Anyways! Sandy Beach has some problems that needs to be solved! :
Sand being removed:
Sand being added:
A. How much sand will the tide remove from the beach during this 6-hour period?
We use the equation R(t) for this one! Integral of 2+5sin(4pit/25) from 0 to 6. It should output an approximation of 31.816 cubic yards of sand removed.
B. Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
Since the sand is being eaten [R(t)] at the same time it's being put on the beach [S(t)], we need to combine them to find out how much of the sand remains! Our new equation will be:
Y(t) = [S(t) - R(t)] + 2500
C. Find the rate at which the total amount of sand on the beach is changing at time t=4.
I don't think you include the +2500 in Y(t) for this one because it's asking for the rate of change of the sand... :/ Bleh, confused. Either way, if you plug in 4 into Y(t) without the +2500 then you'll get -1.908750647 which should be rounded out to about -1.909 cubic yards...??
D. For
, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.Given that the boundary is 0 to 6, those are our endpoints i assume? I'm too lazy and confused to find the critical points... so using my magic calculator of magic, I found that our critical points are t= 0, 5.118, and 6! ... But I still don't know whether 5.118 is my minimum...
Now then~
*Brings out ipod and begins singing to Help! by the Beatles...*
For part b, you forgot to add the integral and the boundaries, (fnInt, and ,x,0,t). You get the right answer for c because you didnt get the right equation lol. You should be taking the derivative of Y(t) either way, which is just your S(t)-R(t). To check if 5.118 is a minimum, you have to graph S(t)-R(t), which you can do by:
ReplyDeleteplotting 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". Now into "Y3" just press "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). You hit graph and there is your graph! then if it changes from negative to positive, you realize that it is a minimum (this S(t)-R(t) is the graph of the derivative already, so dont look at the SLOPE of this graph, or you will be looking at the second derivative!)
Hey Letty
ReplyDeleteFor b) they are asking for y(t) which is the amount of sand (yrs^3)
S(t) and R(t) are measured in yrs^3/ hrs(amount/ time)
to get the amount (yrs^3) you need to find the integral of S(t) and R(t)then add 2500 which is the starting amount of sand
For part d, 5.118 is the minimum because, if you graph the equation, you see that the two endpoints 0 & 6, are maximums. Since there are only 3 critical points, this suggests that 5.118 is the minimum.
ReplyDelete