6
| 2 + 5sin(4pie/25) dx
0
fnInt ( 2+ 5sin(4pieX/25) , x , 0 , 6)
~ 31.815
About 32 yds^3/Hour
2) Since sand is being distributed and lessened at the same time , we need a Y(t) formula that will represent the amount of sand distrubuted after the removal of sand by sea.
So
y(t) =
6
| ( r(t) - s(t) ) dt + 2500
0
3) To find the rate at which the sand is changing at T =4 . We plug in for t and solve.
So..
2 + 5sin(4pie4/25) dx - 15(4)/ 1+3(4)
= 6.524 - 4.615
~ 1.909
The sand is changing by around 2 yds/Hour at T = 4
4) To find the amount of sand on the beach we find the critical points by finding the derivative of the y(t) formula and solving for possible minimums.
Well , Am feeling lazy lol. So.. Ill conclude that (0,6) are endpoints.
So .. Ill plug these into the y(t) are see which one is the most reasonable answer.
T=0
Rate of change is 2
T=6
Rate of change is -2.116
T=6 is the minimum.
Minimum value is 2506.725 yrds^3/hour at T= 6
For #3 aren't we supposed to leave the negative sign or it doesn't matter? :D
ReplyDeleteI have no idea lol. :D
ReplyDeleteFor part d, how do you define "the most reasonable answer?"
ReplyDeletejust check all of them !
for part C, where did (0,6) come from?
ReplyDeleteexplain.. haha
yea where did (0,6) come from? did u get it from a plugging something in?
ReplyDelete