Related rates Balloon question. Thanks

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SUMMARY

The discussion centers on calculating the rate of change of volume for a spherical balloon as air is pumped into it, given a surface area increase rate of 20 cm²/s when the radius is 4 cm. The correct formula for the volume change rate, dv/dt, is derived from the relationship between surface area and volume of a sphere. The user initially misapplies the formula, leading to incorrect calculations. The correct approach involves using the formulas for dV/dt and dA/dt in terms of the radius and its rate of change.

PREREQUISITES
  • Understanding of related rates in calculus
  • Familiarity with the formulas for the surface area and volume of a sphere
  • Knowledge of differentiation techniques
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the formulas for the surface area (A = 4πr²) and volume (V = (4/3)πr³) of a sphere
  • Learn how to apply the chain rule in related rates problems
  • Practice solving related rates problems involving different geometric shapes
  • Explore the concept of implicit differentiation in calculus
USEFUL FOR

Students studying calculus, particularly those focusing on related rates, as well as educators looking for examples to illustrate these concepts in a classroom setting.

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Homework Statement



Air is being pumped into a spherical balloon. Suppose we know the surface area of the balloon increases at a rate of 20cm^2/s when its radius is 4cm. What is the rate its volume is changing at that instant?

Homework Equations


The Attempt at a Solution



I went...

dv/dt = 4∏ (4)^2 * 20/1 = 1280∏

I know its wrong. Anyone help me out? What am i doing wrong?
 
Last edited:
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Wait no, I went:

dv/dt = (1/4pi(4)^2) *20/1

gives me 5 / 16pi
 
I think that's still wrong. Write out generic formulae for dV/dt and dA/dt as functions of r and dr/dt..
 

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