A spherical balloon has a radius of 20 cm.It starts losing air and its radius shrinks at a rate of 1 cm/s.What is the rate of change of the balloon's volume when its radius reaches 5 cm? A)-3.14 * 10^-4 m³/s B)-3.14 m³/s C)-1 * 10^-4 m³/s D)-5 * 10^-4 m³/s E)-3.14 * 10^-2 m³/s

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Quizplus · Accepted Answer

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. The rate of change of volume with respect to time ($\frac{dV}{dt}$) can be found by differentiating both sides with respect to $t$, using the chain rule: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Given $\frac{dr}{dt} = -1$ cm/s and $r = 5$ cm at the moment of interest, substituting these values gives $\frac{dV}{dt} = 4\pi (5)^2 (-1) = -100\pi$ cm$^3$/s, which matches the format of choice A.

A Spherical Balloon Has a Radius of 20 Cm