Find the volume of a sphere with respect to time

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SUMMARY

The discussion focuses on finding the volume of a sphere with respect to time, specifically addressing the equation \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\). The known volume is 100 cm³, leading to the realization that when \(dV/dt = 100\) and \(r = 25\), the relationship becomes clearer. Participants express confusion regarding the application of the chain rule and the inverse of \(4\pi r^2\), but ultimately clarify the correct approach to solving for \(\frac{dr}{dt}\).

PREREQUISITES
  • Understanding of calculus, specifically the chain rule.
  • Familiarity with the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\).
  • Knowledge of derivatives and rates of change in relation to volume.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Study the application of the chain rule in calculus.
  • Learn how to derive the volume formula for a sphere.
  • Explore related rates problems in calculus.
  • Investigate the implications of differentiating volume with respect to time.
USEFUL FOR

Students studying calculus, particularly those tackling related rates problems, and educators looking for examples of applying the chain rule in real-world scenarios.

bobsmith76
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Homework Statement



Screenshot2012-01-27at25235PM.png


I guess I should point out that the volume was known, which is 100 cm.

The Attempt at a Solution



I don't see why using the chain rule compels you to take the inverse of 4∏r^2. Any help would be appreciated.

I have a feeling that the solution is something like

100 = (4pir^2)/x

but that still is not right.
 
Last edited:
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bobsmith76 said:
Screenshot2012-01-27at25235PM.png


I don't see why using the chain rule compels you to take the inverse of 4∏r^2. Any help would be appreciated.

They're simply solving \displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\ \text{ for }\ \frac{dr}{dt}\,.
 
Ok, now I got it. The way the book proceeded was a bit bewildering. I didn't see how dV/dt = 100, and r = 25, but now I do.
 

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