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Rate of Change With Volumes

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A trough in the shape of an isosceles trapezoid is 30 cm wide on the bottom, 80 cm on the top, 50 cm tall, and 10 cm long. It is being filled with water at 0.2 cm3/min. How fast is the water level rising when the water is 30 cm deep.

    3. The attempt at a solution

    Well I used a similar triangle deal to eliminate the 'base' of the triangles on each side of the rectangle that make up the trapezoid. Then I wrote this:

    v=((30h)+(.5h)(h)(2)) * 10

    but when I go to derive it, it all hits zero...something is wrong here. What is it?
     
  2. jcsd
  3. Nov 8, 2009 #2
    Hi, I do not think your equation relating volume to height is correct, try working it out again.

    Substitute the equation you obtain and the values given into dh/dt=dh/dV*dV/dt to find your answer.

    Hope this helps.
     
  4. Nov 8, 2009 #3
    Uh...ok. I'll give it a try.
     
  5. Nov 8, 2009 #4

    Mark44

    Staff: Mentor

    That's a very odd looking trough if it's only 10 cm long. Are you sure it isn't 100 cm long instead?
     
  6. Nov 8, 2009 #5
    Whoops, yeah. It's 1 m long, so 100 cm. My bad.
     
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