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Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. We know that: $$V = \frac{4}{3}\pi r^3$$ $$S = 4\pi r^2$$ The problem states that the rate of decrease of volume is proportional to the surface area. Let $k$ be a positive constant of proportionality: $$-\frac{dV}{dt} = kS$$
Differentiating $V$ with respect to $t$ using the chain rule: $$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \frac{dr}{dt}$$
Substitute the expression for $\frac{dV}{dt}$ into the given proportionality equation: $$-(4\pi r^2 \frac{dr}{dt}) = k(4\pi r^2)$$ Assuming $r \neq 0$, we can divide both sides by $4\pi r^2$: $$-\frac{dr}{dt} = k$$ $$\frac{dr}{dt} = -k$$
Since $k$ is a constant, $\frac{dr}{dt}$ is a constant. The negative sign indicates that the radius is decreasing at a constant rate.
Final Answer: The rate of change of radius is -k, which is constant.
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