2005 FR 5

Both R(t) and S(t) have units of cubic yards per hour and t is measured in hours for . At time t=0, the beach contains 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.

integral of R(t), which is integral (2+5sin(4 pie t/ 25)from 0,6, and the amount of sand removed comes out to be 31.816. the reason for finding the integral of R(t) is because it outputs how much sand was removed from 0,6, which is 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) is the rate of how much sand is being removed and S(t) is how much sand is being added, and 2500 is how much sand there is at the start, you would have to subtract how much is removed from how much is added, then add it to 2500 since that's how much sand there is, Y(t)= 2500 + (S(t)- R(t)) or Y(t)= 2500 + ((15t/1+3t) - (2+5sin((4 pie t)/25)

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

since we want to find the total rate of how much the sand is changing at t=4, then we would plug in 4 into Y(4)=((15(4))/1+3(4)) - (2+5sin((4 pie (4))/25) without 2500, and we would get the total rate of how much sand is changing because we subtract how much sand is being removed which is 6.524 from how much sand is being added which is 4.615, and get -1.909, but since we want the total, doesn't matter if its negative or positive, its 1.909 cubic yards of sand


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

At t=5.1, because that's where it intercepts and it looks that both equations are Even, the rate is nearly the same. the minimum value would be Y(5.1)= 2500 +((15(5.1))/(1+3(5.1))- (2 + 5sin((4 pie (5.1)/25) which equals 2499.961 cubic yards of sand which is the minimum value.