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Math Help - differential modelling

  1. #1
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    differential modelling

    Can someone pls help me out with this question.
    As water floes out from a straight cylindrical vessel, the rate of change of the height of the free surface level is directly proportional to the square root of the height.It os observed that the height at t=o is h02.25m and at t=90s h=1.75m.When will be height be h=1.25m
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by lakuzy100 View Post
    Can someone pls help me out with this question.
    As water floes out from a straight cylindrical vessel, the rate of change of the height of the free surface level is directly proportional to the square root of the height.It os observed that the height at t=o is h02.25m and at t=90s h=1.75m.When will be height be h=1.25m
    The statement

    the rate of change of the height of the free surface level is directly proportional to the square root of the height
    tells us how to set up the differential equation.

    It tells us that the rate of change of the height \left(\frac{\,dh}{\,dt}\right) is directly proportional to the square root of the height \left(\sqrt{h}\right)

    Thus, our DE is \frac{\,dh}{\,dt}=k\sqrt{h}, where k is some constant of proportionality.

    To solve, apply the separation of variables technique:

    \frac{\,dh}{\,dt}=k\sqrt{h}\implies \frac{\,dh}{\sqrt{h}}=k\,dt

    Integrating both sides, we see that

    2\sqrt{h}=kt+C\implies h=\left(\tfrac{1}{2}kt+C\right)^2

    Use the first condition to find C.

    Then use the second condition to find k.

    Finally, find the time it takes to reach a height of 1.25.

    Can you take it from here?

    --Chris
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