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Related Rates

  1. May 12, 2005 #1
    am i doing this right?
    A cylindrical tank with a radius of 5 ft and a height of 20ft is filled with a certain liquid chemical. A hole is punched in the bottom. At that moment the chemical drains ut of the tank at the rate of 2ft^3/min. At what rate is the height of liquid in the tank changing?

    V = 2(pi)rh
    dV/dt = [2(pi)r]dh/dt
    2 = [2(pi)5]dh/dt
    dh/dt = 1/5(pi)

    The height of liquid in the tank is changing at the rate is 1/5(pi) /min
     
  2. jcsd
  3. May 12, 2005 #2
    Looks good to me.
     
  4. May 12, 2005 #3
    thanks.....
     
  5. May 12, 2005 #4
    Except thats not the volume of a cylinder. :)
     
  6. May 12, 2005 #5

    HallsofIvy

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    [tex] 2\pi rh[/tex] is the area of the curved portion of the cylinder.
     
  7. May 12, 2005 #6
    Such a type of text book questions never include the real effects viz., changing flowrate with respect to height of liquid column in the tank etc. Otherwise, your procedure is ok except that the volume equation of the cylinder as suggested already.
     
  8. May 12, 2005 #7
    V = (pi)r^2h


    V = 25(pi)h

    dV/dt = 25(pi)(dh/dt)
    -2 = 25(pi)(dh/dt)
    -2/(25)(pi) ft/min = dh/dt

    better?
     
  9. May 13, 2005 #8

    HallsofIvy

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    Much better!
     
  10. May 14, 2005 #9
    just for any question like that.....when it is asking what the rate of changing is....u need a negative sign....
    when it is asking for the rate of decreasing.....u don't need a negative sign....is that correct?
     
  11. May 14, 2005 #10
    That depends on the context of the question. Usually when something is decreasing you will use a negative sign. For example, if the height is decreasing at a rate of 2 meters per second, you would write that like this.

    [tex]\frac{dh}{dt} = -2m/s[/tex]
     
  12. May 14, 2005 #11
    but when the question ask what is the rate of decreasing....do u still need a negative sign because we know that it is decreasing already....
     
  13. May 14, 2005 #12
    I would put the negative to indicate decreasing. If you did not put the negative sign and applied the positive value, you would have an incorrect answer.
     
  14. May 14, 2005 #13
    i mean in the answer....do u need negative signs to indicate decreasing when the question asks for decreasing already....
     
  15. May 14, 2005 #14
    I would put, [tex]\frac{dh}{dt} = - \mbox{rate}[/tex] just to make sure there's no mix up.

    If the question specifically asks, "what rate is the height decreasing?" then I guess it's acceptable to put a positive answer, although that could be ambiguous.

    I would put my first answer just to make clear you know what you are talking about.
     
  16. May 14, 2005 #15
    okay....i see...thanks...
     
  17. May 14, 2005 #16
    If it gives you find the rate at which h is decreasing, I would report my answer as positive, since when read with the question it makes more sense that way. "h is decreasing at a rate of 3 m/s." Not "h is decreasing at a rate of -3m/s"
     
  18. May 15, 2005 #17
    Right, if it is read that way then the negative is implied by the word decreasing. But I was saying that safest way to make sure there was no mix up is to put

    [tex]\frac{dh}{dt} = ...[/tex]

    where the sign is correctly placed. If you keep the decreasing rate positive because of a wording you could forgot to change it to a negative when working with other things using that value.
     
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