A cylindrical tank has a base radius of 5 cm and a height of 9 cm and is initially filled with water. The water flows out through a hole in the bottom of the tank at a rate of 5$ \sqrt[]{h} $cm3/min, where h cm is the height of the water in the tank at time t. The time taken in minutes for the tank to empty is given by:

Question

A cylindrical tank has a base radius of 5 cm and a height of 9 cm and is initially filled with water. The water flows out through a hole in the bottom of the tank at a rate of 5$ \sqrt[]{h} $cm3/min, where h cm is the height of the water in the tank at time t. The time taken in minutes for the tank to empty is given by:

BlackTom AI · Accepted Answer

We start by identifying the relationship between volume and water height. For a cylindrical tank with radius r = 5 cm, the base area is A = πr^2 = 25π cm^2, so the volume V is V = A h = 25π h. The water leaves through a hole at a rate dV/dt = -5 √h cm^3/min, where h is the current water height.

Since V = 25π h, differentiate to get dV/dt = 25π dh/dt. Equate the two expressions for dV/dt: 25π dh/dt = -5 √h. Solve for dh/dt: dh/dt = -(1/(5π)) √h.

Separate variables: dh / √h = -(1/(5π)) dt. Integrate both sides from the initial height h = 9 cm at t = 0 to the final height h = 0 cm at time t = T:
∫_{9}^{0} h^{-1/2} dh = ∫_{0}^{T} -(1/(5π)) dt.

The left integral evaluates to [2√h]_{9}^{0} = 2(0 − 3) = −6. The right integral evaluates to −(1/(5π)) T. Thus −6 = −(1/(5π)) T, so T = 6 · 5π = 30π minutes.

Therefore, the time taken for the tank to empty is 30π minutes. (Note: If any options were provided, this would correspond to the value 30π in minutes.)

MUF0102 Adv. Mathematics Unit 2 - Semester 2, 2025 3.8 Applications of calculus quiz

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