FR 5

A)R(t)=rate of which sand is removed. The is how much is being removed. so take the integral. So to see how much the tide take away the first 6 hrs set the integral from 0-6 making the equation:
fnInt(2+5sin(4nt/25),T,0,6) = 31.81593137 simplified it is 31.816 cubic yards

B)To fond the amount of sand at a certain time you must subtract S(t)-R(t)since there is no exact time but a starting point the integral will be from 0-x. At t=0 there is already 2500 sand so it must be added so the equation is:
Y(t)=2500+fnInt (S(t)-R(t), t, 0, x)

C)Y(t)=total number of cubic yards of sand on the beach. Y'(t)=rate at which the total number of sand is changing. so the derivative is Y'(t) is just S(t)-R(t). now we just have to find it at t=4 giving the equation Y'(4) = 15(4)/1+3(4) - 2 + 5sin(4(4)π/25). After the simplifying it equals Y'(4) ≈ -1.909 cubic yards/hour.

D)To find the minimum amount at the equation at part A don't you graph it and find the intersection. If so then the intersection is at (5.1178, 4.6943)so at 5.1178 it equals 4.6943 + 2500=2504.6943 cubic yards.

DONE???