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Math Help - Unusual use for integral, explanation why?

  1. #1
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    Unusual use for integral, explanation why?

    Hi,

    So I have the following problem:

    A vessel is in the form of an inverted circular cone with a semi-vertical angle of 30 degrees. Water is poured in at 5cm^3/min and leaks out from a hole at the bottom at a rate of \frac{\sqrt{h}}{30}cm^3/min. Set up the integral that gives the time for the depth to change from 0.5cm to 5cm.

    I have got the (correct) integral:

    \frac{dt}{dh} = \frac{10*\pi*h^2}{150-\sqrt{h}}

    Then the answers say I should take the integral of the above equation, with respect to h, from h=0.5 to h=5.

    I'm wondering, why? Doesn't taking that integral just find the area of the curve? How does it relate to the question? Or does doing that subtract the time at t=5 from the time at t=0.5?

    Just confused about the why. Thanks!
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  2. #2
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    You are right! It does find the area UNDER the curve. However, it is not just any ordinary area. The x-axis (or h-axis) measures distance (that is length). The y-axix measures dt/dh which is the rate of change of time with respect to distance (h). That is the y axis measures time/length. If you have a square with corners (0,0),(a,0),(0,b),(a,b) in the dt/dh versus h plane, it's area is just a*b and the units are length*time/length = time. So the area actually signifies time!
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  3. #3
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    Ah, much clearer now. Thanks heaps!
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