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) -_-
Dont forget to put that your answers are approximations! If you were writing this on paper, it would be that "squiggly" equal sign haha.
ReplyDeleteAnd I think that for part d, you should round to the thousandths place to be more accurate, even though this too is still an approximation.
Well what I did for part d, I first plotted the 2 equations of S(t) and R(t) in the "y=" in the calculator. I put R(t) into the "Y1" and the S(t) into "Y2". Next, I just pressed "math+9", then i clicked "vars"-->"Y-vars" --->Function---> "Y2". Then I added the subtraction sign and added "Y1" using the same method. Now I have Y2-Y1, which is equal to S(t)-R(t). Then you just add the conditions ",x,0,5.118)", NOT up to 6 like you put. Then just hit enter and add "2500" to get your answer.! TADA!
i think dianna said all ur mistakes, it seems like u understand the general idea of the question. but what dianna said is important there worth points. yea doing D in paper is hard i tried it. the calculator takes few minutes
ReplyDeleteCalculator is very useful for solving part d
ReplyDelete