How to Solve a Conical Tank Calculus Problem

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Homework Help Overview

The discussion revolves around a calculus problem involving a conical tank that is draining water. The tank has specific dimensions, and the question focuses on determining the rate at which the water level is dropping when the radius of the water is 3 meters.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the radius and height of the water in the tank, with some attempting to derive the volume of water in terms of radius and height. There is also a discussion about the correct variable to differentiate, questioning whether the rate of change of radius or height is being sought.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the correct approach to find the rate of change of the water level. Some participants have provided guidance on the relationships between the variables involved, while others have pointed out potential errors in the original calculations.

Contextual Notes

There is a noted discrepancy in the interpretation of the answer, with different participants suggesting different forms of the result. Additionally, the problem's setup and the specific variables to differentiate are under scrutiny.

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I have a calculus question and was wondering if some could help me see what I am doing wrong in this question. Thank you

A conical tank with an altitude of 10m and whose base has a radius of 4m is mounted with its vertex down. The tank is full of water which is draining through the vertex at the rate of 5 m^3/min. how fast is the level of the water dropping when the radius is 3?

the answer is 5pi/9

this is how i did it:

r/h=4/10
h=5pi/2

V=[(pi)(r^2)(h)]/3
V=[(pi)(r^2)(5r/2)]/3
V=[pi5r^3]/6
dv/dt=[15pi(r^2)]/6] * dr/dt
5=[(15pi(3^2)]/6] * dr/dt
dr/dt= (2pi)/9
 
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Firstly, the answer is actually [tex]\frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}[/tex], not [tex]\frac{5\pi}{9}\frac{\mbox{m}}{\mbox{min}}[/tex].

Secondly, are you sure [tex]\frac{d r}{d t}[/tex] is what the question is looking for? Re-read it.
 
Last edited:
punjabi_monster said:
I have a calculus question and was wondering if some could help me see what I am doing wrong in this question. Thank you

A conical tank with an altitude of 10m and whose base has a radius of 4m is mounted with its vertex down. The tank is full of water which is draining through the vertex at the rate of 5 m^3/min. how fast is the level of the water dropping when the radius is 3?
the answer is 5pi/9

this is how i did it:

r/h=4/10
h=5pi/2

V=[(pi)(r^2)(h)]/3
V=[(pi)(r^2)(5r/2)]/3
V=[pi5r^3]/6
dv/dt=[15pi(r^2)]/6] * dr/dt
5=[(15pi(3^2)]/6] * dr/dt
dr/dt= (2pi)/9

Notice how the question asks for how fast the level of the water is dropping. This would mean a change in height with respect to time, not the radius. Since it's asking for the rate of change when the radius is 3 then:

[tex]r=\frac{3h}{10}[/tex]
[tex]V=\frac{1}{3} \pi (\frac{3h}{10})^2 h[/tex]
[tex]V= \frac {3}{100} \pi h^3[/tex]
[tex]\frac {dV}{dt}= \frac{9}{100} \pi h^2 \frac{dh}{dt}[/tex]

Solving for dh/dt should get you the answer [tex]\frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}[/tex] as stated above by Data.
 
You actually need to use [tex]r = \frac{2h}{5}[/tex]. But other than that you're completely right.
 
thanks for your help.
 

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