Results 1 to 4 of 4

Math Help - average value word problem

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    249

    average value word problem

    Water is run at a constant rate of 1 ft^3/min to fill a cylindrical tank of radius 3ft and height 5ft. Assuming that the tank is initially empty, make a conjecture about the average weight of the water in the tank over the time period required to fill it, and then check your conjecture by integrating. (Take the weight density of water to be 62.4 lb/ft^3).

    i found the volume of the cylinder to be 235.619 ft^3 which means it will take 235.619 min to fill the tank. so i set up the integral as (1/235.619) integral from (0 to 235.619) of (62.4) dt. i didn't get the correct answer. the correct answer is 1404(pi). please help me set up the correct integral.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,021
    Thanks
    280
    The volume of the tank is:
    <br />
V = \pi R^2h = \pi(3 ft)^2*5ft = 45 \pi ft^3<br />

    Since the tank fills at a constant rate of  1 ft^3/min , it takes the tank 45\pi min to fill, and the weight of water as a function of time is:

    <br />
W = \int _0 ^T ( 1 ft^3/min * 62.4 lb/ft^3) dt = T*62.4 lb/min.<br />

    The "average" weight is simply the weight when the tank is half way through the time required to fill it, which is at T = \frac 1 2 45 \pi \ min:
    <br />
W = \frac 1 2 * 45 \pi \ min * 62.4\ lb/min = 1404 \pi \ lbs<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2008
    Posts
    249
    how come after you take the integral you don't divide by 45pi? isn't the formula for average value the integral divided by (b-a) which in this case is (T-0) = T = 45pi?
    Last edited by oblixps; January 13th 2010 at 08:52 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,021
    Thanks
    280
    If we did this in a more formal way it would be like this:

    <br />
W_{Avg} = \frac 1 T \int _{t=a} ^{t=b} w(t)*dt<br />

    For your problem the weight as a function of time is  w(t) = \rho R t , where  \rho is the density of water = 62.4 lb/ft^3, and R is the fill rate = 1 ft^3/min. The limits on the integral are a = 0 min and b = 45  \pi min. So:

    <br />
W_Avg = \frac 1 {45 \pi} \int _ 0 ^{45 \pi} 62.4 \frac {lb} {ft^3} * 1 \frac {ft^3} {min} t dt = \frac 1 {2* 45 \pi} 62.4 * (45 \pi) ^ 2 = \frac 1 2 62.4 *45 \pi = 1404 \pi \  lb<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Average/Rate/Time Word Problem.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 27th 2011, 11:01 AM
  2. Replies: 5
    Last Post: May 26th 2011, 05:11 AM
  3. Replies: 2
    Last Post: March 13th 2010, 04:49 AM
  4. Average problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 10th 2010, 11:29 AM
  5. Replies: 3
    Last Post: January 31st 2010, 01:18 PM

Search Tags


/mathhelpforum @mathhelpforum