Category Archives: Geometry
the mathematical study of space, shape, and size
Suppose you are in the can manufacturing industry. Your job is to create cans according to the specifications of canned goods suppliers who will put their products in your cans and distribute them for sale. Now suppose one of those canned good suppliers comes to you and says, “We need a can that can contain a 300 mL volume of our product. Joe Shmoe says he can provide us these cans for $X. What can you do?” If you want the canned good suppliers business you will have to underbid Joe. (Provided you can’t sale them on some other facet of your company, i.e. service, quality, etc.) How will you do it? Where can you cut cost? What is the most inexpensive can you can make to satisfy the spec? In the real world this is a very complex question because it involves labor, facility, process, and material expenses along with several others. For the sake of this exercise however, we will use the principle of the derivative to optimize our can design in order to minimize material cost.
Now let’s assume you are set up to produce a standard cylindrical can with a top and a botttom. (I realize the canned goods supplier will have to add their product before the top is sealed and I realize that actual cans have a small lip on the top and bottom, but for the sake of simplicity…) To minimize the cost of the cans we want to minimize the amount of material used. But what shape cylinder will contain the required 300 mL but use minimal material? A short fat can or a tall thin can? How “tall” or how “short”?
First, realize the total material for the can can be determined by finding the area of the metal. For a cylinder this is the area of the bottom plus the area of the top plus the area of the side.
We can rewrite the area of the top and bottom as:
If you imagine removing the top and bottom, cutting the side along a vertical line, and laying it out flat you can see that the area of the side is the area of a rectangle with the length of one side equal to the height of the can and the length of the other side equal to the circumference of the top or bottom. So its area is:
Putting this all together, the total area of a can is given by:
Now the volume of the cylinder is given by:
If we rearrange this expression for the height, h we get:
Now we substitute this into our expression for the area to eliminate the h and make the whole thing in terms of 1 variable. (r – because the volume is given to us in the spec)
This equation tells us how much area a can has of a particular volume with any radius. With this in hand we can now use the derivative to optimize (maximize or minimize) the can design and find out which radius will give us the smallest area and thus the most inexpensive (as far as material costs go) cylindrical can that can contain the prescribed volume. Taking the derivative of the area equation using the power rule we have:
Now we set this equal to zero to find the minimum area. Solving for r we get:
Knowing that the volume is 300 mL our radius is:
Putting this back in our equation for h we get a height of:
These are the dimensions for our can of minimum material. Below I have tabulated and graphed some other values for the radius and associated heights that yield the necessary volume and show that the resulting surface area is larger than our optimized design.
One other beautiful thing about this technique is that I have a general formula for an optimized cylindrical can for any volume. I can just put in whatever the specification says and get my design.
Although this is greatly simplified, I hope this provides another demonstration of the real world applicability of these techniques.
Previously we discussed various 2-dimensional geometric shapes or conic sections like the parabola, hyperbola, and ellipse. Now let us consider the 3-dimensional cylinder or cylindrical surface. In mathematics, a general cylindrical surface is defined as follows:
Let C be a curve in a plane, and let L be a line that is not in a parallel plane. The set of points on all lines that are parallel to L and intersect C is a cylinder.
Notice that the above cylindrical surfaces can be irregular and open or closed. When we extend the parabola, hyperbola, and ellipse directly into three dimensions, we simply get parabolic, hyperbolic, and elliptic cylinders. These cylindrical surfaces have the same algebraic expression as the corresponding 2-dimensional conic sections.
Now if we extend the equations into 3-dimensional space by algebraically adding 3rd dimension terms, we get the analogous surfaces of the paraboloid, hyperboloid, and the ellipsoid. The elliptic paraboloid below is given by the equation:
If we simply change the sign of one of the terms above we get the hyperbolic paraboloid below given by:
The hyperboloid has two general forms and one special degenerate form. The first form seen below is called the hyperboloid of one sheet. It is given by:
The special degenerate form of the hyperboloid of one sheet given by:
Is the double-napped cone.
The second form of the hyperboloid is called the hyperboloid of two sheets and is given by:
Finally we have the ellipsoid which is given by:
A special case of the ellipsoid where a=b=c is the sphere.
Click here to see a plethora of different and interesting algebraic surfaces.
Ready to take a look at the third conic section? (The first two we looked at were the parabola and the ellipse). The hyperbola is created when the plane passes through both nappes of the right circular conical surface at an angle with the surface’s longitudinal axis that is less than the angle made with the axis by the generating straight line on the surface.
The hyperbola is found in many natural phenomena including an open orbit of celestial bodies or simply the curvature of light around a large gravitational body like a star. It is also the path of subatomic particles repulsed by the nucleus. In more everyday terms, the tip of a shadow cast by the sun traces out a hyperbola on the ground as the day progresses. The mathematics of hyperbolas is also used in navigation and global positioning systems.
The definition of a hyperbola that we will use to develop an algebraic expression is similar to that of the ellipse. A hyperbola is all the points in a plane the difference of whose distances from the two foci is a (positive) constant. The ellipse was simply the sum instead of the difference.
We begin with the hyperbola centered on the origin in a Cartesian-coordinate plane. As with the ellipse we use the foci, (c,0) and (-c,0). The distance from each focus to a general point (x,y) is given by the following two expressions using the two-dimensional distance formula:
Now using the above definition we will set the difference between these two to a constant number, 2a, just as we did with the ellipse. One important thing to note is since as we move about the plane the difference between these may change from negative to positive or positive to negative depending on which distance is longer. We will use the absolute value to insure we get the magnitude of the difference in distance.
As with the ellipse, using some elementary algebra we arrive at:
Now “a” represents the x-intercepts or the vertices. (Let y=0 and multiply both sides by “a” squared). With the ellipse, “a” was always greater than “c” so we factored out a negative 1 from the denominator of the second term and changed the sign between the two terms to a positive. Then we used the substitution:
In the present case, “a” will always be less than “c” so we use the substitution:
This yields the final expression:
Notice here that if we try to locate the y-intercepts by setting x=0, we get imaginary results. (Square root of negative “b” squared). Therefore this is the expression for the hyperbola with vertices on the x-axis. The expression for the hyperbola with vertices on the y-axis is:
Just as with the parabola we can use the same substitutions for translation and axes and rotation of axes to develop expressions for more general hyperbolas.
The axis that runs through the vertices is called the transverse axis. The axis that is orthogonal to the transverse axis is the conjugate axis. A hyperbola is often drawn by constructing a rectangle that is 2a by 2b in dimension with the vertices on the 2a length sides. Then diagonals are drawn through the rectangle and used as asymptotes for the hyperbolic curve. The equation for these asymptotes can be derived by solving the main equations for y.
This is the second post of this series on conic sections. This time we will look at the same double-napped right circular conical surface but the intersecting plane will travel completely through one half of the surface and not through the base as seen below.
From a practical standpoint, various celestial bodies are known to travel in elliptical orbits. This includes our own planet with the sun at one of the foci. Also the reflective property of an ellipse where any light or sound emitted from one focus is reflected to the other focus is utilized in optics design. Also this phenomenon can be experienced in what are called “whispering galleries”. These are buildings with elliptical geometry such that a person at one focus can hear very easily someone speaking at the other focus due to the reflective property. One example of this that I have experienced myself is in the rotunda of the US Capital building.
In order to derive an algebraic expression for the ellipse we begin with the definition that an ellipse is the set of points in the plane such that the sum of the distances from the point to each of the two foci is constant for every point. A good way to visualize this is to imagine you have a piece of string, two pushpins, a piece of paper or cardboard, and a pencil. If you use the two pushpins to pin down each end of the string such that the distance between the two pushpins is less than the length of the string. Then use the pencil to pull the string taut in every direction and make a mark with the pencil along this perimeter as seen below you will have constructed an ellipse.
This way of looking at the ellipse probably makes it more intuitively clear what is meant by the above definition. You can see that every point on the ellipse has the same distance from one pin to the pencil and back to the other pin. This is the constant length of the string.
We begin with an ellipse centered on the origin and the two foci on the x-axis. One focus at (c,0) and the other at (-c,0). Then the distance from a point (x,y) in the plane to each focus is given by:
Now by the above definition we know that the sum of these distances should equal a constant for every point of the ellipse. Although it might seem strange we will use the constant 2a. The reason for the 2 will become more evident once we reach the end of our derivation which goes as follows.
The reason for the change in the last step is due to the fact that we know that the length of the string, 2a, must be greater than the distance between the foci, 2c. If 2a is greater than 2c, then a is greater than c and a squared must be greater than c squared. Typically a substitution is made at this point of,
Also notice from the substituting expression that a is greater than b. From this equation we can find the location of the major axis (longer axis) vertices and the endpoints of the minor axis (shorter axis). Simply set y=0 in the above expression and we get:
Since as we said before that a is greater than b, then the minor axis is along the y axis and has a length of 2b and the major axis is along the x axis and has a length of 2a. Conversely the equation for an ellipse with the major axis along the y axis is:
Just as with the parabola we can use the same substitutions for translation of axes and rotation of axes to develop expressions for other more general ellipses.
One final point of interest concerning the relationship between an ellipse, a circle, and a parabola. If the distance between the two foci is reduced to zero (i.e. foci are at the same point) the ellipse becomes a circle. If this same interfocal distance goes to infinity, the ellipse becomes a parabola.
The next post in this series soon to follow on hyperbolas.
The parabola is one of four conic sections that I intend to provide a discourse on. The conic sections are so named because they can be obtained by passing a plane at varying angles through a double-napped right circular conical surface. The intersection of the plane and the surface is the conic section. In the case of the parabola, the plane passes through one half of the surface at an angle such that the plane is parallel to a generating straight line on the conical surface as seen below.
In the real world parabolas are typically seen when any projectile object travels through the air and is subjected to the pull of gravity (if air resistance is negligible). If you have ever seen a long exposure photograph of a bouncing ball that is moving laterally, you have seen a series of parabolas. Or if you have observed a water fountain flow as the water goes up and then down due to gravity. Also, the golden McDonald’s arches are a pair of parabolas. Paraboloids, the three- dimensional cousin of the parabola, is a common shape used in satellite dishes and headlight beam reflective surfaces. In highway engineering, the parabola is used to design vertical curves or hillcrests.
We can derive an algebraic expression for the parabola using the definition that a parabola is the set of all ordered pairs or points that lie in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix that also lie in the plane. (Provided the focus does not lie on the directrix) A line in the plane that is perpendicular or orthogonal to the directrix that passes through the focus is called the axis of symmetry or simply the axis. In layman’s terms it divides the parabola into two halves that are mirror images of one another. The point on the parabola that the axis intersects is called the vertex.
Consider a parabola whose vertex is at the origin (0,0) and has a focus at the point (0,f). Intuitively then we know that the distance from the vertex to the focus is f. (This is also called the focal distance.) We also see that this parabola will open along the y-axis. Since the vertex lies on the parabola it must be equidistant from the focus and directrix. Consequently we know that the directrix must be the line y = -f. Next using the distance formula, we can say the distance from a random point in the plane (x,y) to the focus (0,f) is given by:
Next realize that the point on our directrix that falls vertically below the point (x,y) is given by (x,-f). Then using the same methodology as above we can say that the vertical distance from the directrix to our point (x,y) is given by:
By the definition of a parabola we can set these two values equal to one another and resolve to standard form:
Through similar reasoning we could derive the expression for a parabola with a vertex at the origin that opens along the x-axis using a focus point of (f,0) and a directrix of x = -f. As might be expected this will yield:
Next let us consider how we can develop expressions for a parabola with a vertex that is not at the origin. The methodology utilized to consider a vertex at any general point in the plane is usually called translation of axes.
Translation of Axes
Consider the above two axes. The base x and y axes are shifted to a new general location where the new origin, O’, is at the base coordinate position (h,k). The point P can be described in both coordinate systems and the two systems can be related as follows:
Using these short equations in a substitutionary fashion we can go through the same procedure as above and derive an expression for a parabola with a vertical axis with a vertex at O’ or base coordinate (h,k). This yields:
Likewise, for a parabola with a horizontal axis we obtain:
Rotation of Axes
The other transformation we can do is called rotation of axes and it allows us to generate an expression for a parabola with an axis at an angle other than vertical or horizontal.
As with translation we want to develop a relationship between the two coordinate systems as they describe the same point in the plane. Observing the diagram above we will define the length of the blue line from the origin to the point P as having the magnitude d. Next consider the triangle OPQ’. Using this right triangle we can say:
Then look at triangle OPQ. Similarly as above we can say:
Using the trigonometry identities:
We can write an expressions for x and y and substitute using x’ and y’ to obtain expressions of the relationship between the two systems.
Finally using these last two expressions we will solve for x’ and y’. This will allow us to generate a parabola on the X’ and Y’ axes and then make a substitution in order to find the expression for the parabola in the base coordinate system. The derivation is done by solving each of the last two expressions for x’ and y’ respectively. This will result in four equations:
Next we will set x’ from the one equation equal to x’ from the other and solve for y’. Likewise we will set y’ equal to y’ and then solve for x’.
The last step in each of the above derivations is accomplished by the following identity:
In a similar fashion as above with translation we can substitute the two boxed equations into the general parabola equation to derive a general equation for a rotated parabola. An important caveat to remember with regard to this formula is that the angle can only be an acute angle. However this does not pose a problem since any parabola in any quadrant can be obtained by an acute rotation from an adjacent axis. I will not here go through each step to derive the general formula as it is rather lengthy and cumbersome for this format. However here is the end result in implicit form for a parabola whose original axis was vertical:
I hope that this has been helpful and I hope the next time you are using the water fountain you notice the lowly parabola at work.
I would love to hear your questions or comments.