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Maximizing and Minimizing

In the last post about the derivative we emphasized its utility with some examples about how it can be used to find some relationships between position, velocity, and acceleration or to maximize profit and minimize cost in numerous different practical situations. Well now I want to discuss that in more depth and provide an example. Last time we used the following definition for the derivative (or slope) using the limit:

With this we can find the exact slope at any point on a curve. Now consider the following graph. Each of the red lines represents the slope at that point with a tangent line

Consider what the slope of such a tangent line would be at a minimum or maximum on a curve. That’s right, the line would be flat and horizontal like this:

Therefore the slope at these points is zero. With this geometrical understanding of a minimum or maximum we can find the derivative and then determine where the derivative (or slope) is equal to zero.

Now let’s consider an example function that relates the production level of a manufacturer to the potential revenue per unit. Obviously we want to maximize revenue so the key will be to find the maximum. Here is the function:

Now you might ask, “Well how in the world did we know this tells us about this relationship?” The short answer is that after some experimenting by the company at various different production levels and measuring the resultant revenue, the corresponding data points were plotted and the following function was mathematically obtained as a best-fit approximation to describe those data points. (More on this in a future post.) So assuming this is the correct function that we are working with, let’s first find the derivative.

Now this derivative gives us the slope at any point along the curve of the original function above. Where does this new function equal zero? This is the same as finding the roots of the equation which is a typical problem in algebra. Here is how it’s done based on our example.

Now if we put these two production level values into the original function it will tell us the revenue produced. Here are the results.

Therefore production level 2 provides the maximum revenue potential. (The root zero is an interesting point called an inflection point. But that’s another topic.) To demonstrate that this process has in fact located the maximum, here is a table of revenue values at other production levels and a graph of the function.

Next time we will look more in depth at the derivative and some basic rules or short-cuts we can use so that we don’t have to go through the entire limit process every time we want to find the derivative.

The Hyperbola

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.

The Ellipse

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.

As before, the intersection of the plane and the surface is the conic section.

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,

which results in the final standard form 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:

Likewise if we let x=0, 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

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.