Need assistance with Derivative/Intregal 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 good day .

 

[Hootenanny]

Question (b) I think it should be

Y(t) = 2500 + \int S(t) \;\; dt - \int R(t) \;\; dt

As an indefinate integral.

Regards,
~Hoot

 

[HallsofIvy]

R(t)= 15t/1+ 3t? Isn't that the same as 18t? Is it possible that you meant 15t/(1+ 3t)?
In (b) the integral is not from 0 to 6, it's from 0 to t:
Y(t)= 2500+ \int_0^t (S(x)- R(x))dx
or equivalently
Y(t)= 2500+ \int_0^t S(x)dx- \int_0^tR(x)dx

c) Of course, the derivative of the integral is just
S(t)- R(t). Equivalent to your answer is S(4)- R(4)=
2+ 5(sin((4)(4)(\pi)/25)- 60/(1+ 12)
which is approximately 1.91.

d) A maximum or minimum occurs either at an endpoint (t= 0 or t= 6) or where the derivative is 0. Once again, the derivative is S(t)- R(t)=
2+ 5(sin(4t\pi/25)- \frac{15}{1+ 3t}.
You may need to graph that to see where it is 0- that's not going to have a simple "algebraic" solution.

 

[tangents]

Just a few questions. The integral is S(x)-R(x) and not the other way around? also Hallsof Ivy I think you switched the values of s(t) and r(t) in parts C and D =). Thank you guys are helping me out I kind of knew what to do but just couldn't get it on paper. ^_^

 

[tangents]

Ok for Part D I graphed S(t)-R(t). At x=0 y=-2, which doesn't make sense how much sand there is. At x=6 I get y=2.1 and x= 5.1 is a zero. But these values are too small as the original amount of sand is 2500 subic yards. Am I supposed to plug those three values into the original equation of part b and see which ones gives the lowest value?

 

[HallsofIvy]

Yes, S(t) is sand being removed so it contributes to the 'change in the sand' -S(t). R(t) is sand being added so it contributes to the 'change in the sand' + R(t) dY/dt= R(t)- S(t).

 

[tangents]

oh ok I get what you're saying now.
Just one last question if anyone can confirm but for part D, I found the zero of teh derivative to be 5.13, I know it's a minimum because the graph changes from negative to positive. So Now I plug that into my original equation like this?: y(x)=2500+Integral [S(t) from 0 to 5.1(?)] - [integral R(x) from 0 to 5.1(?)]? I just want to make sure this is correct because it's the only way I can think of solving it