Results 1 to 5 of 5

Math Help - Depth of water when the water level is falling the least rapidly

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    40
    Awards
    1

    Depth of water when the water level is falling the least rapidly

    Can someone help me with this problem? Thank you.

    A spherical tank of radius 10ft is being filled with water. When it is completely full, a plug at its bottom is removed. According to Torricelli's law, the water drains in such a way that dV/dt = -k*sqrt(y), where V is the volume of water in the tank and k is a positive empirical constant.

    a) Find dy/dt as a function of the depth y
    b) Find the depth of water when the water level is falling at least rapidly (you will need to compute the derivative of dy/dt with respect to y)

    Regarding a), I have done this:
    V=1/3*pi*y^2(3a-y) where a is the radius
    dV/dt=10*pi*2y(dy/dt) - (1/3)*pi*3y^2(dy/dt)
    -k*sqrt(y) = (dy/dt)(20*pi*y - pi*y^2)
    dy/dt=-[(k*sqrt(y))/pi*y(20-y)]

    And this is what I came up for dy/dt as a function of the depth y

    But I am not sure how to proceed with point b) Any help will be apprecited.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Maybe I am misunderstanding, but it appears you're using the equation for the volume of a cone........the tank is spherical.

    dV equals the area of the liquid multiplied by dy.

    Try to equate the area of the liquid in terms of h.

    By the diagram, since the tank is a sphere, the surface area of the liquid is a circle with radius

    \sqrt{25-(y-5)^{2}}

    The area of a circle is {\pi}r^{2}, so we have the area of the surface of the liquid at time t as

    {\pi}(\sqrt{25-(y-5)^{2}})^{2}=10{\pi}y-{\pi}y^{2}


    So, we have (10{\pi}y-{\pi}y^{2})\frac{dy}{dt}=-k\sqrt{y}

    \frac{10{\pi}y-{\pi}y^{2}}{\sqrt{y}}dy=-kdt

    Now, we have a separated DE.
    Last edited by galactus; November 24th 2008 at 06:38 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2007
    Posts
    40
    Awards
    1
    V=1/3*pi*y^2(3a-y) is the volume of a spherical tank that is partly filled - this way I have the depth of the sphere in the equation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Oh, I see. At first glance it looked like the cone formula. My bad.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2007
    Posts
    40
    Awards
    1
    I just saw the rest of your reply. I see that I am on the right way.

    I now wonder how to find the depth of water when the water level is falling at least rapidly. They suggest to compute the derivative of dy/dt with respect to y. Does this mean that I have to take the derivative of the derivative and why it will give me the depth of water when the water level is falling at least rapidly.

    I just did the b). For those interested I took the derivative of dy/dt and found one critical value. Thank you for the help!
    Last edited by hasanbalkan; October 22nd 2007 at 06:06 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. change in depth of water
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 31st 2010, 09:26 PM
  2. Rate of Change: Depth of Water in a Tank
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 17th 2010, 03:34 PM
  3. Replies: 3
    Last Post: February 15th 2010, 12:53 PM
  4. Replies: 6
    Last Post: December 8th 2009, 08:44 PM
  5. calculate water depth rate..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 19th 2009, 10:26 AM

Search Tags


/mathhelpforum @mathhelpforum