Oh damn... I need paper
And calculator...
Okay, (a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.[I'm not writing the functions, it's early and I'm lazy, ha.]
fnInt(R(t),x,0,6)≈ 31.816 cubic yards
(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
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.
What is this? rate of change... instantaneous? I think I forgot how to do this.
Wait a minute. Change of y over change of t at 4
So... Y(4)/4≈2498.091/4=624.52275 cubic yards per hour
(d) For 0≤t≤6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.
Find critical points S'(t)-R'(t)=0 at t=3.007 and t=.505
t=3.007, S(t)-R(t)=-2.49
t=.505, S(t)-R(t)=-.253
endpoint values
t=0, S(t)-R(t)=-2
t=6, S(t)-R(t)=2.11
So absolute minimum at t=3.007
I think I got most of them wrong... If they are, someone explain please?
Well now, off to do the rest of the calculus homework... woohoo... homework...
HI for C its asking the rate at which sand changes at time t=4
ReplyDeleteYou have your two equations and both of them are rates for pumping sand and removing sand. So if you want to know the rate at t=4 you just plug in t=4 into
S(t) - R(t)=
4.615 - 6.524 = -1.909 cubic yards/hour
and you have the rate at which is changing at that time. hope it helps
Yes and because you made that error stated my Miriam in part c, it messed up your part d. So make S(t)-R(t)=0 , not the derivative of it! (:
ReplyDeleteHey Cynthia
ReplyDeleteFor b) they are asking Y(t),it being the amount of sand (yrs^3)
You put Y(t)=S(t)-R(t)+2500
S(t) and R(t) is measured in yrs^3/ hrs so to get yrs^3 you have to find the integral of S(t) and R(t).
Y(t)= [∫S(t)-R(t)dt]+2500 you exclude 2500 when finding the integral because that is your starting amount of sand
For part c, you take the derivative !
ReplyDeleteSo, you use S(t)-R(t).
Using this information, you can now solve part d correctly :]