a)How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.
S(t)= yrs^3/ hrs - adds
R(t)= yrs^3/ hrs - removes
t=0 there is 2500 yrs^3 of sand
0∫6 R(t)= 31.816 yrs^3 of sand is removed
b)Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
Y(t)= yrs^3 (the amount of sand not the rate or acceleration)
Y(t)= [∫S(t)-R(t)dt]+2500
*S(t)is being add and R(t)is being removed therefore you subtract
*you add 2500 once you find the integral because 2500 yrs^3 is the starting amount
c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
Y(t)= yrs^3 of sand (outputs amount not the rate)
Y'(t)= yrs^3/ hr (outputs rate of which the sand on the beach is changing over time)
Y'(t)= S(t)- R(t)
Y'(4)=S(4)- R(4)
Y'(4)=4.6154 - 6.5241
Y'(4)= -1.9087 yrs^3/ hrs
d)For [0, 6] at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.
* mins at critical points and/or ending points
Critical Points
Y(t)= yrs^3
Y'(t)= yrs^3/ hrs
Y'(t)= 0
Y'(5.118) =0 t=5.118 hrs
Before Y'(5.118) Y'(t)<0 and After Y'(5.118) Y'(t)>0
Y(5.118)= 2492.37 yrs^3 of sand
Y(0)= 2500 yrs^3 of sand
Y(6)= 2493.28 yrs^3 of sand
The minimum is at t=5.118 with a total of 2492.37 yrs^3 of sand
So concise !
ReplyDeletevery good job, i like your reasoning and little * that call attention to critical details. :]
For part c, however, you do not explain why you used the equation Y'(t)...
I love how your answers are straightforward! Im not so sure if part a is negative, since sand cant be negative, (: I suggest you fix that. and for part b, dont forget to add "dt", and if you missed it anywhere else, add it (:
ReplyDeleteOh and denise i think her explanation is "Before Y'(5.118) Y'(t)<0 and After Y'(5.118) Y'(t)>0"
and I think that is enough, yet I am not sure,
Good Job Sandra!