and t=6 are both maximums. Before t=5.118, the outputs of y'(t) are negative, while after t=5.118, the outputs of y'(t) are positive. This means t=5.118 is definitely a minimum.
The amount of sand on Sandy Beach reaches a minimum at t=approx.5.118 hours
To find the actual minimum value, we must plug in t=5.118 into Y(t)= fnInt(S(t)-R(t),T,0,x) +2500.
Calculator: fnInt(S(t)-R(t),T,0,5.118) +2500= 2492.369483=approx. 2492.369 cubic yards