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Related Rates height of rising water

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data
    A rectangular swimming pool 16m x 10m is being filled at a rate of .8m^3/min. How fast is height of the water rising?



    2. Relevant equations
    V = LWH


    3. The attempt at a solution

    H = V/LW
    (dh/dt) = LW(dv/dt)
    (dh/dt) = (10)(16)(.8)


    This question looks so easy and I realize Im probablly making a stupid mistake somewhere, could someone point it out for me please?
     
  2. jcsd
  3. Mar 31, 2008 #2
    V=abh----(*), where a=16, b=10, h is the height and it is changing so we don't know it.

    we need to know

    [tex]\frac{dh}{dt}=??[/tex] we also know that

    [tex]\frac{dV}{dt}=.8\frac{m^{3}}{min}[/tex]

    Now differentiate implicitly (*) and plug in the information you were provided. Does this help?
     
  4. Mar 31, 2008 #3
    Ok, where u went wrong is that LW should be on denominator, and not on the numerator.

    (dh/dt) = (dv/dt)/(LW)
     
  5. Mar 31, 2008 #4
    Thanks for the help. One thing I don't understand though..
    I rearrange V = LWH to solve for H.
    That gives me H = V/LW.
    Now when I take the derivate implicitly, wouldnt I need to use the quotient rule to get the derivative of V/LW? That is how I got LW(dv/dt) and not the correct answer of (dv/dt)/(LW)
     
  6. Mar 31, 2008 #5
    BUt notice that Length (L) and Width(W) are constants. In general:

    [tex]\frac{d(Cf(x))}{dx}=C\frac{d(f(x))}{dx}[/tex] C is a constant
     
  7. Mar 31, 2008 #6
    So since the denominator is a constant it isn't necessary to use the quotient rule?
     
  8. Mar 31, 2008 #7
    Well, even if you used the quotient rule, you will get the exact same answer, if you do the calculations right at first place.
     
  9. Mar 31, 2008 #8
    See what happenes if you really want to apply the quotient rule

    [tex](\frac{V}{LW})'=\frac{V'LW-(LW)'V}{(LW)^{2}}=\frac{V'LW}{(LW)^{2}}=\frac{V'}{LW}[/tex]

    SO, the answer is the same, that's why i love math, because it is all consistent, no matter which way u decide to pursue.
     
  10. Mar 31, 2008 #9
    If I tried this using the quotient rule, would this be the correct way to go about it? (I know it isnt..)
    (dh/dt) = (160)(dv/dt)-(v)(0)
     
    Last edited: Mar 31, 2008
  11. Mar 31, 2008 #10
    Well, i think you answered it yourself.

    Do you know the quotient rule??????
     
  12. Mar 31, 2008 #11
    (dh/dt) = (160)(.8)

    = 128 ...this isn't the correct answer?
     
  13. Mar 31, 2008 #12
    No.

    it should read sth like

    .8/160= whatever

    You need to review the quotient rule!!!
     
  14. Mar 31, 2008 #13
    V= L*w*h
    dh/dt=?
    Derivative of V = K h

    dv/dt =K dh/dt
    .8m^3/min = 160 dh/dt
    (.8m^3/min)/160m^2 = dh/dt
    Notice we know it is the correct answer because the Units cancel out this is very interesting fact that even if you dont really know what your doing as long as the unit makes sense then your good remeber this for your next test lol!


    ( the length is not changin the width is not changing only the height is changin as you fill the pool)
     
  15. Mar 31, 2008 #14
    kjlhdfkljdshfkljnvckjnd pretty ugly mistake. I guess I haven't done quotient rule in awhile hah. I didn't pay any attention when you showed me how to do it with the quotient rule. Sorry for wasting your time bro, but thanks for all the help.
     
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