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Thread: "Related Rates" Problem #1

  1. #1
    s3a is offline
    Super Member
    Nov 2008

    "Related Rates" Problem #1

    I read my book and understood what the book was saying but when I try to do the problems, I just can't seem to get them right and I am having a lot of trouble with them. Here is 1 of my 3 questions (the 2 other questions were posted each in their own thread):

    "A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of (12 ft^3)/min, how fast is the water level rising when the water is 6 inches deep?"

    If someone could explain how to solve this problem in detail, it would be greatly appreciated! I attached my work (even though I think that I am so lost that it's kind of pointless to attach).

    Thanks in advance!
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  2. #2
    Super Member
    Jun 2009
    First thing is that your diagram is wrong. This is simply a 'v' shaped trough. That is, if you take a vertical section through the trough at right angles to its length, what you see is a v shape, an isosceles triangle.

    Next, pretty much all of these related rates problems use the function of a function rule.
    For differentiation purposes the rule is usually written as \frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx},, but in 'practical' examples, it doesn't matter what letters are used.
    In this example the variables are, V, the volume of water, h, the height of the water and t, time.
    You are given the rate of change of volume, \frac{dV}{dt}=12.
    What you are required to calculate is the rate of change of height, \frac{dh}{dt}, (when h=0.5 ).
    You now have to figure out a suitable function of a function relationship linking these quantities.
    It is : \frac{dV}{dt}=\frac{dV}{dh}.\frac{dh}{dt}.
    In order to make use of this, you need to calculate \frac{dV}{dh}, and in order to calculate that, you need a relationship between Vand h.

    Now, when the height of the water in the trough is hft., the cross-sectional area of the water will be 3h^{2}/2, (use similar triangles to work out the 'width' of the water, then it's simply base times height divided by 2), in which case V=15h^{2}.

    So, use this to calculate \frac{dV}{dh}, put the whole lot together and calculate \frac{dh}{dt}.
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