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Math exam question(s).

  1. Jun 21, 2005 #1
    My final math exam is tomorrow, and I am having trouble with some questions as I go through this review package.

    I have the answers so its very frustrating to arrive at the wrong answer over, and over.

    Anyways the question is:

    The position of a bucket on a circular tar sands excavator is given by the equation: p(t)= 10sin(6pit + 3pi) + 20

    Where t is the time in minutes. There are 20 buckets on the excavator and each holds 500 kg

    How many kilograms of tar sands are excavated in one hour?

    The answer is 1800000 kg

    This is what I did: p(t)=10sin(6pi(60)+3pi) + 20
    P(t)= 28.69

    from here I am lost so I just did 500 x 20= 10000
    10000 x 28.69= 286900 kg

    Can someone please help me? Thanks
  2. jcsd
  3. Jun 21, 2005 #2
    How many times have did one excavator come to its starting place? How many times have did 20 excavators come to its starting place? can you answer these questions?
  4. Jun 21, 2005 #3


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    Are you to a calculus level yet? Can you take the derivative of the function you stated? The derivative of position is velocity...angular velocity in this case.
  5. Jun 21, 2005 #4
    Well I think the period is .0055555556. I assume its that since im using 6pi(60) as b. So 2pi/b gives me .0055555... When I plug in 60 minutes into the equation I get like 28.69. So I divided 28.69 by the period and got 5164.199959. Then I multiplied 5164.2 by 20 since there is 20 other buckets, and got 103284. I then multiplied this by 500 kg and got 51651999.6 kg.

    Sorry Fred I havn't been exposed to any sort of calculus yet.
  6. Jun 21, 2005 #5
    Think of the equation this way. The constant c does not change anything actually so we need not pay any heed to it, right? That's because we are interested in the period of this function. now take [itex]sin(6\pi*x+3\pi)[/itex] like this [itex]sin(a+b)[/itex]

    Can you conclude anything about the function from here? The given function is actually equal to [itex]-10sin(6\pix)[/itex]. What can you do now?

    Fred, I don't understand, what does this quesation have to do with the derivatives? We do not need to find the velocity. Or am I missing something?
  7. Jun 21, 2005 #6
    It seems I cannot edit my posts for some unknown reason. Anyway the last latexed equation should be

  8. Jun 21, 2005 #7
    I am really confused now. I don't know how you went from sin(6pix+3pi) to
    -10sin(6pix). It seems I can't even answer your simple questions. What chance do I have in finding the answer.
  9. Jun 21, 2005 #8
    Do you know the sum formula for sine function that says


    All I did was replacing

    [tex]a=6\pi*x, b=3\pi[/tex]

    and while doing the calculations heed that

    [tex]cos(3\pi)=-1, sin(3\pi)=0[/tex] .

    Do you understand it now?
  10. Jun 22, 2005 #9


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    I took the derivative of the position function with respect to t and used that and the fact that there are 20 buckets and they hold 500 kg. My answer rounded to the one given. Seemed to work ok for me.
  11. Jun 22, 2005 #10
    Well from using the period of the function I get the exact answer. As I said, the function is actually


    The period of this function is 1/3. so, until x=60 this function repeats itself 180 times.

    Multiply this by 50*20, and you get the answer.
    Could you help me understanding your solution? I still have no idea how to use the deriative.
  12. Jun 22, 2005 #11


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    Like I said, there's a rounding issue, but I did this:

    [tex]P(t) = 10 \sin (6 \pi t + 3 \pi) + 20[/tex]

    [tex]P'(t) = (10)(6 \pi) \cos (6 \pi t + 3 \pi)[/tex]

    Evaluate the derivative at t=60 to calculate the number of times the wheel has gone around:

    [tex]P'(60) = 188.45[/tex]
    I used absolute value here since the negative value really doesn't mean anything.

    To get the total # of kg:
    [tex]Total = (188.45 revs)(20 \frac{buckets}{rev})(500 \frac{kg}{bucket})[/tex]

    [tex]Total = 1.88 x 10^6 kg[/tex]
  13. Jun 22, 2005 #12
    I don't see the point still. I mean P'(t) is the velocity and pluggin 60 into this we get the speed of one bucket at time t=60. As far as i am concerned this is not the number of times the wheel has gone around.

    Am I wrong?
  14. Jun 22, 2005 #13


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    You got me to see I forgot a part of my solution!

    [tex]P(t) = 10 \sin (6 \pi t + 3 \pi) + 20[/tex]

    [tex]P'(t) = (10)(6 \pi) \cos (6 \pi t + 3 \pi)[/tex]

    Evaluate the derivative at t=60 to calculate the number of ANGULAR VELOCITY of the wheel:

    [tex]\omega = P'(60) = 188.45 \frac{rad}{min}[/tex]
    I used absolute value here since the negative value really doesn't mean anything.

    Now calculate the number of revolutions the wheel has gone thru:
    [tex]\theta = (188.5 \frac{rad}{min})(60 min)(\frac{1 rev}{2 \pi rad})[/tex]
    [tex]\theta = 1800 rev[/tex]

    To get the total # of kg:
    [tex]Total = (1800 revs)(20 \frac{buckets}{rev})(500 \frac{kg}{bucket})[/tex]

    [tex]Total = 18.00 x 10^6 kg[/tex]

    Thanks for keeping me honest!
  15. Jun 22, 2005 #14
    Anytime :)

    Mozart, now you have two solutions to choose from. Mine is just a simple trigonometric solution, Fred's is a much more explanatory of the motion because he uses the derivative to find the angular velocity, which gives a lot more info about the info. Both of the solutions give the correct answer so here ya go!

  16. Jun 24, 2005 #15
    Hey thanks guys, and sorry for the late response, but some genius decided to put my exam on the day of the prom, and well I was kind of out of it for a few days. Anyways the exam turned out to be quite simple. Nothing was as hard as this question. I know it's not really a hard question for you guys but for me its pretty complicated. Well thanks again.
  17. Jun 24, 2005 #16


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    Actually, Fred's answer is high by a factor of 10.

    And you don't need to take the first deriviative to find [itex]\omega[/itex]. The equation you have is of the form :

    [tex]x = R sin(\omega t + \phi ) [/tex]

    You just have to read off the coefficient of 't' to find the angular velocity [itex] \omega [/itex]. In this case, [itex]\omega = 6 \pi [/itex]. The rest follows as Fred described.

    Fred, you found [itex]\dot{x} = v = R \omega[/itex]. You must divide by R = 10, to get the right answer.
  18. Jun 25, 2005 #17


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    ahhhhhhhhhhhhhhhhh......crud. You're right.
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