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Conical tank calc help

  1. Mar 16, 2005 #1
    I have a calculus question and was wondering if some could help me see what I am doing wrong in this question. Thank you

    A conical tank with an altitude of 10m and whose base has a radius of 4m is mounted with its vertex down. The tank is full of water which is draining through the vertex at the rate of 5 m^3/min. how fast is the level of the water dropping when the radius is 3?

    the answer is 5pi/9

    this is how i did it:

    r/h=4/10
    h=5pi/2

    V=[(pi)(r^2)(h)]/3
    V=[(pi)(r^2)(5r/2)]/3
    V=[pi5r^3]/6
    dv/dt=[15pi(r^2)]/6] * dr/dt
    5=[(15pi(3^2)]/6] * dr/dt
    dr/dt= (2pi)/9
     
  2. jcsd
  3. Mar 16, 2005 #2
    Firstly, the answer is actually [tex]\frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}[/tex], not [tex]\frac{5\pi}{9}\frac{\mbox{m}}{\mbox{min}}[/tex].

    Secondly, are you sure [tex]\frac{d r}{d t}[/tex] is what the question is looking for? Re-read it.
     
    Last edited: Mar 16, 2005
  4. Mar 17, 2005 #3
    Notice how the question asks for how fast the level of the water is dropping. This would mean a change in height with respect to time, not the radius. Since it's asking for the rate of change when the radius is 3 then:

    [tex] r=\frac{3h}{10} [/tex]
    [tex] V=\frac{1}{3} \pi (\frac{3h}{10})^2 h [/tex]
    [tex] V= \frac {3}{100} \pi h^3 [/tex]
    [tex] \frac {dV}{dt}= \frac{9}{100} \pi h^2 \frac{dh}{dt} [/tex]

    Solving for dh/dt should get you the answer [tex] \frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}[/tex] as stated above by Data.
     
  5. Mar 17, 2005 #4
    You actually need to use [tex]r = \frac{2h}{5}[/tex]. But other than that you're completely right.
     
  6. Mar 18, 2005 #5
    thanks for your help.
     
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