How to Calculate Water Level and Boat Speed in Conical Reservoir Problems?

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SUMMARY

This discussion focuses on solving two problems related to fluid dynamics and geometry involving a conical reservoir and a dinghy. The first problem calculates the rate at which the water level is falling and the change in the radius of the water's surface when the water is 5 meters deep, using the formula for the volume of a cone. The second problem involves determining the speed at which a dinghy approaches a dock and the rate of change of the angle theta when 10 feet of rope are out. Key equations include the volume of a cone, represented as V = (1/3)πr²h.

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  • Understanding of calculus, specifically related rates
  • Familiarity with the geometry of cones
  • Knowledge of fluid dynamics principles
  • Ability to apply the chain rule in differentiation
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  • Study related rates in calculus, focusing on conical shapes
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  • Explore the principles of fluid dynamics, particularly in conical reservoirs
  • Investigate trigonometric relationships in right triangles for problems involving angles and distances
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Students in calculus or physics courses, educators teaching fluid dynamics, and engineers working with conical structures or fluid systems.

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Homework Statement


A draining conical reservoir. Water is flowing at the rate of 50 m^3/min from a shalloe concrete conical reservoir (vertex down) of base radius 45m and height of 6m.

a. How fast (centimeters per minute) is the water level falling when the water is 5m deep?

b. How fast is the radius of the water's surface changing then? Answer in centimeters per minute.

2. Hauling in a dinghy. A dinghy is pulled toward a dock by a rope from the bow through a ring to the dock 6ft above the bow. The rope is hauled in at the rate of 2 ft/sec.

a. How fast is the boat approaching the dock when 10ft of rope are out?

b. At what rate is the angle theta changing then (see the figure)?

Homework Equations


1.

a. Not given, but cone volume = 1/3*pi*r2*h

The Attempt at a Solution


1.

a. dv/dt = 50/(pi)(45)2(5) or dv/dt = 50/(pi)(45y/6)2(6)

b. Nothing.

2.

a. Nothing.

b. Nothing.
 

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