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Derivatives: Swimming Pool

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A rectandular swimming pool is 10m wide and 20m long. The bottom of the pools is a sloping plane with the depth of the pool varying along the length of the pool from 1m at the shallow end to 5m at the deep end. Water is being pumped into the pool at the rate of [tex]30 m^3 / min[/tex]. Shat that the total volume of water in the pook when the depth of thewater at the deep end is hm (where [tex]0 \leq h \leq 4[/tex]) satisfies

    [tex] V = 25h^2 [/tex]

    Hence find the rate at which the water level is rising when the water is 3m deep at the deep end.

    2. Relevant equations
    n/a
    3. The attempt at a solution

    Equation to describe volume at h...

    When 4m full, the 'length' filled by the water is 20m, therefore..

    [tex]
    \begin{align*}
    V &= \frac{1}{2}LHW\\
    &=\frac{1}{2}(5h)(h)(10)\\
    &=(2.5h^2)(10)\\
    &=25h^2\\
    \end{align*}
    [/tex]

    Then, i tried to differentiate this expression to find h' (difference in height), but this came to 0.

    Ideas?
     
  2. jcsd
  3. Apr 8, 2009 #2

    Tom Mattson

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    Staff Emeritus
    Science Advisor
    Gold Member

    Say what? If [itex]V(h)=25h^2[/itex] then [itex]\frac{dV}{dh}=50h[/itex]. So [itex]\frac{dV}{dt}=?[/itex]
     
  4. Apr 8, 2009 #3
    Is the final answer 1/150 m/s?
     
  5. Apr 8, 2009 #4
    I'm a newb but I'll try to help -_-.

    Try differentiating [tex] V=25h^2 [/tex] with respect to [tex]t[/tex]. Since you know [tex]dV/dt[/tex] and [tex]h[/tex], you can figure out the answer.
     
    Last edited: Apr 8, 2009
  6. Apr 9, 2009 #5
    [tex]V=25h^2[/tex]

    [tex]
    \begin{align*}
    differentiated
    &=\frac{dV}{dt} (25h^2)\\
    &=\frac{dV}{dh}(25h^2) \frac{dh}{dt}\\
    &=(50h) \frac{dh}{dt}\\
    \frac{dh}{dt} &= 50h\\
    sub. in value for h to find dh/dt\\
    \frac{dt}{dh} &= 50(3)\\
    \frac{dt}{dh} &= 150\\
    \frac{dh}{dt} &= 1/150\\
    \end{align*}
    [/tex]
     
  7. Apr 9, 2009 #6
    Does that look even a little bit right to anyone???
     
  8. Apr 9, 2009 #7
    Not right.

    For one thing, it looks like you are writing [tex]\frac{dV}{dt}(25h^2)[/tex] when you mean [tex]\frac{d}{dt}(25h^2)[/tex].

    Here is better notation.

    [tex]
    \begin{align*}
    \frac{dV}{dt}&=\frac{dV}{dh}\frac{dh}{dt}\\
    &=\frac{d}{dh}(25h^2) \frac{dh}{dt}
    \end{align*}
    [/tex]

    What's next?
     
  9. Apr 9, 2009 #8
    [tex]
    \begin{align*}
    &= 50h \frac{dh}{dt} \\
    \end{align*}
    [/tex]

    Then move dh/dt over to the other side, where it is 'dividing', i.e [tex] \frac{1}{\frac{dh}{dt}} = 50h[/tex]

    so

    [tex]\frac{dt}{dh} = 50h[/tex]

    but we want dh/dt, so needs to be rearranged,

    [tex]\frac{dh}{dt} = \frac{1}{50h}[/tex]

    then I subbed in 3 for H to get 1/150
     
  10. Apr 9, 2009 #9
    Yes, now write it as part of an equation. Then what?
     
  11. Apr 9, 2009 #10
    I editted above post ^^ Accidently hit reply before I was finished.
     
  12. Apr 9, 2009 #11
    No, in your first step, the other side is not "1." What is the other side?
     
  13. Apr 9, 2009 #12
    dV/dt? and does that = 30?
     
  14. Apr 9, 2009 #13
  15. Apr 9, 2009 #14
    Thanks for all the help!

    Ta muchly!
    x
     
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