Question

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

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

Step-by-step solution

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

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: \(15 \mathrm{~L} / \mathrm{s} \times 1000 \mathrm{~cm}^3 / \mathrm{L} = 15000 \mathrm{~cm}^3 / \mathrm{s}\)

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.

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 \(V = s^3\).

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: \(\frac{d(s^3)}{dt} = 3s^2\frac{ds}{dt}\)

Substitute the rate of height increase

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

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: \(s = \sqrt{3333.33}\) \(s \approx 57.74\)

State the answer

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

Key concepts

Cubical Tank

A cubical tank is a container shaped like a cube, meaning all its sides are of equal length. Cubes are unique because of their symmetry and uniformity, making them a simple yet effective shape for many storage purposes. In mathematics and science, a cube is defined by the length of one of its edges, often referred to as \(s\). A cube's geometric properties make it a perfect choice for this problem, as the volume and surface calculations are straightforward once you know one side's length. Understanding the basic shape of a cubical tank helps with visualizing the problem and carrying out calculations.

Volume Rate

Volume rate refers to the speed at which a fluid changes its volume over time. In this problem, water flows into the cubical tank at a rate of 15 L/s, which informs us about how quickly the water level in the tank is rising.

• Initially, the volume rate is given in liters, a common measure of capacity for fluids.
• To align units, we convert this to cubic centimeters, leading to a rate of 15000 cm³/s (since 1 L = 1000 cm³).

This conversion is crucial for solving the problem, as it ensures that all measurements are in compatible units, allowing for accurate calculations.

Height Increase

The term height increase describes how fast the water level in the tank rises. It is crucial to understand this because it determines how the volume changes over time. In this case, the water height is increasing by 1.5 cm per second.
Since the tank is cubical, as water is added, each additional centimeter of height involves a volume of water spread evenly over the base. This makes the calculations simpler because you can directly relate the increase in height to increases in volume, thanks to the uniform shape of the cube.

Volume Formula

The volume formula for a cube is one of the simplest in geometry: it is the cube of the side length, \(V = s^3\). This formula is foundational for solving this type of problem as it helps relate the physical characteristics of the tank to the mathematical expressions.
Using the volume formula, the relationship between the changing height and the overall volume increase can be expressed through differentiation, which reveals how we can compute the side length of the cubical tank when given both the rate of volume increase and the rate of height increase.

• By understanding \(s = \sqrt[3]{V}\), you can deduce each side's length if the total volume is known.
• This cubic relationship underscores how any change in the side affects the volume—exponentially due to the power of three.