Multiple choice

The volume of a spherical balloon is increasing at the rate of 20 cm3/s. The rate of change of its surface area at the instant when its radius is 8 cm is:

  1. 5 cm2/s

  2. 10 cm2/s

  3. 4 cm2/s

  4. 2.5 cm2/s
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A Correct answer
Explanation

Volume V = (4/3)*pi*r^3. dV/dt = 4*pi*r^2 * dr/dt = 20. At r=8, 4*pi*64 * dr/dt = 20, so dr/dt = 20 / (256*pi) = 5 / (64*pi). Surface area S = 4*pi*r^2. dS/dt = 8*pi*r * dr/dt. Substituting: dS/dt = 8*pi*8 * (5 / (64*pi)) = 64*pi * (5 / (64*pi)) = 5 cm^2/s.

AI explanation

Using the formula for the volume of a sphere V = 4/3 * pi * r^3, the rate of change of volume is dV/dt = 4 * pi * r^2 * (dr/dt). Substituting dV/dt = 20 and r = 8 gives 20 = 4 * pi * 64 * (dr/dt), so dr/dt = 5 / (64 * pi). The surface area formula is S = 4 * pi * r^2, and its rate of change is dS/dt = 8 * pi * r * (dr/dt). Substituting the values gives dS/dt = 8 * pi * 8 * (5 / (64 * pi)) = 5. The rate of change of its surface area is 5 cm2/s.

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