Question

A water heater that has the shape of a right cylindrical tank with a radius of 1 ft and a height of 4 ft is being drained. How fast is water draining out of the tank (in ft³/min) if the water level is dropping at 6 min/in?

Accepted Answer

First, identify the shape of the tank. It is a right cylindrical tank, which means its volume can be calculated using the formula for the volume of a cylinder: $$ V = \pi r^2 h $$, where $$ r  is $$the radius and $$ h  is $$the height. Given that the radius $$ r  is 1 ft $$and the height $$ h  is 4 ft$$, the volume of the tank is $$ V = \pi (1)^2 h = \pi h . $$The problem states that the water level is dropping at a rate of 6 min/in. To find the rate at which the volume is changing, we need to relate the change in height to the change in volume. Use the chain rule to differentiate the volume with respect to time: $$ \frac{dV}{dt} = \pi \frac{dh}{dt} . $$Convert the rate of change of the water level from min/in to ft/min. Since 1 inch is $$ \frac{1}{12}  ft$$, the rate $$ \frac{dh}{dt}  is$$ $$ 6 	ext{ min/in} 	imes \frac{1}{12} 	ext{ ft/in} = \frac{1}{2} 	ext{ ft/min} . $$Substitute $$ \frac{dh}{dt} = \frac{1}{2} 	ext{ ft/min} $$ into the differentiated volume formula: $$ \frac{dV}{dt} = \pi 	imes \frac{1}{2} . $$This gives the rate at which the water is draining out of the tank in ft³/min.

A water heater that has the shape of a right cylindrical tank with a radius of 1 ft and a height of 4 ft is being drained. How fast is water draining out of the tank (in ft³/min) if the water level is dropping at 6 min/in?

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Key Concepts

Volume of a Cylinder

Related Rates

Differentiation