 # Gas is escaping from a spherical balloonat the rate of 2cm3/min. How fast is thesurface area changing when the radius is12 cm? [V = 43πr3] and S = 4πr2 This is another related rates problem. We know the rate of change of the balloon's volume with respect to time (dV/dt), and we want to find the rate of change of the balloon's surface area with respect to time (dS/dt) when the radius r = 12 cm.

We're given that the volume V of a sphere is given by V = 4/3πr³ and the surface area S is given by S = 4πr². Also, we know that dV/dt = -2 cm³/min (the negative sign indicates the volume is decreasing).

Differentiating both V and S with respect to time t gives:

dV/dt = 4πr² * dr/dt (from V = 4/3πr³)

and

dS/dt = 8πr * dr/dt (from S = 4πr²).

We can solve the first equation for dr/dt, the rate of change of the radius with respect to time:

dr/dt = (dV/dt) / (4πr²) = -2 cm³/min / (4π12² cm²) = -2 / (4π144) cm/min ≈ -0.0011 cm/min.

Then substitute this value into the second equation to find dS/dt:

dS/dt = 8πr * dr/dt = 8π*12 cm * (-0.0011 cm/min) = -0.334 cm²/min.

So, when the radius of the balloon is 12 cm, the surface area is decreasing at a rate of approximately 0.334 cm² per minute.

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