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Homework Help: Need assistance with Derivative/Intregal Problem

  1. Apr 17, 2006 #1
    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

    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 good day .
  2. jcsd
  3. Apr 17, 2006 #2


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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.

  4. Apr 17, 2006 #3


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    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.
  5. Apr 17, 2006 #4
    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. ^_^
  6. Apr 17, 2006 #5
    Ok for Part D I graphed S(t)-R(t). At x=0 y=-2, which doesnt 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 in to the original equation of part b and see which ones gives the lowest value?
  7. Apr 17, 2006 #6


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    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).
  8. Apr 17, 2006 #7
    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
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