2005 FR 5

Rate sand is removed: R(t) = 2+5sin(4πt/25)
Rate sand is added: S(t) = 15t/(1+3t)
Initial: 2500 cubic yards of sand

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

₀∫⁶ 2+5sin(4πt/25)dt = (2t + (125/4π)cos(4πt/25))]⁶₀ ≈ 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.

Y(t) = 2500 + 0 ∫ 6 (2+5sin(4πt/25)) - (15t/(1+3t))dt

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

Rate sand is added - Rate sand is removed, when t=4
15t/(1+3t) - 2+5sin(4πt/25)
15(4)/(1+3(4)) - 2+5sin(4π(4)/25)
4.615 - 6.524
- 1.909 cubic yards of sand per hour

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

Y'(t) = 0 and when t = 0 or 6 are critical points where as a minimum or a maximum may occur.
Y'(t) = S(t) - R(t) = 0
Y'(5.118) ≈ 0
Y(5.118) ≈ 2492.369 cubic yards of sand