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Math Help - Related Rates

  1. #1
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    Related Rates

    Hello fellow mathletes!

    I'm not sure if this is in the right forum but I have come across a related rates question in my textbook and just cannot figure it out.

    A spherical storage tank has a radius of 15m. Water is being pumped in at a rate of 36pi m^3/min. Find the rate of depth increase at the instant that the water level is 6m below the very top of the sphere.

    HINT: rate of depth increase = (rate of inflow)/(surface area of liquid)

    Thanks in advance and look forward to solving your problems,
    Katy
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  2. #2
    MHF Contributor

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    Re: Related Rates

    Generally speaking you find "related rates" by starting with a formula relating the things them selves, then differentiating with respect to time, t, to get a formula relating there rates.

    Here, you are asked about the rate of change of the volume in a sphere with respect to height (or depth). So: if you have a spherical "goldfish bowl", of radius R, filled with water up to height z, what is the volume of the water? One way to do that is to start with the fact that the boundary of the sphere can be modeled as x^2+ y^+ z^2= R^2 and convert to cylindrical coordinates: [itex]r^2+ z^2= R^2[/itex] and integrate from z=0 to z= Z: \int_{\theta= 0}^{2\pi}\int_{z=0}^Z\int_{r= 0}^{\sqrt{R^2- z^2} rdrdzd\theta.
    Last edited by HallsofIvy; April 13th 2014 at 02:47 PM.
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