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A cylindrical swimming pool (whose center axis is vertical) is being filled from a fire hose at a rate of 5 cubic feet per second. If the pool is 40 feet across, how fast is the water level increasing when the pool is one third full? 12 hours ago
well you know the rate of change in the volume with respect to time \[\frac{dV}{dt} = 5\] you know the volume of the cylinder \[V = \pi r^2 h\] then rewriting the equation with h as the subject \[h = \frac{V}{2 \pi r^2}\] so then the rate of change in height with respect to time should be \[\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}\] and thinking about the problem... I think its a constant rate of change
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