A pumping station adds sand to the beach at a rate modeled by the function S, given by
Both R(t) and S(t) have units of cubic yards per hour and t is measured in hours for 0 greaterthanorequalsto x lessthanorequalsto 6. 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.
fnInt(2+5sin(4piet/25),x,0,6)= 31.816 cubic yards
---We use function R(t) because it just asking about the sand that is being removed.
(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
Y(t)= 2500+fnInt( S(t)-R(t), x,0,6)
-----It's the integral of function S(t) minus R(t) from 0 to 6 because we want to know how many cubic yards are going to be plus the ones that are already at time t=0
(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
Plug in t=4 into
S(t) - R(t)=
4.615 - 6.524 = -1.909 cubic yards/hour
----- We plug t=4 into the rate functions so we know the rate at which it's changing at that time.
(d) For 0 greaterthanorequalsto x lessthanorequalsto 6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.
------We set S(t) - R(t) = 0 , so we can find the critical points. These rate functions are the derivative of the volume of sand on the beach.
The critical points are
X=0
X= 6
X= 5.12, here the functions change sign from negative to positive so there is a minimum.
So at time t= 5.12 there is a minimum amount of sand on the beach.
To know exactly how much it's there at that time, we have to plug in t= 5.12 into Y(t) likethis
Y(t)= 2500+fnInt( S(5.12)-R(5.12), x,0,6)
* I couldn't find it cuz i have to know the exact antiderivative for the integral and it's hard (ithink,,__ help here pleasee) -_-
