Find the derivative of a balloon's circumference

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SUMMARY

The discussion focuses on finding the rate of change of a balloon's circumference (dc/dt) in relation to its volume (dV/dt) and radius (r). The relevant equations include the volume of a sphere, Vsphere = (4/3)(π)(r^3), and the circumference, C = (2)(π)(r). The user derives the relationship (1/2r^2)(dv/dt) = (dc/dt) and explores graphing the function, mistakenly believing that circumference decreases while the balloon inflates. Clarification is provided that dc/dt is positive, indicating that circumference increases over time, albeit at a decreasing rate as the radius grows.

PREREQUISITES
  • Understanding of derivatives and related rates
  • Familiarity with the equations for volume and circumference of a sphere
  • Basic knowledge of graphing functions
  • Concept of integration (for advanced exploration)
NEXT STEPS
  • Study the relationship between volume and surface area in spherical objects
  • Learn about derivatives in calculus, specifically related rates
  • Explore integration techniques to express circumference as a function of time
  • Utilize graphing utilities to visualize functions and their derivatives
USEFUL FOR

Students in pre-calculus or calculus courses, educators teaching related rates, and anyone interested in the mathematical modeling of physical phenomena such as balloon inflation.

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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