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Math Help - Rate Of Change of Cone

  1. #1
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    Rate Of Change of Cone

    A cone-shaped coffee filter of radius 6cm and depth 10cm contains water, which drips out through a hole at the bottom at a constant rate of 1.5cm^3 per second.

    a) If the filter starts out full, how long does it take to empty?


    v = (1/3)(pi)(r^2)h - 1.5t
    0 = (1/3)(pi)(6^2)(10) - 1.5t
    0 = 376.99 - 1.5t
    376.99 = 1.5t
    t = 251.33 sec

    b) Find the volume of water in the filter when the depth of the water is h cm.

    v = (1/3)(pi)(r^2)h - 1.5t
    v + 1.5t = (1/3)(pi)(r^2)h
    h = (v + 1.5t) / ((1/3)(pi)r^2)

    c) How fast is the water level falling when the depth is 8cm?

    8 = (v + 1.5t) / ((1/3)(pi)r^2)

    How do I solve this with so many unknown variables (volume, time and radius)? My suspicion is that I did part (b) incorrectly. Can anyone please show me where I went wrong? Thanks!
    Last edited by Jessem; December 1st 2009 at 08:00 PM.
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  2. #2
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    I apologize, I really hate to bounce but the problem seems pretty basic and I would really like to get it figured out so I can get some sleep. I'm still trying to figure it out but can't get anywhere. Any help would be greatly appreciated! Thanks!
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  3. #3
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    (a)

    Rate=1.5cm^{3}/s
    Volume=\frac{1}{3}\pi(6)^{2}(10)=1200\pi
    So Time\;taken=\frac{1200\pi}{1.5}=2513.27s

    (b)

    By similar triangles, \frac{h}{10}=\frac{r}{6}\Rightarrow r=\frac{3}{5}h
    So \displaystyle V=\frac{1}{3}\pi\left(\frac{3}{5}h\right)^{2}h=\fr  ac{3}{25}\pi h^{3}

    (c)

    By Chain Rule, \frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}
    Given \frac{dV}{dt}=-1.5cm^{3}/s
    Differentiate V with restpect to h, \frac{dV}{dh}=\frac{9}{25}\pi h^{2}
    At h=8, \frac{dV}{dh}=\frac{9}{25}\pi (9)^{2}=\frac{729}{25}\pi
    Thus, \frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}\Rig  htarrow -1.5=\frac{729}{25}\pi \times\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=?
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  4. #4
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    Quote Originally Posted by acc100jt View Post
    (a)

    Rate=1.5cm^{3}/s
    Volume=\frac{1}{3}\pi(6)^{2}(10)=1200\pi
    So Time\;taken=\frac{1200\pi}{1.5}=2513.27s
    I do believe that (1/3)(6^2)(10) = 120, not 1200. Which means that t=251.33s

    Quote Originally Posted by acc100jt View Post
    (b)

    By similar triangles, \frac{h}{10}=\frac{r}{6}\Rightarrow r=\frac{3}{5}h
    So \displaystyle V=\frac{1}{3}\pi\left(\frac{3}{5}h\right)^{2}h=\fr  ac{3}{25}\pi h^{3}
    I've never learned this 'similar triangles' thing, but I see that you're just finding the ratio between r and h of the cone so we can express r in terms of h. Sounds good, this is what I was missing.

    Quote Originally Posted by acc100jt View Post
    (c)

    By Chain Rule, \frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}
    Given \frac{dV}{dt}=-1.5cm^{3}/s
    Differentiate V with restpect to h, \frac{dV}{dh}=\frac{9}{25}\pi h^{2}
    At h=8, \frac{dV}{dh}=\frac{9}{25}\pi (9)^{2}=\frac{729}{25}\pi
    Thus, \frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}\Rig  htarrow -1.5=\frac{729}{25}\pi \times\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=?
    The height of the water should be 8cm, but this was also very helpful. I got -0.0164 = dh/dt for yours (using h = 9cm), and I got dh/dt = -0.021 (using h = 8cm).

    Thanks for all your help!
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