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Anna35 [415]
4 years ago
11

A cylindrical water trough has a diameter of 6 feet and a height of 2 feet. It is being filled at the rate of 4 ft^3/min. How fa

st is the water level rising when the water is 1 foot deep?
Mathematics
1 answer:
viva [34]4 years ago
8 0
Volume of Cylinder V = πr²h

diameter = 6 feet, radius = 6/2 = 3 feet

dV/dt =  (dV/dh) * (dh/dt)  by change rule.

V =  <span>πr²h

dV/dh =  </span>πr² = π*3² = 9π  ft²<span>

dV/dt = 2 ft</span>³/min

<span>dV/dt =  (dV/dh) * (dh/dt)
</span>
2 ft³/min = 9π ft²   * (dh/dt)

9π ft²   * (dh/dt) = 2 ft<span>³/min
</span>
(dh/dt) = (2 ft³/min)/(9π ft<span>²)   

(dh/dt) = (2/</span>9π)  ft/min

So the water level is rising at  (2/9<span>π)  ft/min.</span>
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