2026 RI P1 Q3

2026 RI P1 Q3

Junior College 2
6 marks

Water is poured at a rate of \(0.5\text{ m}^3\) per minute into an empty container in the form of an open hemisphere with inside radius \(3\text{ m}\) (see diagram). At time \(t\) minutes after the start, the radius of the water surface is \(r\text{ m}\) and the volume of the water in the hemispherical container is \(\frac13\pi h^2(9-h)\), where \(h\text{ m}\) is the depth of the water. When the depth of the water is \(1\text{ m}\),

  1. find the rate of change of the depth of water,[3]
  2. find the rate of change of the radius of the water surface.[3]

Solution:

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Answer:(a) \(\frac1{10\pi}\text{ m min}^{-1}\) (b) \(\frac1{5\pi\sqrt5}\text{ m min}^{-1}\)

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