The College Mathematics Journal, Vol. 32, No. 5. (Nov., 2001), pp

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Dipsticks for Cylindrical Storage Tanks-Exact and Approximate Pam Littleton; David A. Sanchez The College Mathematics Journal, Vol. 32, No. 5. (Nov., 2001), pp. 352-358. Stable URL: http://links.

1 Dipsticks for Cylindrical Storage Tanks-Exact and Approximate Pam Littleton; David A. Sanchez The College Mathematics Journal, Vol. 32, No. 5. (Nov., 2001), pp Stable URL: The College Mathematics Journal is currently published by Mathematical Association of America. Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at JSTOR's Terms and Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you may use content in the JSTOR archive only for your personal, non-commercial use. Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission. The JSTOR Archive is a trusted digital repository providing for long-term preservation and access to leading academic journals and scholarly literature from around the world. The Archive is supported by libraries, scholarly societies, publishers, and foundations. It is an initiative of JSTOR, a not-for-profit organization with a mission to help the scholarly community take advantage of advances in technology. For more information regarding JSTOR, please contact support@jstor.org. Wed Aug 22 13:28:

2 Dipsticks for Cylindrical Storage Tan ks-exact and Approximate Pam Littleton and David A. Sanchez Pam Littleton tarleton.edu) is professor of mathematics at Tarleton State University. Her primary area of interest is secondary teacher preparation. David Sanchez (dsanchez@math.tamu.edu)did his undergraduate studies at the University of New Mexico and received his Ph.D. from the University of Michigan. He has been a professor of mathematics at UCLA and the University of New Mexico. His major field of research is ordinary differential equations. For nine years he acquired a severe case of administrativitis, but recovered and is happily teaching at Texas A&M University. A problem often found in pre-calculus and calculus textbooks, of importance in agriculture, is the calibration of a dipstick to measure the volume of liquid in a cylindrical tank lying on its side. The dipstick is inserted into the top of the tank, as in Figure 1. Figure 1. This gives rise to two different problems: Problem 1. Determine the volume corresponding to uniformly spaced calibration marks on the dipstick. Problem 2. Determine the calibration marks on the dipstick corresponding to uniformly spaced volume levels. In the first case the dipstick might have volume levels marked every six inches whereas in the second case the marks might represent fifty gallon increases in volume. THE MATHEMATICAL ASSOCIATION OF AMERICA

3 Let r be the radius of the cylinder and t its length, and suppose that the tank is no more than half full, so the height 11 of the liquid is at most r. See Figure 2 for a cross section and let A/, denote the area of the shaded sector. Figure 2. The volume of liquid in the tank is then V = k!a/, where k is the conversion factor from volume in cubic units to liquid volume, e.g., 7.48 gal./ft3. If the tank is more than half full we can invert the above picture and the shaded portion then represents the empty part of the tank, so V = k!(nr2 - A,,). Thus we need only consider the case h 5 r. Two quantities are of significance in the calculation of the area A/,: the central angle 0 in radians subtended by the chord representing the level of the liquid, and the height 11 (see Figure 3). Figure 3. The area of the pie-shaped circular sector is ir20, and the two right triangles have two possible interpretations (Figure 4): r sin 812 (T' - (r - h)')"' Figure 4. VOL. 32, NO. 5, NOVEMBER 2001 THE COLLEGE MATHEMATICS JOURNAL

4 The area of the isosceles triangle in terms of 6' is or, in terms of h, Subtracting this area from that of the circular sector gives two expressions for Al,,one in terms of 6' and one in terms of 11: where 0 = 2 cos-' (y), or h = r - r cos 6'12,or The last formula is simplified by the rescaling p = $, which will be useful in later calculations, and becomes Using (1) and (3)we have the following solutions to the given problems: Problem 1. a) Given a height hofor a dipstick mark, find 4= 2 cospl (+) and compute All, = r2a(qo).then Vo= ktai,, is the volume level associated with ho,or b) Given a height ho for a dipstick mark, let pa = $ and compute All, = r2b(po).then Vo= klai,, is the volume level associated with ho. Problem 2. a) Given a volume Vosolve Vo= klr2a(6')for 8 = 6'0, 0 < Qo < n.then ho = r - r cos 6'012 is the height of the dipstick mark associated with volume Vo,or b) Given a volume Vosolve Vo= klr2b(p)for p = p0,o < 10< 1. Then ho = rpo is the height of the dipstick mark associated with volume Vo. 0THE MATHEMATICAL ASSOCIATION OF AMERICA

5 Example. Let r = 2 ft. and t = 10 ft., so the total liquid volume of tank is V = (7.48)(10)~(2)~ E 940 gallons. Solution (Problem 1). If we mark the dipstick every six inches then marks at height h, = 72/2 ft., rz = 1,...,4, would correspond to p,, = 1214,and volumes 0,, - sin 0, V,, = (7.48)(10)(2)~ The first four marks on the dipstick would be approximately 68, 184, 322, and 470 gallons. Since h4 = r it is a simple arithmetical calculation to compute the remaining marks; for instance the next mark would be ( ) = 618 gal. Solution (Problem 2). If we mark the dipstick at 100 gallon intervals, then V,, = loon, 72 = 1, 2,...,9 and to obtain the first four marks we must solve 100n El,, - sin 0, -, h,, = 2-2 cos 0,,/2 (7.48)(10)(2)2 2 The first four one hundred gallon marks would be , , , and feet. Since the next mark will correspond to a volume of 440 gallons in the empty part of the upper half of the tank, the remaining marks will have heights 4 - h,, where h,, corresponds to volumes , n = 0, 1,...,4. While the transcendental equations involving 0 in Problems 1 and 2, parts a) are much easier to solve than parts b), the natural parameter is the height h, or the normalized height p = hlr. The simplest but often impractical solutions to the problems are: Problem 1. Make the desired calibration marks on the dipstick, insert the stick, and begin filling the tank until the level reaches the first mark. Measure the volume poured and mark the stick; continue filling until the second mark is reached, etc. Problem 2. Begin filling the tank until the first volume level is reached, insert the stick and mark it. Continue filling until the second volume level is reached, insert the stick, etc. VOL. 32, NO. 5, NOVEMBER 2001 THE COLLEGE MATHEMATICS JOURNAL 355

6 Readers satisfied with these solutions should probably not have started reading this paper and should stop now. What is needed is an accurate but simpler expression that approximates B(p) = cospl(l - p) - (1- p)(2p - p2)'f2. Its second term is p 2 (4) The maximum error in replacing (1 - p/2)'i2 by 1 - ip/2 is approximately foro~p5 1. A difficulty with the first part of the expression is that cos-'(1 -p) has an infinite derivative at p = 0 SO a Taylor series expansion fails. We use instead an integral representation: Combining (4) and (5) gives the approximation which is simpler to compute and is very accurate as the following table indicates: P B(p) B(p) approx. Error The obvious advantage of the approximation is that there are no transcendental expressions to evaluate and the calculation can be done with a simple calculator. Then the approximate solution to Problem 1 is, given a height ho < r, the corresponding volume is For instance, in the example above with r = 2 ft., Y = 10 ft. the approximate volume corresponding to a height of 1 ft. is 318 gallons compared to the actual answer of 322 gallons. THE MATHEMATICAL ASSOCIATION OF AMERICA

7 Problem 2 is a much tougher nut to crack, but the above approximation does provide a polynomial to solve instead of a transcendental equation. Given a fixed volume Vo < V/2then we must solve for p. Let z = = (!)'I2 and M = Vo/klr2,hence 0 < M < n/2,then the previous equation becomes the polynomial equation where 0 5 z < 1. A straightforward analysis of P(z) and its derivative shows that there is only one root in the interval 0 < z < 1. For example, if a tank is 114 full then Vo= ktnr2/4 so M = in and P(z) = 572% 3202' + 60" with root z E Then h/r = z so h 2 & r. The exact answer using A(%)and solving % - sin 6' n/4 =, h=r-rcos%/2 2 is h = r. The approximation given above for B (!) is useful and has some computational advantage. The reader or students might wish to consider Problems 1 and 2 for a spherical tank, or a cylindrical tank with an elliptical cross section lying on its side. For the spherical tank the exact expression for the volume is a cubic in h, whereas for the elliptical tank the solution is the same expression as (2)with r2replaced by ab and!by ha 7 where 2b and 2a are the major and minor axes, respectively, of the ellipse. Another application would be in solving a problem like When h = ho 5 r the liquid level in the tank is rising at a rate of q ft.1min. How fast is the volume increasing at this moment? Using the two exact expressions ( I) and (2)for V we have where But replacing B(p) with its approximation results in a simpler differentiation. We obtain VOL. 32, NO. 5, NOVEMBER 2001 THE COLLEGE MATHEMATICS JOURNAL

8 For instance, in the previous example (r = 2 ft., t = 10 ft.) if ho = 312 ft. and q = 2 dh in./min. = ft./min. = ii;, then = 0.75 and The exact answer, obtained after considerably more calculation, is gal./min. Acknowledgments. The authors wish to express their appreciation to Dave Engelhard, Michael Kallaher, and Jack Robertson of Washington State University whose wonderful collection of agricultural related mathematics problems, Tools of the Trade, (COMAP, 1999) provided a source for the analysis above. They also wish to point out an entertaining discussion of the dipstick problem in the book, Calculus Mysteries and Thrillers, (MAA, 1998) by R. Grant Woods. A New Duality In "Java Man," (New K~rker,July 30, 2001, pp ) its author. Malcolm Gladuell. quotes a list from Tlle World cf C(iflerile by Bennett Alan Weinberg and Bonnie K. Bealer (Routledge, 2001) that contrash coffee and tea which, Weinberg and Beales say, ha\e come to represent almost entirely opposite sensib~lities: Coffee Asl~ect Ma1 e Boisterous Indulgence Hardheaded Topology Heidegger Beethoven Libertarian Promiscuous Trrl AF~I~c~ Feniale Decorous Te~npesance Rornantic Geometry Carnap Mozart Statist Pure Readers of the.iourn~ilinay not find the contrast between the items on the fifth line of the list quite as marked as that between some of the others. They may be too close to the subjects. 0THE MATHEMATICAL ASSOCIATION OF AMERICA

The College Mathematics Journal, Vol. 24, No. 4. (Sep., 1993), pp

Taylor Polynomial Approximations in Polar Coordinates Sheldon P. Gordon The College Mathematics Journal, Vol. 24, No. 4. (Sep., 1993), pp. 325-330. Stable URL: http://links.jstor.org/sici?sici=0746-8342%28199309%2924%3a4%3c325%3atpaipc%3e2.0.co%3b2-m