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The radius of the cone is 14 cm and water height is 9 cm and rate change volume dV/dt = - 2.5 cm³/sec.

\ Here water is leaking (negative value), so the rate of change of volume dv/dt = - 2 ft³/hour. \

The volume of cone is .

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Differentiate implicitly with respect to t to obtain the related - rate equation

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The above equation represent the rate of change of V is related to the rates of change of both h and r.

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Here to find rate is the radius of the top of the water in the tank changing when the depth of the water is 6ft.

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So, the value of h is convert into the r values in the volume equation. Here two cones are available, first one is original cone and second one is shape of cone with water height 6 ft.

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Two similar triangles then ratios of any two sides will be equal.

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. Volume :

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Differentiate implicitly with respect to t.

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If h = 6 ft then and dv/dt = - 2.

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The rate of change of the radius of the top of the water in the tank changing when the depth of the water is 6ft is dr/dt = - 0.04949 ft/hour.

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