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Related Rates Problem

  1. Feb 27, 2009 #1
    There is a trough that is 10 meters long, 6 meters wide, and is in the shape of an equalateral triangle. The volume is changing at a rate of 0.2 m/s a second. The goal is to find the rate at which the height is changing at 2m.

    So I initially set the problem up as V=0.5bhl (where b=width, h=height, and l=length).

    Then I took the derivative of both sides & found
    dv/dt=0.5bl(dh/dt)

    From here I plugged in the numbers 0.2=0.5(10)(6)(dh/dt)
    & found dh/dt=1/150.

    However, I don't have a place to plug height in for 2m. So would it be correct to reason dh/dt is independent from the height? Or do I have something incorrect in my setup?
     
  2. jcsd
  3. Feb 27, 2009 #2

    Mark44

    Staff: Mentor

    You have a mistake.
    Your formula for dV/dt is incorrect, since it assumes that only h and V are changing. If you draw a picture of the trough with some water in it, you should see that b is not constant.

    Your formula for volume V is correct, but it doesn't show the relationship between b and h, where b is the width of the water across its top edge, and h is the depth of the water. The only quantity that isn't changing in this problem is the length of the water, 10 meters.

    You are given that the cross section of the trough is an equilateral triangle, which tells you that the interior angles of the triangle are all equal. Use that knowledge to get a relationship between the height h of water, and b, the width across the top edge of the water. Then write your formulas for volume V and dV/dt.
     
  4. Feb 27, 2009 #3

    J$C

    User Avatar

    Remember that when the volume is changing, the height and width are also changing. The size of the trough will be the same, but since the amount of liquid is changing the height of the liquid in the trough is a function, and so is the width. If you take some liquid out of the trough the width is going to be smaller, right? Only the length stays constant.

    So you need to consider that you have more than one function on the right hand side of your volume equation when differentiating. Try expressing the width as a function of height to leave yourself with only one function on the right hand side. This time however it will not be a linear function and when you take the derivative remember to use implicit differentiation.
     
  5. Feb 27, 2009 #4

    J$C

    User Avatar

    Well it looks like Mark beat me to it. What he said!
     
  6. Feb 27, 2009 #5
    >.< I forgot about that. Thanks it makes sense now. :)
     
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