Need assistance with Derivative/Intregal Problem

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Homework Help Overview

The discussion revolves around a calculus problem involving the rates of sand being removed from and added to a beach. The functions provided describe these rates over a specified time interval, and the original poster seeks assistance with various parts of the problem, including integrals and derivatives.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the total amount of sand on the beach using integrals of the given functions and raises questions about the correctness of their expressions.
  • Some participants question the setup of the equations, particularly the order of the integrals and the interpretation of the functions.
  • Others suggest examining the derivative to find critical points for determining minima or maxima of the sand amount.
  • There is discussion about the implications of the values obtained from the graphs and whether they make sense in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, offering corrections and alternative interpretations of the equations. There is a focus on understanding the relationships between the functions and their derivatives, with some guidance provided on how to approach finding critical points. However, no consensus has been reached on the final answers or methods.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for solving the problem. There are also discussions about potential misunderstandings regarding the functions and their integration limits.

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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 .
 
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Question (b) I think it should be

[tex]Y(t) = 2500 + \int S(t) \;\; dt - \int R(t) \;\; dt[/tex]

As an indefinate integral.

Regards,
~Hoot
 
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:
[tex]Y(t)= 2500+ \int_0^t (S(x)- R(x))dx[/tex]
or equivalently
[tex]Y(t)= 2500+ \int_0^t S(x)dx- \int_0^tR(x)dx[/tex]

c) Of course, the derivative of the integral is just
S(t)- R(t). Equivalent to your answer is S(4)- R(4)=
[tex]2+ 5(sin((4)(4)(\pi)/25)- 60/(1+ 12)[/tex]
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)=
[tex]2+ 5(sin(4t\pi/25)- \frac{15}{1+ 3t}[/tex].
You may need to graph that to see where it is 0- that's not going to have a simple "algebraic" solution.
 
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. ^_^
 
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?
 
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).
 
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 the 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