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Relative rates help need.

  1. Dec 13, 2012 #1
    A trough is 8 ft long and its ends have the shape of isosceles triangles that are 2 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 11 ft3/min, how fast is the water level rising when the water is 9 inches deep?


    I got


    V=2h^2 (8)/2
    V=8h^2
    dv/dx=16h
    11=16h

    how do I get the height Im a confused and is everything else right? and can someone switch this to the homework thread just noticed
     
  2. jcsd
  3. Dec 14, 2012 #2

    HallsofIvy

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    The crucial point is that what ever the height of the water a cross section will have the same "shape". The base will be twice the height. Since the area of a triangle is "1/2 base times height", the are of a cross section will be "1/2 times twice the height times the height" or just [itex]h^2[/itex].

    Yes, this is correct.

    There is no "x" in the problem! What is true is that dv/dh= 16h.

    But dV/dh is not 11. You told that "the trough is being filled with water at a rate of 11 ft3/min". The denominator is in minutes- dv/dt= 11 cubic feet per minute where t is the time. What you need to do is to differentiate both sides of V= 8h^2 with respect to time, using the chain rule.

    You don't get the height- that is given as 9 inches. You are asked "how fast is the water level rising" which is a rate of change: dh/dt.
     
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