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Construction of a Storage Tank A storage tank for propane gas has the shape of a right circular cylinder with hemispherical ends. The length of the cylinder is \(6 \mathrm{ft}\), and the radius of each hemisphere is \(r \mathrm{ft}\) Show that the volume of the tank is \(\frac{2}{3} \pi r^{2}(2 r+9) \mathrm{ft}^{3}\). If the tank were constructed with a radius of \(4.1 \mathrm{ft}\) instead of a specified radius of \(4 \mathrm{ft}\), what would be the approximate percentage error in its volume?

Short Answer

Expert verified
The total volume of the storage tank is given by the formula \(V_\text{total} = \frac{2}{3}\pi r^2(2r + 9)\). When the radius is changed from 4 ft to 4.1 ft, the percentage error in its volume is approximately 4.48%.

Step by step solution

01

Find the volume of a hemisphere#Endtag#

To calculate the volume of a hemisphere (\(V_\text{hemis}\)), we need to know the formula for the volume of a sphere, which is given by \(V = \frac{4}{3}\pi r^3\). Half of this value will be the volume of a hemisphere: \(V_\text{hemis} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3\). #Endtag#
02

Find the volume of a cylinder#Endtag#

The volume of a cylinder (\(V_\text{cyl}\)) is given by the formula \(V_\text{cyl} = \pi r^2 h\), where \(r\) is the radius of the cylinder and \(h = 6 \: ft\) is its height (length). So, the volume of the cylinder is: \(V_\text{cyl} = \pi r^2 \cdot 6 = 6\pi r^2\). #Endtag#
03

Add the volumes of the hemispheres and cylinder#Endtag#

To find the total volume of the storage tank, we need to add the volumes of the two hemispheres (\(2 \cdot V_\text{hemis}\)) and the cylinder (\(V_\text{cyl}\)): \(V_\text{total} = 2 \cdot (\frac{2}{3}\pi r^3) + 6\pi r^2\). #Endtag#
04

Simplify the volume formula#Endtag#

Now, let's simplify the volume formula for the storage tank: \(V_\text{total} = \frac{4}{3}\pi r^3 + 6\pi r^2 = \frac{2}{3}\pi r^2(2r + 9)\). #Endtag#
05

Calculate the percentage error#Endtag#

To find the percentage error in volume when the radius is changed from 4 ft to 4.1 ft, first calculate the volume for both radii, and then find the difference between the volumes. Finally, divide the difference by the volume of the tank with a radius of 4 ft and multiply by 100 to get the percentage error. Original Volume (\(r = 4\)): \(V_\text{orig} = \frac{2}{3}\pi (4^2)(2 \cdot 4 + 9) = \approx 602.88 \: ft^3\) New Volume (\(r = 4.1\)): \(V_\text{new} = \frac{2}{3}\pi (4.1^2)(2 \cdot 4.1 + 9) = \approx 629.87 \: ft^3\) Difference in Volumes: \(\Delta V = V_\text{new} - V_\text{orig} = \approx 27.00 \: ft^3\) Percentage Error: \(\text{Error \%} = \frac{\Delta V}{V_\text{orig}} \times 100 = \frac{27.00}{602.88} \times 100 \approx 4.48\%\). #Endtag#

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Calculation
Calculating volumes is an essential skill in geometry and calculus, particularly when dealing with complex shapes. The volume of an object refers to the space it occupies. In this exercise, we deal with a storage tank, which combines a cylindrical shape with hemispherical ends. To find the total volume, you need to calculate the volume of the cylinder and the hemispheres individually, then add them together.

For the cylinder, the formula is straightforward: \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height or length. For a hemisphere, use half of the sphere's volume: \( V = \frac{2}{3}\pi r^3 \).

Combining these calculations helps ensure that you understand how much space the entire structure occupies.
Cylindrical Shapes
The cylinder is a familiar geometric shape characterized by two parallel circular bases and a consistent radius. When calculating the volume of a cylindrical shape, it’s crucial to identify the correct measurements. The essential parameters are the radius, \( r \), and the length or height, \( h \).

In this tank example, the cylinder forms the middle part, with hemispheres attached at both ends. The volume of just the cylinder is given by \( V = \pi r^2 h \). Here, the radius is consistent with that of the hemispheres, and the height is given as 6 ft. Understanding how these measurements influence the overall volume is vital, particularly when part of a more complex structure.
Percentage Error
When constructing real-world objects like tanks, calculating the percentage error can be useful, especially when precision is required. Percentage error helps you understand the discrepancy between the intended and actual measurements.

In our example, the radius adjustment from 4 ft to 4.1 ft leads to a different volume. First, find the difference between the two calculated volumes: \( \Delta V = V_\text{new} - V_\text{orig} \). Then, determining the percentage error involves dividing this difference by the original volume and multiplying by 100: \( \text{Error \%} = \left( \frac{\Delta V}{V_\text{orig}} \right) \times 100 \). Understanding and calculating this error is crucial for accurate construction and resource management.
Integration in Geometry
Integration in geometry allows us to find areas, volumes, and other physical properties by summing infinitesimally small quantities. It plays a pivotal role when handling complex shapes, like our propane tank, consisting of a cylinder and hemispheres. Although the exercise explicitly does not require integration, understanding this concept is beneficial. Complex shapes can often be broken down and handled by integration, especially in calculus. It helps deal with non-uniform shapes or when deriving formulas, like the volume of hemispheres (as part of a sphere).

Thus, integration acts as a powerful tool in mathematical modeling and problem-solving beyond simple arithmetic calculations.

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Most popular questions from this chapter

In Exercises, (a) find the equations of the tangent and the normal lines to the curve at the indicated point. (The normal line at a point on the curve is the line perpendicular to the tangent line at that point.) (b) Then use a graphing utility to plot the curve and the tangent and normal lines on the same screen. $$ x^{2}+y^{2}=9 ; \quad(-1,2 \sqrt{2}) $$

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=x^{3}+x+2 ; \quad(1,4) $$

In Exercises, (a) find the equations of the tangent and the normal lines to the curve at the indicated point. (The normal line at a point on the curve is the line perpendicular to the tangent line at that point.) (b) Then use a graphing utility to plot the curve and the tangent and normal lines on the same screen. $$ 4 x y-9=0 ; \quad\left(3, \frac{3}{4}\right) $$

Two families of curves are orthogonal trajectories of each other if every curve of one family is orthogonal to every curve in the other family. In Exercises \(93-96\), (a) show that the given families of curves are orthogonal to each other, and (b) sketch a few members of each family on the same set of axes. $$ x^{2}+y^{2}=c x, \quad x^{2}+y^{2}=k y, \quad c, k \text { constants } $$

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves. In Exercises \(89-92\), show that the curves with the given equations are orthogonal. $$ x^{2}+3 y^{2}=4, \quad y=x^{3} $$

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