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Chapter 1: Problem 55

Water flows into a cubical tank at a rate of \(15 \mathrm{~L} / \mathrm{s}\). If the top surface of the water in the tank is rising by \(1.5 \mathrm{~cm}\) every second, what is the length of each side of the tank?

Short Answer

Expert verified
Answer: The length of each side of the cubical tank is approximately 57.74 cm.

Step by step solution

01

Write down the given information

We are given: - The rate of water flowing into the tank, which is 15 L/s - The rate of increase in the height of the water surface, which is 1.5 cm/s
02

Convert units for consistency

First, we need to convert the given rates to consistent units. We convert the volume rate from liters to cubic centimeters (1 L = 1000 cm³). So, the rate of water flowing into the tank is: \(15 \mathrm{~L} / \mathrm{s} \times 1000 \mathrm{~cm}^3 / \mathrm{L} = 15000 \mathrm{~cm}^3 / \mathrm{s}\)
03

Calculate the rate of increase in volume

The rate of increase in the volume of water in the tank (dV/dt) is equal to the rate at which water flows into the tank. So, \(\frac{dV}{dt} = 15000 \mathrm{~cm}^3 / \mathrm{s}\) Now we need to find the relationship between the height increase rate and the volume increase rate.
04

Use the volume formula of a cube

Let the length of each side of the cubical tank be \(s \mathrm{~cm}\). The volume of a cube is given by the formula \(V = s^3\).
05

Calculate the rate of change of volume with respect to height

Differentiating the volume formula with respect to time \(t\) gives: \(\frac{dV}{dt} = \frac{d(s^3)}{dt}\) Since the height is increasing uniformly at a rate of 1.5 cm/s, we can write \(s = h + ct\) where \(h\) is the initial height of the water, \(c\) is the height increase rate, and \(t\) is the time in seconds. Now, we can differentiate \(s^3\) using the chain rule: \(\frac{d(s^3)}{dt} = 3s^2\frac{ds}{dt}\)
06

Substitute the rate of height increase

Substitute the height increase rate for \(\frac{ds}{dt}\): \(\frac{dV}{dt} = 3s^2(1.5)\) \(\frac{dV}{dt} = 4.5s^2\)
07

Equate the rate of change of volume and solve for s

Since \(\frac{dV}{dt}\) is equal to the rate at which water flows into the tank, we can equate the expressions for dV/dt and solve for the side length s: \(15000 = 4.5s^2\) Dividing by 4.5, we get: \(s^2 = \frac{15000}{4.5}\) \(s^2 = 3333.33\) Taking the square root of both sides: \(s = \sqrt{3333.33}\) \(s \approx 57.74\)
08

State the answer

The length of each side of the cubical tank is approximately 57.74 cm.

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