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
. 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(R(t),t,0,6)
fnInt(2+5sin(4pi t/25),t,0,6)=approximately 31.815 or 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)= fnInt(S(t)-R(t)dt)+2500
Y(t)=fnInt((15t/1+3t)-(2+5sin(4pi t/25)),t,0,x)+2500
(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
So you need the rate of change of the equation in part b, at t=4, since the equation is the total amount of sand on the beach, and you need to know the rate at which this is changing.
If you dont want to find the derivative, just RECALL that...
When told the given equations, they are already in the units cubic yards per hour, which is the rate at which it is changing., therefore you just need to subtract the rate at which the tide removes sand from the rate at which the tide adds sand. (S(t)-R(t))
But if you want to take the derivative of Y(t)...
Since you know that Y(t) is the antiderivative, that means that the derivative would just equal the original equation "f(x)", which in this case is S(t)-R(t). <--- It is ALREADY the derivative (Rate of Change)
same thing if you do....
Y'(4)=nDeriv(fnInt(S(t)-R(t))+2500... The integral (fnInt cancels out with the nDeriv, because if you take the antiderivative and then the derivative, then you are left with S(t)-R(t), because of the Fundamental Theorem of Calculus Part 1. Taking the derivative of 2500 is just 0, since it is a constant.
so S(4)-R(4)= 4.6154-6.5241= approximately -1.908 or -1.909 cubic yards/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)=S(t)-R(t)
0=S(t)-R(t)
(The GRAPH of Y(t))
Using the Calculator, you find that the critical point is t=5.118 (the 2 curves intersect at this point twice if looking at both equations, but if you're looking at the graph of S(t)-R(t), the zero is t=5.118) and the endpoints are t=0 and 6. By looking at the graph of Y'(t)=S(t)-R(t), its is clear that x=0 is a maximum because Y'(t) is negative after that point and it is also obvious that x=6 is a maximum because Y'(t) is positive before that point. By looking at x=5.118, you see that Y'(t) changes from negative to positive at that point, and therefore it is a minimum.
The Minimum Value of sand can be found by plugging in the minimum value to the equation that outputs "the total number of cubic yards of sand on the beach at time t.", also known as Y(t), the equation we found in part b. (Wow this equation really comes in handy! (:).
So Y(5.118)= fnInt(S(5.118)-R(5.118),t,0,5.118) +2500. (You want the area of the graph from 0 to 5.118 in order to get the amount of sand at the time t.)
Y(5.118)= -7.630517431+2500 = approximately 2492.369 cubic yards of sand.
Well what I did for part D to find Y(5.118), 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" to add "fnInt(", 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. Then just hit enter and add "2500" to get your answer.! TADA!
Im not so sure if this is correct after seeing that some have the same answers, but many others do not. If I am mistaken, feel free to correct me, since it is best to learn from failed attempts (:
