Multiple choice

Water is flowing out at the rate of $\displaystyle 6::m^{3}/min$ from a reservoir shaped like a hemispherical bowl of radius$ R = 13$ m The volume of water in the hemispherical bowl is given by $\displaystyle v=\frac{\pi }{3}.y^{2}\left ( 3R-y \right )$ when the water is $y$ meter deep Find a t what rate is the water level changing when the water is $8$ m deep.

  1. $\displaystyle -\frac{1}{12\pi }m/min$
  2. $\displaystyle -\frac{1}{18\pi }m/min$
  3. $\displaystyle -\frac{1}{24\pi }m/min$
  4. $\displaystyle -\frac{1}{30\pi }m/min$
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C Correct answer
Explanation

Given V = (pi/3) * y^2 * (3R - y). Differentiate with respect to t: dV/dt = (pi/3) * [2y * (3R - y) * dy/dt + y^2 * (-dy/dt)]. Given dV/dt = -6, R = 13, y = 8. -6 = (pi/3) * [16 * (39 - 8) * dy/dt - 64 * dy/dt]. -6 = (pi/3) * [496 - 64] * dy/dt = (pi/3) * 432 * dy/dt = 144 * pi * dy/dt. dy/dt = -6 / (144 * pi) = -1 / (24 * pi).

AI explanation

Differentiate the volume formula V = (π/3)y^2(3R - y) with respect to time to get dV/dt = (π/3)(6Ry - 3y^2)(dy/dt), which simplifies to dV/dt = πy(2R - y)(dy/dt). Given dV/dt = -6, R = 13, and y = 8, substitute the values to find -6 = π(8)(26 - 8)(dy/dt), which becomes -6 = 144π(dy/dt). Solving for dy/dt gives -6 / (144π), resulting in a rate of -1/(24π) m/min.

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