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Math Help - Related rates....a little harder

  1. #1
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    Related rates....a little harder

    Water is leaking out from a spherical tank of radius 13 feet such that the water level in the tank is falling at a constant rate of 2 ft/hr. At what rate is the radius of the top surface of the water decreasing when the depth of the water in the tank is 8 feet?

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  2. #2
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    Cross section radius:

    Let R=13 ft be the radius of the spherical tank, h be the depth of the water in the tank, and r be the radius of the (circular) surface of the water. Drawing a vertical cross-section of the tank, we see that the distance from the surface to the center of the sphere is R-h, and that a radius r of the surface and the center of the sphere form a right triangle with legs R-h, r, and hypoteneuse R
    Thus r^2 + (R-h)^2 = R^2 \implies r = \sqrt{R^2 - (R-h)^2}

    We find \frac{dr}{dt} using the chain rule:

    \frac{dr}{dt} = \frac{dr}{dh}\frac{dh}{dt}

    We know \frac{dh}{dt} is -2 ft/hr, so we can obtain \frac{dr}{dt} in ft/hr as a function of h from the above. Then just put in R=13 ft and h=8 ft to get the answer.


    --Kevin C.
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