Solve Rate of Change for Spherical Balloon Radius: 500 cc/min

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SUMMARY

The problem involves calculating the rate of change of the radius of a spherical balloon being inflated at a constant volume rate of 500 cubic centimeters per minute. The volume formula for a sphere, V = (4/3)πr³, is utilized to derive the relationship between the volume change and the radius change. By applying the chain rule, the equation dr/dt = (1/(4πr²))(dV/dt) is established, allowing for the calculation of dr/dt at specific radii of 30 cm and 60 cm. The user successfully resolves the issue by correctly substituting the values into the derived formula.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with the volume formula for a sphere
  • Knowledge of the chain rule in calculus
  • Basic algebra for manipulating equations
NEXT STEPS
  • Calculate the rate of change of the radius at r = 30 cm using dr/dt = (1/(4πr²))(dV/dt)
  • Calculate the rate of change of the radius at r = 60 cm using the same formula
  • Explore applications of related rates in real-world scenarios
  • Review advanced topics in calculus, such as implicit differentiation
USEFUL FOR

Students studying calculus, particularly those focusing on related rates, as well as educators looking for practical examples of volume and differentiation concepts.

bjr_jyd15
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Hi everyone. I'm stuck on this problem in calc:

A spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is a)30 cm and b)60 cm?

Here's what I have so far.

V=4/3*pi r^3

dV/dt=500

dr/dt=?

dV/dt=4/3*pi (3r^2) dr/dt

Where can I go from here? Any help would be great. Thanks
 
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OOPS, MAJOR EDIT:
"dV/dt=4/3*pi (3r^2) dr/dt"
You've got this equation, right?
So, regrouping you have:
[tex]\frac{dr}{dt}=\frac{1}{4{\pi}r^{2}}\frac{dV}{dt}[/tex]
 
Last edited:
got it

thanks, i got it. The 4/3 just screwed me up. i just plugged dv/dt from original info.
 

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