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