2005 Free Response 5

(a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.

R(t) represents the rate of sand being removed. The integral of R(t) will represent the amount of sand being removed.

6
∫R(t)= 2 + 5sin(4πt/25) dx
0

fnInt(2+5sin(4πt/25),t,0,6) ≈ 31.816 cubic yards of sand

(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.

Since R(t) represents the rate of sand being removed, subtract R(t) from S(t), which represents the rate of sand being pumped.
*Find the integral of both to find just the amount of sand.
**Add 2500 because that was the amount of sand originaly there at t=0.

Y(t)= 2500 + fnInt( S(t)-R(t) ),x,0,6

(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.

Y(t) represents the total number of cubic yards of sand on the beach (cubic yards), so the derivative of Y(t) will represent the rate at which the total number of sand is changing (cubic yards/hour)

Y'(t) is just S(t)-R(t).

Y'(4) = 15(4)/1+3(4) - 2 + 5sin(4(4)π/25)
Y'(4) = 60/13 - 2 + sin(16π/25)
Y'(4) = 4.615 - 6.524
Y'(4) ≈ -1.909 cubic yards/hour

(d) For 0≤t≤6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.

From what I remember from class, you:
Graph both S(t) and R(t). Use your graphing calculator for this :p

Then, find the point where they intersect.
*The point of intersection shows where the amount of sand being removed is the same as the sand being pumped at 0≤t≤6.

The point of intersection is at (5.1178, 4.6943)
*4.6943 is the amount of sand at t=5.1178
**Add 2500 because that was the amount of sand originaly there at t=0.

So, the grand total cubic yeards of sand is 2504.6943