Related Rates-Conical Reservoir

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Homework Help Overview

The problem involves a conical reservoir with specific dimensions and a scenario where water is being filled at a constant rate. Participants are tasked with determining the rate at which water is entering the reservoir based on the changing depth of the water.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss expressing the volume of the cone in terms of a single variable, specifically the height of the water. There is a focus on the relationship between the radius and height of the cone, with attempts to derive a formula for volume based on these variables.

Discussion Status

Some participants have provided guidance on expressing the volume in terms of height and have confirmed the relationship between radius and height. There is ongoing exploration of how to differentiate the volume with respect to time and apply the given depth of water.

Contextual Notes

Participants are working under the constraints of the problem statement, including the fixed dimensions of the conical reservoir and the rate of change of water depth. There is an emphasis on ensuring the correct relationships and formulas are established before proceeding further.

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Homework Statement


A conical reservoir has a depth of 24 ft and a circular top of radius 12 ft. It is being filled so that the depth of the water is increasing at a constant rate of 4ft/hour. Determine the rate of which the water is entering the reservoir when the depth is 5 ft.


Homework Equations


V=(1/3)pi r^2 h


The Attempt at a Solution


I'm not really sure where to start even, I drew a picture but that's it.
 
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The rate of water coming in is dV/dt, right? Your first job is probably to express V in terms of a single variable, say h. Given the proportions of your cone, r and h are proportional. Can you write an expression relating them?
 
Dick said:
The rate of water coming in is dV/dt, right? Your first job is probably to express V in terms of a single variable, say h. Given the proportions of your cone, r and h are proportional. Can you write an expression relating them?
So, is it just r=h/2 and V=(1/3)pi (h/2)^2 h, which is (pi h^3)/12?
Than V'(dh/dt) would be dV/dt right?
or am I missing something still?
 
Sure, r=h/2 and V(t)=pi*h(t)^3/12. Now find dV/dt in terms of h(t) and h'(t) and solve for dV(t)/dt given h(t)=5 ft.
 

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