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

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

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