Difference between Maximum and Minimum Heights for Both the Roller Coasters - Speech or Presentation Example

Mathematical Modeling Teacher’s Institute Mathematical Modeling Introduction Mathematical modeling is an approach to describe different phenomenon of nature, society and human activities using various mathematical formulas. Mathematical modeling describes real world issues. These issues may relate to science, engineering, economics, sports, financing and many other disciplines. Using mathematical formulas, people express the cause and effect relationship of a phenomenon. Mathematical modeling is nothing but finding relationship among different variables that describe a phenomenon or object (Simon Fraser University n.d.). The process of mathematical modeling is like conducting a lab experiment in physics, chemistry, biology, medicine, and other disciplines. In the laboratory, one tries to establish cause and effect relationship of a phenomenon. Due to various objective difficulties, many relationships cannot be established through laboratory experiments. This is when the relationship is established through mathematical formulas that model a phenomenon or object. Then by changing the values of one or more variables one can get the answers needed in the real world problems. This assignment does not report either development of a mathematical model or present a mathematical model. It uses a mathematical model to derive certain cause and effect relationship. The assignment uses two different models that describe at what height from the ground a coaster will be located at a certain period. Model 1: Maximum and Minimum Location The model expressed by the equation h (t) = - 2 t3 + 34 t2 – 128 t + 96 describes the location of a coaster from the ground for 12 seconds after it comes out of the loop. The given equation is a third degree polynomial equation where the dependent variable is height and the independent variable is time. The scope is to determine maximum and minimum heights from the ground level. The maximum and minimum will be found through the derivatives of the equation. The condition for maximum is h’ (t) = 0, and h” (t) is negative. The condition of minimum is h’ (t) = 0, and h” (t) is positive (Lawrence Spector n.d.). Figure 1 shows the movement of the coaster for 12 seconds (Microsoft Corporation n.d.,). Figure 1. Coaster movement for 12 seconds h’ (t) = - 2* 3t (3-1) + 34 * 2t (2-1) - 128 + 0. Thus, h’ (t) = - 6t 2 + 68 t - 128 + 0 For h’ (t) = 0 ; - 6t 2 + 68 t - 128 = 0 The roots of the above equation are found from the relation; t = [{- 68 +/- (68^2-4*-6*-128)} / 2* -6]. The two solutions are t1 = 2.38 and t2 = 8.95. The second derivative of the equation h (t) = - 2 t3 + 34 t2 – 128 t + 96 is h” (t) = - 12t + 68. For t = 2.38, h” (t) = 39.4, which is positive. For t = 8.95, h” (t) = - 39.4, which is negative. Hence the coaster is at the minimum of the loop at t = 2.38 sec, and at maximum at t = 8.95 sec. Minimum height; h min = - 2*2.38^3 + 34*2.38^2 - 128*2.38 + 96 = 43.01 Maximum height; h min = - 2*8.95^3 + 34*8.95^2 - 128*8.95 + 96 = 240.05 Model 1: Solution of t using Factor theorem When the coaster is at the ground level, the height is zero. So, the equation becomes - 2 t3 + 34 t2 – 128 t + 96 = 0. This is a cubic equation; it will have three roots. The solution is achieved using factoring theorem (Purple Math n.d.). - 2 t3 + 34 t2 – 128 t + 96 = 0 (t-4) (-t^2+13t-12) = 0 (t-4) (-t^2 + t + 12t – 12) = 0 (t-4) (-t+12) (t-1) =0 Three roots of the equations are t1 = 4 ; t2 = 1, and t3=12. For these values h(t) = 0. Model 2: Maximum and Minimum Location Model is represented by equation; h(t) = t^3 – 14t^2 + 40 t. First derivative, h’(t) = 3t^2 – 28t + 40 Second derivative, h” (t) = 6 t – 28 Roots of equation 3t^2 – 14t + 40 = 0 are t1 = 1.76, and t2 = 7.57 For t1 = 1.76, h” (t) = - 17.44, which is negative. For t2 = 7.57, h” (t) = 17.42, which is positive. Hence the coaster is at the maximum of the loop at t = 1.76 sec, and at minimum at t = 7.57 sec. Maximum height; h min = 1.76^3 - 14*1.76^2 + 40*1.76 = 32.48 Minimum height; h min = 7.57^3 – 14*7.57^2 + 40*7.57 = -65.67 The ride start at t = 0 sec; h (0) = 0^3 – 14*0^2+40*0 = 0; at the ground level Cylindrical Can A can consist of two circular areas; top and bottom = 2 π r2, and a cylindrical perimeter = 2 π r h. Volume of the can, V = π r2 h = 333; or h = 333 / π r2 . Metal used in the can, M (r) = 2 π r2 + 2 π r h = 2 π r2 + 2 π r (333/ π r2 ) = 2 π r2 + 666 r -1. M ‘ (r) = 4 π r – 666 r -2 For minimum metal M’(r) = 0 ; 4 π r – 666 r -2 = 0 ; or 4 π r3 = 666; or r3 = 666 / 4 π; or r = 5.50 Therefore, minimum area needed= 2 π r2 + 2 π r (333/ π r2 ) = 2* π*(5.50)^2+666*(5.50) -1= 311.15 cm2 Box The height of the box = x, and the area of the box is (60-2x)*(40-2x). Volume of the box, V (x)= x *(60-2x)*(40-2x) = 4x3 – 200 x2 + 2400x V’(x) = 12x2 – 400 x + 2400 At maximum V’(x) = 0; therefore, 12x2 – 400 x + 2400 = 0. This equation gives two roots; x1 = 7.85, x2 = 25.48; however, x cannot be more than 20 because of restriction of the side 40-2x. Hence; x = 7.85 V = 4 * 7.85^3 – 200 * 7.85 ^ 2 + 2400 * 7.85 = 8450.45 cm3 x 60 – 2x x 40 - 2x Conclusion This assignment used the following mathematical models: h (t) = - 2 t3 + 34 t2 – 128 t + 96 h(t) = t^3 – 14t^2 + 40 t M (r) = 2 π r2 + 2 π r h = 2 π r2 + 2 π r (333/ π r2 ) V (x)= x *(60-2x)*(40-2x) = 4x3 – 200 x2 + 2400x. The above models found different answers that relate to real world issues. Mathematical model is a tool that can be used to solve various practical issues. Reference List Lawrence Spector n.d., Maximum and Minimum Values. [ONLINE] Available at: http://www.themathpage.com/acalc/max.htm. [Accessed 20 June 14]. Microsoft Corporation n.d., Basic Tasks in Excel 2013. [ONLINE] Available at: http://office.microsoft.com/en-001/excel-help/basic-tasks-in-excel-2013-HA102813812.aspx. [Accessed 20 June 14]. Purple Math n.d., The Factor Theorem. [ONLINE] Available at: http://www.purplemath.com/modules/factrthm.htm. [Accessed 20 June 14]. Simon Fraser University n.d., What is Mathematical Modeling. [ONLINE] Available at: http://www.sfu.ca/~vdabbagh/Chap1-modeling.pdf. [Accessed 20 June 14]. Read More