A "Sandy" Free Response 2005FR 5

The tide removes sand from Sandy Point Beach at a rate modeled by the function R, given by

A pumping station adds sand to the beach at a rate modeled by the function S, given by
a) The integral of R(t) within the interval
will output the amount of sand that the tide removed.

PLUG IN f Int (2+5sin(4πt/25) , x, 0, 6)

Answer is approximately 31.815 cubic yards of sand.



b) The total number of cubic yards of sand on the beach would have to include the pumping AND removing of sand AND the initial 2500 cubic yards of sand at t=0, So...

y(t) = f Int( S(t)-R(t) dt ) +2500
= f Int( 15t/(1+3t) - 2+5sin(4πt/25) dt ) + 2500



c)TOTAL amt of sand at t=4.

f Int ( S(4)-R(4) dt), x, 0, 4) +2500
= f Int [ 15(4)/1+3(4) - 2+5sin(4(4)π/25 dt] +2500
= 4.6154 - 6.5241
= -1.908 cubic yards of sand

d) On my graphing calc, i input-ed Y1 as R(t) and Y2 as S(t). Then, i hit 2nd calc and pressed 5 to find the intersect of both graphs. The intersect will give me the minimum.
The two graphs intersect at 5.118, but this is not yet the answer. To find the exact value it must be plugged into our Y(t).

=f Int ( S(5.118)-R(5.118), x, 0, 5.118) + 2500
=20.93254581 - 28.56306324 + 2500
= -7.63051743 + 2500
= 2492.369 cubic yards of sand = MINIMUM!!!


Sorry if im not clear in some areas, its 1 AM!!