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Related rates: dh/dt given time

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A solution is being poured into a conical filter in a chemistry experiment at a rate of 5cm^3/min. The filter is 15 cm high with a diameter of 10 cm at the top. The solution is dropping out of the filter at a rate of 1 cm^3/min. Determine the rate at which the height of the solution is increasing 3 minutes in this process.

    [tex] h = 15 cm [/tex]
    [tex] r = \frac{10cm}{2} = 5 cm [/tex]
    [tex] \frac{dV}{dt} = \frac{5cm^3}{min} - \frac{4cm^3}{min} = \frac{1cm^3}{min} [/tex]
    [tex] t = 3 minutes [/tex]
    [tex] \frac{dh}{dt} = ? [/tex]


    2. Relevant equations

    [tex] V = \frac{1}{3}\pi r^2h [/tex]
    [tex] \frac{r}{h} = \frac{5cm}{15cm} => r = \frac{h}{3} [/tex]

    3. The attempt at a solution

    Substitute.

    [tex] V = \frac{1}{3}\pi (\frac{h}{3})^2h [/tex]
    ...
    [tex] V = \frac{\pi}{27} h^3 [/tex]

    Differentiate.

    [tex] \frac{dV}{dt} = \frac{\pi}{27} 3h^2 \frac{dh}{dt} [/tex]


    At this point I feel I went wrong somewhere because I have not incorporated t so I am unsure of where to go from there. Countless searches brought me no results which is even more confusing.

    I feel I have to get a formula for h given t but that is also a problem, I think?

    Thanks in advance!
     
  2. jcsd
  3. Mar 26, 2012 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    welcome to pf!

    hi virulent! welcome to pf! :smile:
    that looks fine :smile:

    but after all that work, you've lost the plot :redface:

    you've forgotten that dV/dt is known! :wink:
     
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