Problem 76 - Calculus Early Transcendentals

Question

Water in a gas tank Before a gasoline-powered engine is started, water must be drained from the bottom of the fuel tank. Suppose the tank is a right circular cylinder on its side with a length of \(2 \mathrm{ft}\) and a radius of 1 ft. If the water level is 6 in above the lowest part of the tank, determine how much water must be drained from the tank.

Step-by-Step Solution

Answer

Based on the given problem, we found that the volume of water to be drained from the tank is \(\text{Water Volume} = \arccos\left(\frac{3}{4}\right) - 0.25\) cubic feet. Please provide the numerical value of the volume of water to be drained from the tank, rounded to 2 decimal places.

Step 1: Convert all measurements to the same units

Since the length of the tank and its radius are given in feet, we should convert the water level height (6 inches) to feet as well. There are 12 inches in a foot, so we divide 6 by 12. \(6\,\text{in} \cdot \frac{1\,\text{ft}}{12\,\text{in}} = 0.5\,\text{ft}\) The water level is 0.5 feet above the lowest part of the tank.

Step 2: Find the angle representing the water level in the tank

Since the water level forms a sector in the circular face of the tank, we will need to find the angle of that sector. To do this, let us work with the right triangle formed by the radius, the water level, and the remaining distance from the water level to the top of the cylinder. This remaining distance is the difference between the radius (1 ft) and the water level (0.5 ft), which is also 0.5 ft. Using the cosine rule, we can find the angle \(\theta\) opposite the 0.5 ft segment in the triangular section: \(\cos(\theta) = \frac{0.5^2 + 1^2 - 0.5^2}{2 \cdot 0.5 \cdot 1}\) \(\cos(\theta) = \frac{3}{4}\) \(\theta = \arccos\left(\frac{3}{4}\right)\)

Step 3: Calculate the area of the sector

Now we must find the area of the sector that represents the water level in the circular face of the tank. We can use the following formula: \(\text{Sector Area} = \frac{\theta}{2\pi} \cdot \pi r^2\) Plugging in the values for \(\theta\) and \(r\): \(\text{Sector Area} = \frac{\arccos\left(\frac{3}{4}\right)}{2\pi} \cdot \pi (1)^2\) \(\text{Sector Area} = \frac{\arccos\left(\frac{3}{4}\right)}{2}\)

Step 4: Calculate the area of the triangular section

We also have to find the area of the triangular section where the water level is present in the circular face of the tank. We can calculate this area using the following formula for the area of a triangle: \(\text{Triangle Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}\) In our triangular section, the base is 0.5 ft, and the height is also 0.5 ft. Thus, the area of the triangular section is: \(\text{Triangle Area} = \frac{1}{2} \cdot 0.5 \cdot 0.5 = 0.125\)

Step 5: Calculate the area of the water portion in the circular face

The area of the water portion in the circular face is the difference between the sector area and the triangle area: \(\text{Water Area}_{\text{circular face}} = \text{Sector Area} - \text{Triangle Area}\) \(\text{Water Area}_{\text{circular face}} = \frac{\arccos\left(\frac{3}{4}\right)}{2} - 0.125\)

Step 6: Calculate the volume of the water to be drained

Lastly, to find the volume of the water to be drained, we need to multiply the area of the water portion in the circular face by the length of the tank: \(\text{Water Volume} = \text{Water Area}_{\text{circular face}} \cdot \text{length}\) \(\text{Water Volume} = \left(\frac{\arccos\left(\frac{3}{4}\right)}{2} - 0.125\right) \cdot 2\) Therefore, the volume of water to be drained from the tank is: \(\text{Water Volume} = \arccos\left(\frac{3}{4}\right) - 0.25\) cubic feet.

Key Concepts

Cylindrical Tank Volume

Understanding how to calculate the volume of a cylindrical tank is essential in various fields, from engineering to environmental science. A cylinder's volume is given by the product of its base area and height. The base of a right circular cylinder, like a fuel tank, is a circle, and thus the formula is:
\[\text{Volume} = \pi r^2 h\]
where \(r\) represents the radius, and \(h\) represents the height or length of the cylinder. For a tank lying on its side, the height in the volume formula corresponds to the length of the tank.

Sector Area Calculation

A sector is a portion of a circle bounded by two radii and an arc. Calculating the area of a sector is a fundamental arithmetic skill required in many calculus problems that involve circular motion or geometry. The sector area is determined by the formula:
\[\text{Sector Area} = \frac{\theta}{2\pi} \cdot \pi r^2\]
Here, \(\theta\) is the central angle in radians, and \(r\) is the radius of the circle. In practical situations like measuring the water level in a tank, this calculation helps ascertain the portion of the cylinder occupied by the liquid.

Trigonometry in Calculus

Trigonometry is an indispensable tool in calculus, particularly when dealing with problems that involve angles and lengths. It becomes crucial when we need to find relationships between different parts of a geometric shape, such as a sector formed by a fluid in a cylindrical container. We often use trigonometric functions like sine, cosine, and tangent along with their respective arcs (inverse functions) to solve for unknown angles or lengths. For instance, the cosine rule or the law of cosines allows us to compute an angle given the lengths of the sides of a triangle, central to finding the quantity of water in a tank:
\[\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}\]
Here, \(\theta\) is the angle opposite to the side of length \(c\), and \(a\) and \(b\) are lengths of the other two sides.

Integrating Geometric Shapes

Integration is used in calculus to combine simple geometric shapes to determine areas, volumes, and other properties. When handling complex shapes, we often break them down into simpler components, calculate the respective areas or volumes, and then integrate these to form the whole. The process involves summing infinite infinitesimally small particles, which exactly represent the desired shape. For the given problem of water in a tank, integration isn't directly needed, but steps resemble the integration process by decomposing the tank's circular cross-section into a sector and a triangle to find the water volume.
However, in calculus, if we were to calculate the water volume across varying tank cross-sections or variable water levels, we would use definite integration, applying limits corresponding to the boundaries of the fluid.