Using implicit differentiation, it is found that the radius is increasing at a rate of 0.0049 centimeters per second.
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
Applying implicit differentiation, the rate of change is given by:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
- A spherical balloon is inflated with gas at the rate of 300 cubic centimeters per minute, hence [tex]\frac{dV}{dt} = 300[/tex].
- Radius of 70 centimeters, hence [tex]r = 70[/tex]
Then
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]300 = 4\pi (70)^2\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt} = \frac{300}{4\pi(70)^2}[/tex]
[tex]\frac{dr}{dt} = 0.0049[/tex]
The radius is increasing at a rate of 0.0049 centimeters per second.
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