Related rates question conical reservoir

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Homework Help Overview

The problem involves a conical reservoir with specific dimensions and rates of water being pumped in and drained out. Participants are tasked with determining the change in diameter of the water surface when the height is at a certain level.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the height and radius of the cone, using the equation for volume. There are attempts to differentiate the volume equation with respect to time and apply the rates of change of volume. Questions arise regarding specific calculations and the interpretation of values used in the equations.

Discussion Status

Some participants have provided calculations and identified potential errors in reasoning. There is acknowledgment of a mistake in the original calculations, and a suggestion to consider the change in diameter rather than just the radius. The discussion appears to be moving towards clarification of the approach taken.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on ensuring the correct interpretation of the problem setup and the relationships between variables.

shanshan
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Homework Statement


A conical reservoir is 50m deep and 200m across the top. Water is being pumped in at a rate of 7000m^3/min, and at the same time, water is being drained out at a rate of 9000 m^3/min. What is the change in diameter when the height of the water in the reservoir is 20m?


Homework Equations


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


The Attempt at a Solution


h=(.5r)
v=(1/3)(pi)(r^2)(.5r)
=(1/6)(pi)(r^3)
v'=(.5)(pi)(r^2)(r')

7000=.5pi(40^2)(r')
= 2.785
-9000= .5pi(40^2)(r')
=-3.581

2.785-3.581 = -0.796 m/min

my answer, however, is supposed to be -1.6m/min
 
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shanshan said:

Homework Statement


A conical reservoir is 50m deep and 200m across the top. Water is being pumped in at a rate of 7000m^3/min, and at the same time, water is being drained out at a rate of 9000 m^3/min. What is the change in diameter when the height of the water in the reservoir is 20m?


Homework Equations


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


The Attempt at a Solution


h=(.5r)
v=(1/3)(pi)(r^2)(.5r)
=(1/6)(pi)(r^3)
v'=(.5)(pi)(r^2)(r')

7000=.5pi(40^2)(r')
= 2.785
-9000= .5pi(40^2)(r')
=-3.581

2.785-3.581 = -0.796 m/min

my answer, however, is supposed to be -1.6m/min

What is (40^2) ?
 
I got the 40 from h=.5r and r=20
 
shanshan said:
I got the 40 from h=.5r and r=20

Well then, very well I suppose.

Your question is to find change in diameter, hence multiply it by 2 and you get your -1.592 ~ -1.6
 
AHHH silly mistake. thankyou!
 

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