Calculus Problem: Blowing Up a Spherical Balloon

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SUMMARY

The discussion focuses on solving a calculus problem related to the dynamics of a spherical balloon's volume as it inflates. The key point is that while the volume flow rate, denoted as ##\frac{dV}{dt}##, remains constant, the rate of change of the radius, ##\frac{dr}{dt}##, is variable. The assumption that ##\frac{dr}{dt}## is constant is incorrect; it only holds true at a specific radius of 6.50 cm, where ##\frac{dr}{dt} = 0.900\,\mathrm{cm}/\mathrm{s}##. To fully understand the problem, one must rearrange the equations to derive ##\frac{dr}{dt}## and calculate ##\frac{dV}{dr}## for further insights.

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  • Understanding of calculus concepts, particularly derivatives
  • Familiarity with the formula for the volume of a sphere
  • Knowledge of variable rates of change in physics
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the relationship between volume and radius in spherical geometry
  • Learn how to derive and interpret ##\frac{dV}{dr}## in calculus
  • Explore the implications of variable rates of change in real-world applications
  • Practice solving similar problems involving related rates in calculus
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Students studying calculus, particularly those focusing on related rates, as well as educators looking for examples of applying calculus to physical scenarios like inflating balloons.

Idan9988
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Homework Statement
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Relevant Equations
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IMG_20230527_195520.jpg

I'm struggling with section a. This is my calculation:
IMG20230527195328.jpg

The expression remains depend on the variable t, while in the answer is a concrete number:
Screenshot_2023-05-27-19-54-03-99_e2d5b3f32b79de1d45acd1fad96fbb0f.jpg
 
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r = r_0 + 0.9t is only valid if dr/dt is constant.

Why did you assume that dr/dt was constant? The question only tells you that dr/dt = 0.900\,\mathrm{cm}/\mathrm{s} when r = 6.50\,\mathrm{cm}.
 
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Likes   Reactions: Idan9988, malawi_glenn and erobz
Agree,

The answer (a) has all the information. Since the volume flow rate is constant, then ##\frac {dV}{dt}## is a constant.

##\frac {dr}{dt}## is variable.

If you rearrange the expression to solve for ##\frac {dr}{dt}## and you get the answer to (b) and the behavior that explains (c).
 
Calculate ##\frac {dV} {dr}## and use this to inform your answer.
 

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