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Homework Help: Related rates word problem

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A swimming pool is 20 feet wide, 40 feet long, 3 feet deep at the shallow end and 9 feet deep at its deepest point. If the pool is being filled at a rate of 0.8 feet^3/min, how fast is the water level rising when the depth at the deepest point is 5 feet?

    200710191446396332840199968500003988.jpg


    2. Relevant equations

    I believe the formula for volume of a swimming pool is V = lw((h+H/2)) where h is the shallowest depth and H is the deepest depth? Not 100% sure


    3. The attempt at a solution

    I know that dV/dt = 0.8 ft^3/minute. The dimensions of the pool (20 by 40 ft) are constant and do not change. h can be expressed in terms of H (H/3) and subbed into the equation. I need to find dH/dt when H=5 feet. Other than this I'm completely stuck! Thanks for any help!
     
  2. jcsd
  3. May 1, 2010 #2
    The dimensions of the pool are actually NOT constant. Yes, the width stats at 20 feet, but the length actually depends on how much water is in there. For instance, before the water reached 6ft of depth, its length approaches 34 feet (6+12+16). At 6ft of depth, you switch over to the "shallow end", and it immediately jumps to the full 40ft length.
    I'll take a look at this and get back to you.
    (edit...)
    Ok, we're in business.
    The "big picture" is that we want to express the volume in terms of just ONE other variable. Otherwise, we'd have to use the product rule and we'd need more information, etc...

    There may be an easier way to do this, but ... well, nobody has posted it yet!

    Let's scrap the idea of V = L*W*H.
    The volume of a prism is the area of the base * (another dimension = "width" in our case).
    The "base" is the trapezoid that we see in the picture. The area of this depends on how deep the water is (obviously).

    I'm actually going to break-up the trapezoid into three parts:
    a) left triangle. This is the blue region directly above the "6ft" segment.
    b) rectangle of width 12ft, above the "12ft" segment
    c) triangle on the right, above the "16ft" segment.

    The area of the base is the sum of these three parts. Let's denote the depth of the water by "h".

    a) The dimensions of this triangle are x1 and h. This triangle is similar to a 6x6 triangle (since those are the dimensions of the maximum length and height of this triangle). So we setup the proportion x1/h = 6/6. So x1=h.
    The area of this triangle is .5(length)*(height) = .5(x1)*(h) = .5(h)(h) = .5h2.
    b) The area is 12*h
    c) Use similar triangles to determine the area as a function of h ONLY.

    Add these up. You will get an expression in terms of h ONLY.
    Take the derivative. Don't forget the chain rule (which will give you a factor of dh/dt).
    Plug.
    Chug.
     
    Last edited: May 1, 2010
  4. May 4, 2010 #3
    This helped me solve it correctly! Thank you The Chaz :D
     
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