A spherical balloon is being inflated at $900,\text{cm}^3/\text{s}$. Find the rate of change of radius when $r = 15,\text{cm}$.

Q: A spherical balloon is being inflated at $900,\text{cm}^3/\text{s}$. Find the rate of change of radius when $r = 15,\text{cm}$.

Explanation: Volume of a sphere: $V = \frac{4}{3}\pi r^3$ \[\] Differentiating: \[\frac{dV}{dt} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\] Given $\frac{dV}{dt} = 900$: This implies \[900 = 4\pi (15)^2 \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{900}{4\pi (225)} = \frac{1}{\pi} \text{cm/s}\]

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Explanation: Volume of a sphere: $V = \frac{4}{3}\pi r^3$

Differentiating:

$$\frac{dV}{dt} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$$

Given $\frac{dV}{dt} = 900$: This implies

$$900 = 4\pi (15)^2 \cdot \frac{dr}{dt}  \Rightarrow \frac{dr}{dt} = \frac{900}{4\pi (225)} = \frac{1}{\pi} \text{cm/s}$$

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