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Math Help - Rates Of Change Question

  1. #1
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    Rates Of Change Question

    "Water is being poured into a conical vessel of semi vertical angle of 30 degrees at the rate of 100Pi cm cubed per minute. find the rate at which the depth pf water is increasing when the water is 5cm deep."


    I can't really seem to get my head around this question.
    As far as I can make out it involves a little bit of trigonometery which I'm terrible at. The only thing I can make out so far is that I'm looking for dV/dT.


    All help really appreciated!
    The question is worth quiet a bit of my exam.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Fionnan View Post
    "Water is being poured into a conical vessel of semi vertical angle of 30 degrees at the rate of 100Pi cm cubed per minute. find the rate at which the depth pf water is increasing when the water is 5cm deep."


    I can't really seem to get my head around this question.
    As far as I can make out it involves a little bit of trigonometery which I'm terrible at. The only thing I can make out so far is that I'm looking for dV/dT.


    All help really appreciated!
    The question is worth quiet a bit of my exam.
    If the depth of water is d, then the radius of th surface is r=d/sqrt(3) since
    the semi vertical angle is 30 degrees. So the volume of water is:

    V = (1/3) pi r^2 d = (1/3) pi d^3/ sqrt(3)

    (volume of cone formula used)

    Now we are told that dV/dt = 100 pi cc/min. so we differentiate the formula
    for V wrt t:

    dV/dt = [1/(3 sqrt(3)] pi 3 d^2 dd/dt

    so when d=5 cm we have:

    dd/dt = 4 sqrt(3) cm/min

    which is the rate of change of depth.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    If the depth of water is d, then the radius of th surface is r=d/sqrt(3) since
    the semi vertical angle is 30 degrees. So the volume of water is:

    V = (1/3) pi r^2 d = (1/3) pi d^3/ sqrt(3)

    (volume of cone formula used)

    Now we are told that dV/dt = 100 pi cc/min. so we differentiate the formula
    for V wrt t:

    dV/dt = [1/(3 sqrt(3)] pi 3 d^2 dd/dt

    so when d=5 cm we have:

    dd/dt = 4 sqrt(3) cm/min

    which is the rate of change of depth.

    RonL
    Amazing! There's just one simple thing I don't understand if you could explain it to me it'd be brilliant. When you say "3 sqrt(3)" is that 3 (square root sign)3
    And when you say "(1/3) pi d^3/ sqrt(3)" is that 1/3 (multiplied by) pi d(to the power of 3) ALL OVER the square root of 3?

    Sorry, I know I sound like an idiot but I'm not used to Maths Speak on the internet..
    Thanks a million for the help.
    Last edited by Fionnan; May 17th 2007 at 02:57 PM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    If the depth of water is d, then the radius of th surface is r=d/sqrt(3) since
    the semi vertical angle is 30 degrees.
    The above is because the sides of a 30,60,90 triangle are in the ratio
    1:sqrt(3):2

    So the volume of water is:

    V = (1/3) pi r^2 d = (1/3) pi d^3/ sqrt(3)

    (volume of cone formula used)
    volume of a cone is 1/3 of the area of the base times the height, and
    r=d/sqrt(3), so:

    V = (1/3) pi d^3/3 = pi d^3/9

    So there is a mistake in the formula for V

    Now we are told that dV/dt = 100 pi cc/min. so we differentiate the formula
    for V wrt t:

    dV/dt = [1/(3 sqrt(3)] pi 3 d^2 dd/dt
    So this last should be:

    dV/dt = [1/9] pi 3 d^2 dd/dt

    so:

    100 pi = (1/3) pi d^2 dd/dt

    or:

    300/d^2 = dd/dt

    so when d=5 cm we have:

    dd/dt = 4 sqrt(3) cm/min

    which is the rate of change of depth.

    RonL
    So the final answer is:

    dd/dt = 12 cm/min

    RonL
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