Results 1 to 3 of 3

Math Help - Integration help needed

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    17

    Integration help needed

    Hey guys, I need some help with some integration problems. Some I could solve already but there are two where I need help. If you can help me, however, please don't just say me: integrate this, but also show me how to do it, that would be really nice. Thanks a lot for everyone who helps me.


    1: A large bottle with, 4m high and with a radius of 2m, loses water. Originally, the bottle was full and lost water at a rate proportional to the square root of the depth of the remaining water.

    I'm given that the depth of the water is 1m after 2 hours. The question is after how many hours the tank is empty. Please help me with this question.

    2. In a freezing machine the temperature is constantly 5 degree. An 100 degree hot object is put in it. One minute later the object is only 80 degree hot. When will it reach a temperature of 10 degree?

    Thanks a lot for every advice. But please also give me explanations if you can. I very much appreciate any help. Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2007
    Posts
    17
    Can nobody help me with these problems?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,685
    Thanks
    616
    Hello, Instigator!

    I have a long, looong solution . . . maybe someone can shorten it.


    1) A large bottle, 4m high and with a radius of 2m, loses water.
    Originally, the bottle was full and lost water at a rate proportional
    to the square root of the depth of the remaining water.

    The depth of the water is 1m after 2 hours.
    After how many hours the tank is empty?

    The volume of a cylinder is: . V \:=\:\pi r^2h

    Since r = 2, the volume of the water is: . V \:=\:\pi(2^2)h \:=\:4\pi h
    . . Then: . \frac{dV}{dt} \:=\:4\pi\left(\frac{dh}{dt}\right)

    But we are told that: . \frac{dV}{dt} \:=\:k\sqrt{h}

    So we have: . 4\pi\left(\frac{dh}{dt}\right) \:=\:k\!\cdot\!h^{\frac{1}{2}} . . . a differential equation

    . . Separate variables: . 4\pi h^{-\frac{1}{2}}dh \:=\:k\,dt

    . . Integrate: . 8\pi h^{\frac{1}{2}} \:=\:kt + C


    When t = 0,\;h = 4
    . . We have: . 8\pi\sqrt{4} \:=\:k(0) + C\quad\Rightarrow\quad C \,=\,16\pi

    The equation (so far) is: . 8\pi\sqrt{h} \:=\:kt + 16\pi

    When t=2,\:h=1
    . . 8\pi\sqrt{1} \:=\:k\!\cdot\!2 + 16\pi\quad\Rightarrow\quad k \,=\,-4\pi

    The equation is: . 8\pi\sqrt{h} \;=\;-4\pi t + 16\pi\quad\Rightarrow\quad\sqrt{h} \:=\:2 - \frac{t}{2}

    Hence, the height function is: . h \;=\;\left(2 - \frac{t}{2}\right)^2


    If the tank is empty, then h = 0.

    We have: . 0 \:=\:\left(2 - \frac{t}{2}\right)^2\quad\Rightarrow\quad t \,=\,4

    Therefore, it takes 4 hours to empty the tank.



    But check my reasoning and my work . . . please!
    .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 19
    Last Post: July 12th 2010, 12:40 AM
  2. Integration help needed, thanks
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 27th 2010, 12:41 AM
  3. basic integration help needed
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 8th 2010, 11:14 AM
  4. Substitution integration help needed
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 27th 2009, 06:16 AM
  5. Help on indefinite integration needed!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 2nd 2008, 04:06 AM

Search Tags


/mathhelpforum @mathhelpforum