Math  /  Calculus

QuestionA conical tank (vertex down) is 7 meters across the top and 9 meters deep. If the depth of the water (the height) is decreasing at 6.6 meters per minute, what is the change in the volume of the water in the tank when the height of the water in the tank is 4 meters?
Include units on your final answer, and your answer must be entered as number (not 5*7+3).

Studdy Solution
Substitute h=4 h = 4 meters and dhdt=6.6 \frac{dh}{dt} = -6.6 meters per minute into the differentiated equation:
dVdt=36.75π24316(6.6)\frac{dV}{dt} = \frac{36.75 \pi}{243} \cdot 16 \cdot (-6.6)
Calculate:
dVdt=36.75π24316(6.6)50.2655 cubic meters per minute\frac{dV}{dt} = \frac{36.75 \pi}{243} \cdot 16 \cdot (-6.6) \approx -50.2655 \text{ cubic meters per minute}
The rate of change of the volume of water in the tank is approximately:
50.27 cubic meters per minute \boxed{-50.27 \text{ cubic meters per minute}}

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