Need assistance with Derivative/Integral Problem

[tangents]

Hey guys, I was doing some practice free response questions for my upcoming AP

Calc AB test but was somehwhat unsure of this one.
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Sand is removed from the beach at a certain rate by this function
R(t)= 2+5SIN(4t(pi)/25)

Sand is pumped into the beach with this equation
S(t)=15t/1+3t

Both functions are in cubic yards per hour and the interval of t is from [0,6].
A t=0, there is 2500 cubic feet of sand

a) how much sand will be removed in the 6hr period?
Here I took the integral of R(t) from 0 to 6 which came out to be 31.816 subic yards

b) write an expression for Y(t), the total number of sand on the beat at time t
I came up with Y(t)=2500-(intregal of R(T)+s(t)) from 0 to6

c) find the rate of change of sand at t=4
this was taking the derivative of the equation in part b which came to be about -1.91 cubic yd/hr

d) for [0,6], at what time t is the amount of sand on the beach minimum? what is that value? justify answer
This one I am having trouble with but I graphed the dervative and found x=3 to be when the graph was at its lowest point, although I'm not sure if that is correct.
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Any help is appreciated/assistance is greatly appreciated.

Hope you guys have a great day. [/b]

 

[daon]

Hey guys, I was doing some practice free response questions for my upcoming AP

Calc AB test but was somehwhat unsure of this one.
-------------------------------------------------------------------------

Sand is removed from the beach at a certain rate by this function
R(t)= 2+5SIN(4t(pi)/25)

Sand is pumped into the beach with this equation
S(t)=15t/1+3t

Both functions are in cubic yards per hour and the interval of t is from [0,6].
A t=0, there is 2500 cubic feet of sand

a) how much sand will be removed in the 6hr period?
Here I took the integral of R(t) from 0 to 6 which came out to be 31.816 subic yards

b) write an expression for Y(t), the total number of sand on the beat at time t
I came up with Y(t)=2500-(intregal of R(T)+s(t)) from 0 to6

c) find the rate of change of sand at t=4
this was taking the derivative of the equation in part b which came to be about -1.91 cubic yd/hr

d) for [0,6], at what time t is the amount of sand on the beach minimum? what is that value? justify answer
This one I am having trouble with but I graphed the dervative and found x=3 to be when the graph was at its lowest point, although I'm not sure if that is correct.
----------------------------------------------------------------------

Any help is appreciated/assistance is greatly appreciated.

Hope you guys have a great day. [/b]

a) The total rate of removal of sand will be R(t)-S(t) assuming R(t) > S(t) for all t 0<t<6. So the total sand remvoed over the 6 hour period will be \(\displaystyle \int^6_0 R(t)-S(t) dt\)

b) Y(t) = 2500 - \(\displaystyle \int^6_0 R(t)-S(t) dt\)

c) Your theory is correct.

d) From part c, find the critical points and use your 1st/2nd derivative tests.

 

[skeeter]

(a) correct

(b) Y(t) = 2500 + INT{0 to t}[S(x) - R(x)] dx

(c) Y'(t) = S(t) - R(t) ... Y'(4) = S(4) - R(4) = -1.909 yd³/hr
(three decimal places, remember?)

(d) graph Y'(t) in the interval [0,6] ... when Y'(t) changes form (-) to (+), that will indicate a minimum value of sand on the beach.

minimum sand on the beach occurs at t = 5.118 hrs

amount of sand on the beach at that time is Y(5.118) = 2492.369 yd³