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Math Help - Help with application: 'Cylinder filling with water, rate of sqrt(x), derive formula'

  1. #1
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    Help with application: 'Cylinder filling with water, rate of sqrt(x), derive formula'

    Hey guys, I'm completley stuck on question 2)a)i).

    Everything I do leads me in a circle, of either the x's cancelling, or me getting no where. Can anyone direct me the right way, how should I set up the problem? Thanks.

    Here's the question, I need help on Q2)a)i):
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  2. #2
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    Hello Phatmat
    Quote Originally Posted by Phatmat View Post
    Hey guys, I'm completley stuck on question 2)a)i).

    Everything I do leads me in a circle, of either the x's cancelling, or me getting no where. Can anyone direct me the right way, how should I set up the problem? Thanks.

    Here's the question, I need help on Q2)a)i):
    When the depth of water is x metres, suppose that the volume of water is V \text{m}^3. Then if the time t is measured in minutes, the rate of increase of V with respect to t is \frac{dV}{dt} \text{m}^3/\text{min}. This is the rate at which water is filling the tank.

    So \frac{dV}{dt}=\sqrt x

    The volume of a cylinder = \pi r^2 h. So when r = 2 and h = x, V = 4\pi x

    \Rightarrow \frac{dV}{dx}=4\pi

    \Rightarrow \frac{dV}{dt}=\frac{dV}{dx}\cdot\frac{dx}{dt}=4\pi  \frac{dx}{dt}

    \Rightarrow 4\pi\frac{dx}{dt}=\sqrt x

    Can you continue from here?

    Grandad
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