Dear Miss Majestica,

I am writing about the letter you sent me concerning the wire you used in your show. You are in luck, Miss Majestica. I have been successful in determining that your wire is made of titanium, not aluminum. I would like to take this opportunity to remind you about those complimentary tickets you promised. You can forward them to my Calculus III professor, Dr. Annalisa Crannell, and I am sure that she will remember to give them to me.

The problem in your letter intrigued me and, naturally, I enjoyed applying my knowledge of Calculus to solve it. You explained that one of your circus acts uses this wire, supposedly made of titanium. However, it came to your attention that the wire might in fact be made of aluminum instead. The vendor, who sold you the wire, may have lied to you, and before you order any further merchandise, you need to know whether the wire is made of titanium, or whether the vendor deceived you.

You provided me with numerous measurements of not only the wire, but of the structure holding the wire as well. You said the wire hangs in the form of a catenary curve with the variables of height, weight, and tension. The pillars holding the wires are 75 feet high, 200 feet apart, and the middle of the cable hangs 50 feet above the ground. The cable has a diameter of half a foot and a horizontal tension of 11,219.5 pounds. Finally, you so graciously provided me with the density of titanium, 281 pounds/cubic foot.

I presume you not only want a confirmation of the metallic component in the wire, but you are also interested in how I solved this matter using Calculus. Because I have never seen your wire, I am assuming that it is a flexible wire hanging from two parallel pillars, in which the diameter remains the same throughout the entire pillar.

To begin solving this problem, I first found the equation of a catenary curve on page 280 of Calculus Single Variable, Second Edition ©1999 , Hughes - Hallett Gleason, et al. Let me first begin by showing you a visual diagram of your incredible titanium wire. In Figure 1, I labeled the height of the pillars y, and the horizontal distance from the midpoint of the wire x.

Figure 1. Graph of catenary curve produced by hanging cable and support pillars.

As I stated earlier, I found the equation for the catenary curve. This equation,

,

shows, given y, x, and a tension, T, of 11,219.5 pounds, we can determine w, the weight of one foot of the cable. By looking at the graph in Figure 1, we see y is the height of the pillars in feet, and x is the horizontal distance, in feet, from the midpoint of the wire to the point at which we are measuring. The constant, C, allows the shape of your cable to stay the same, even if the height changes.

From this catenary equation, we see the only variable we need to solve for is weight, w, if we have y, x, and C values. The x and y values help us solve for the constant. For example, the graph shows when x = 0 ft., then y = 50 ft. By plugging these values into the catenary equation we have,

.

Because the cosh (0) equals 1, we can determine what C, the constant, would equal with the equation,

.

Furthermore, the we can apply the same procedure when x = 100 and y = 75, as the graph in Figure 1 shows. These substitutions, including the substitution for C, produce the following equation:

This equation seems complicated. However, closer analysis reveals the only variable we have to solve for is w, or weight. If we plug in the weight of a foot of titanium wire, and the right side of the equation equals the left side (75 ft. = 75ft), then we know the wire is made of titanium. If the right side does not equal 75 feet, then we know the wire is not made of titanium.

We can determine the weight of one foot of titanium wire by multiplying the density of titanium and the volume of one foot of the cable. To find the volume, you must first know the cross-sectional area and the length of the cable that we are measuring. Squaring the radius of the cable and multiplying this value by p determines the cross-sectional area of the cable. Given a radius of 1/4 of a foot, our area equals p/16 ft², and our volume equals p/16 ft³. Using our equation of weight equals density multiplied by volume, we get

Therefore, the weight of titanium equals 55.17 lbs.

When we substitute 55.17 lbs for w into the right side of our catenary equation when x = 100ft. and y = 75ft., we see 75 feet equals 75 feet. Thus, the cable you use in your show, is in fact made of titanium. If you substitute the weight of aluminum into this equation, the right side of the equation would not equal 75 feet.

I hope this explanation clears up the issue concerning the titanium wire. I must say I would no longer believe anything Dick Dastardly says (especially since he just cost you complimentary tickets to your show). I am so glad I could be of assistance to you.

I must add, however, I did not work independently on this project . Matt and Jacqueline of the Franklin and Marshall Math Center guided me to the equation of a catenary curve. My Calculus III professor , Dr. Annalisa Crannell, helped me evaluate the catenary equation, and also edited my letter. Finally, my fellow Calculus III student, Luke Oeding and I worked together in determining how the weight of a foot of titanium wire would be applied to the catenary equation.

So, you see Matilda, it was not an easy problem to solve. But, nevertheless, you made the right decision in coming to me for help. Feel free to contact me in the future if you have anymore mathematical dilemmas. Until then, I look forward to seeing your spectacular show (preferably front row seats, please).