How Fast Does a Balloon's Radius Expand When Inflated?

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Discussion Overview

The discussion revolves around the problem of determining the radius of a sphere-shaped rubber balloon as it is inflated, given specific rates of volume addition and radius expansion. The context includes mathematical reasoning related to calculus and related rates.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving a balloon being inflated at a volume addition rate of $$40cm^3/s$$ and a radius addition rate of $$20 cm/s$$, seeking hints for solving it.
  • Another participant suggests taking the derivative of the sphere volume formula with respect to time as a potential approach.
  • A subsequent reply provides a calculation using the volume formula for a sphere, leading to a proposed radius of $$\frac{1}{\sqrt{2\pi}}$$ cm based on the given rates.
  • Another participant confirms the calculation and provides a detailed differentiation process, arriving at the same radius conclusion.
  • There is a discussion about the classification of the problem as pre-calculus versus calculus, with differing opinions on its educational level.
  • A participant questions the actual limit between pre-calculus and calculus, indicating a broader debate on educational categorization.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical approach to the problem, with some confirming the calculated radius. However, there is disagreement regarding the classification of the problem as pre-calculus or calculus, indicating multiple competing views on educational content.

Contextual Notes

The discussion includes assumptions about the rates of volume and radius expansion and their implications for the calculations. There is also an unresolved debate about the educational classification of the problem.

Who May Find This Useful

Students and educators interested in calculus, related rates problems, and the distinctions between pre-calculus and calculus topics may find this discussion relevant.

Monoxdifly
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Icha is going to blow a sphere-shaped rubber balloon. She uses a pump to infuse the air with the volume addition rate $$40cm^3/s$$. If the radius addition rate is 20 cm/s, the radius of the sphere after being blown is ...
A. $$\frac{1}{\sqrt\pi}$$ cm
B. $$\frac{1}{\sqrt{2\pi}}$$ cm
C. $$\frac{1}{2\sqrt\pi}$$ cm
D. $$\frac{1}{3\sqrt\pi}$$ cm
E. $$\pi$$ cm

Please give me some hints. Thanks.
 
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Monoxdifly said:
Icha is going to blow a sphere-shaped rubber balloon. She uses a pump to infuse the air with the volume addition rate $$40cm^3/s$$. If the radius addition rate is 20 cm/s, the radius of the sphere after being blown is ...
A. $$\frac{1}{\sqrt\pi}$$ cm
B. $$\frac{1}{\sqrt{2\pi}}$$ cm
C. $$\frac{1}{2\sqrt\pi}$$ cm
D. $$\frac{1}{3\sqrt\pi}$$ cm
E. $$\pi$$ cm

Please give me some hints. Thanks.

take the derivative of the sphere volume formula w/respect to time ...
 
$$40cm^3/s=4\pi r^2(20cm/s)$$
$$2cm^2=4\pi r^2$$
$$r^2=\frac{1cm^2}{2\pi}$$
$$r=\frac{1}{\sqrt{2\pi}}cm$$?
 
For a sphere of radius $r$, the volume is:

$$V=\frac{4}{3}\pi r^3$$

Differentiating w.r.t time $t$, we have:

$$\d{V}{t}=4\pi r^2\d{r}{t}\implies r=\frac{1}{2}\sqrt{\frac{\d{V}{t}}{\pi\d{r}{t}}}$$

Plugging in the given numbers, we have:

$$r=\frac{1}{2}\sqrt{\frac{40\frac{\text{cm}^3}{\text{s}}}{20\pi\frac{\text{cm}}{\text{s}}}}=\frac{1}{\sqrt{2\pi}}\text{ cm}\quad\checkmark$$

I will move this thread to our Calculus forum. :)
 
MarkFL said:
I will move this thread to our Calculus forum. :)

I thought this was pre-calculus because it was a question in high school graduation exam.
 
Monoxdifly said:
I thought this was pre-calculus because it was a question in high school graduation exam.

It is a related rate problem involving differential calculus (so that's why I moved it). Many places teach elementary calculus in high school. :)
 
What is the actual limit between pre-calculus and calculus, anyway?
 

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