Results 1 to 4 of 4

Math Help - rate of flowing water

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    16

    Exclamation rate of flowing water

    The ends of a trough are equilateral triangles with one vertex pointing down. The trough is 9 feet long. When the water in the trough is 2 feet deep, its depth is increasing at the rate of 1/2 foot per minute. At what rate is water flowing into the trough at that moment?

    really could use some help getting this problem solved !
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by hotdogking View Post
    The ends of a trough are equilateral triangles with one vertex pointing down. The trough is 9 feet long. When the water in the trough is 2 feet deep, its depth is increasing at the rate of 1/2 foot per minute. At what rate is water flowing into the trough at that moment?

    really could use some help getting this problem solved !
    are the side lengths of the equilateral triangle given?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    16
    No they are not, that is why it is I am kinda stumped on how to do this one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by hotdogking View Post
    No they are not, that is why it is I am kinda stumped on how to do this one.
    now that I think about it, doesn't matter anyway ... all equilateral triangles are similar.

    let s = side length of water triangle

    h = height of water triangle

    for equilateral triangles, \cos(30) = \frac{h}{s} =  \frac{\sqrt{3}}{2}

    s = \frac{2h}{\sqrt{3}}

    Area of an equilateral triangle ...

    A = \frac{\sqrt{3}}{4} s^2 =  \frac{\sqrt{3}}{4} \cdot \frac{4h^2}{3} = \frac{\sqrt{3}}{3} h^2

    V = 9A = 3\sqrt{3} \cdot h^2

    \frac{dV}{dt} = 6\sqrt{3} h \cdot \frac{dh}{dt}

    substitute and calculate the value of \frac{dV}{dt}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 20th 2010, 03:18 PM
  2. Replies: 6
    Last Post: September 20th 2009, 02:53 PM
  3. Replies: 7
    Last Post: February 6th 2009, 01:45 PM
  4. Rate at which water is draining from the tank?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 14th 2008, 09:38 AM
  5. Water flowing into hemisphere rates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 21st 2006, 11:15 PM

Search Tags


/mathhelpforum @mathhelpforum