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Really hard related rates problem

  1. Jan 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A trough has an isosceles trapezoidal cross section as shown in the diagram. Water is draining from the trough at 0.2m^3/s

    At what rate is the surface rea of the water decreasing?

    Dimensions are: base width=0.4m,
    top width= 0.8m
    length=2.5m
    height=0.5m

    2. Relevant equations



    3. The attempt at a solution
    Ok I really really don'tknow where to start, if maybe someone can give me a clue
     
  2. jcsd
  3. Jan 16, 2007 #2

    mjsd

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    ok, I can't be completely certain about the shape, but my guess based on your info, is that it is a lying down "cylinder" of height 2.5m with a isosceles trapezoidal base. So you have no problem working out the volume of this trough... just base x height eh?... next relate rate of change of water volume to rate of change of surface area... now I assume your "surface area" is the area viewing from the top (ie, a rectangle) so what can really change when water level decreases is just width of the rectangle...
    now, that means, you need to first relate water volume change to the width change ...

    since you haven't shown me what you know, it is very hard for me to say things at a level that suits you. so .. that's all I'll say for the moment.
     
  4. Jan 16, 2007 #3

    HallsofIvy

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    Start by drawing a picture. The trough is a trapezoid with those dimensions but the water in the trough is not- it is, however, a trapezoid with that lower base, height a variable, say "y", and top base the distance between the two sides at that y. You ought to be able to calculate that from "similar triangles" or be setting up an appropriate coordinate system and writing the equations of the lines representing the two sides. Once you have the dimensions of the 'water' trapezoid, you can write down the equation for the volume as a function of y and differentiate it- you are given the rate of change, i.e. derivative, of the volume. That will allow you to calculate dy/dt. Now write down the equation of the area of the water surface, as a function of y, and differentiate that to find the rate of change of surface area.
     
  5. Jan 16, 2007 #4
    yea the differentiation part is easy just the similar triangles part i don't get how do i find the volume equation, and i guess area
     
  6. Jan 17, 2007 #5

    HallsofIvy

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    Of course, a trapezoid is NOT a triangle, but you could imagine the two sides continued until they intersect and use that triangle.

    Here was my first thought: set up a coordinate system so that the origin is at the center of the base and the x-axis is along the base. Then, since the base is 0.4 m long, the ends of the base are at (0.2, 0) and (-0.2, 0). Since the top is 0.8 m wide and the height is 0.5 m, the other ends of the lines representing the sides are (0.4, 0.5) and (-0.4, 0.5). What is the equation of the line through (0.2, 0) and (0.4, 0.5)? Find the distance between the x-values of those two lines for a given y (or because of the symmetry, just double x on the line through (0.2, 0) and (0.4, 0.5) as a function of y) which also gives y as a function of x. That gives the height y as a function of the "top" base of the trapezoid of water, x.

    Perhaps even simpler: since every thing here is straight lines, y (height) as a function of x (width) must be linear: y= Ax+ B. When y= 0 (base), x= 0.4. When y= 2.5 (top), x= 0.8. That gives two equations to solve for A and B.

    The area of a trapezoid is [itex]\frac{1}{2}h(b_1+ b_2)[/itex] so, for this problem, the volume, area times length is [itex]V= 2.5\frac{1}{2}y(0.4+ 2x)[/itex] where, again, you will have written y as a function of x. You are told that dV/dt= -0.2 so differentiating that equation will give you dx/dt as a function of x.

    The top surface of the water is always a rectangle with length 2.5 m and height x (again, a function of y). It's area is A= 2.5x so dA/dt= 2.5 dx/dt.
     
  7. Jan 17, 2007 #6
    So what exaactly do I do with the first paragraph?
    And the second paragraph isn't it suppost to be [tex]V=(0.4+0.4+2x)h\frac{1}{2}(2.5)[/tex]
    ???
     
  8. Jan 17, 2007 #7
    CAN SOMEONE PLEASE HELP ME UNDERSTAND THIS PARAGRAPH. I THINK HE HAS THE WRONG DRAWING, SET UP
     
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