Find the derivative of a balloon's circumference

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Homework Help Overview

The discussion revolves around the relationship between the volume of a balloon and its circumference as the volume increases over time. The original poster attempts to express the rate of change of the circumference with respect to time using the formulas for the volume and circumference of a sphere.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the rates of change of volume and circumference, with the original poster attempting to derive an equation relating these rates. Questions arise regarding the interpretation of the graph of circumference versus time, particularly concerning the behavior of the function as the balloon expands.

Discussion Status

Some participants provide clarifications about the nature of the rates of change, noting that while the circumference is increasing, the rate of increase may be decreasing. Others question the assumptions made about the rate of volume increase and the need for additional information to proceed further.

Contextual Notes

There is a noted constraint regarding the lack of information on the specific rate of volume increase (dV/dt), which is essential for further analysis. Additionally, the original poster mentions being in a pre-calculus class, which limits their familiarity with integration, a necessary step for finding circumference as a function of time.

JosephTraverso2
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A balloon's volume is increasing at a rate of dV/dt. Express the rate of change of the circumference with respect to time (dc/dt) in terms of the volume and radius.

Homework Equations


Vsphere = (4/3)(π)(r^3)
C = (2)(π)(r)

The Attempt at a Solution

[/B]
My strategy was to come up with two separate equations of dr/dt and substitute leading to an equation relating dv/dt and dc/dt

dV/dt = (4)(π)(r^2) (dr/dt)
dc/dt = (2)(π)(dr/dt)

(dc/dt) (1/2π) = dr/dt
Substitute...

(4)(π)(r^2) (dc/dt) (1/2π) = dv/dt

Canceling like terms

(2r^2)(dc/dt) = dv/dt

(1/2r^2) (dv/dt) = (dc/dt)

Now I was curious and wanted to graph the function just for fun. I know dv/dt is some number so I assumed it is 3. This yields
3/2r^2 = dc/dt
If I were to graph this function on a graphing utility would I be getting the circumference on the y-axis and time on the x? I ask this because the function is decreasing but the circumference of the balloon is increasing. Please help.
 
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JosephTraverso2 said:
I ask this because the function is decreasing but the circumference of the balloon is increasing. Please help.
Where did you get this idea? dC/dt is positive which means that C increases as time increases. The r2 in the denominator means that C is increasing at a decreasing rate as the balloon grows larger, not that C is decreasing.
If you were to graph this, you need to integrate first to find C as a function of time. You would plot C on the vertical axis and t on the horizontal axis.
 
Last edited:
Clear and concise answer much thanks. But I'm currently in a pre-calculus class and we have not gone over integration yet. If it is not too much trouble can you please show how to do it to obtain circumference as a function of time? Thanks once again
 
JosephTraverso2 said:
Clear and concise answer much thanks. But I'm currently in a pre-calculus class and we have not gone over integration yet. If it is not too much trouble can you please show how to do it to obtain circumference as a function of time? Thanks once again
Is the statement of the problem exactly as was given to you? If so, then you cannot figure out what C as a function of time looks like because you need to know what dV/dt is, i.e. how the volume is increasing. You assumed as an example that it is 3, but is it really? Furthermore, is dV/dt constant or does itself depend on time? So if dV/dt is not given, you cannot do anything more than you have already done.
 
JosephTraverso2 said:
But I'm currently in a pre-calculus class and we have not gone over integration yet.
Thread moved. Problems involving derivatives are definitely in the bailiwick of calculus, regardless of whether the class you are taking is called "precalculus."
Please post problems involving derivatives and related rates in the Calculus & Beyond Homework section - thanks...
 

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