Simple and Complex Mathematical Concepts in Engineering - Term Paper Example

Engineering problems report. By Name Course Professor University City/State Date Engineering concepts in mathematics Content Abstract The engineering field is composed simple and complex mathematical concepts. This report aims at giving an analogy of some of these concepts. The analogy will be accomplished through a consequential analysis of question 1 to 3.Application of mathematical graphs, formulas and results discussion of the three questions is prove of understanding of engineering mathematical concepts. More concepts are tested through a flow of communication of the above concepts in this report. Introduction The report gives an understanding of mathematical concepts through the accomplishment of three tasks (Gong, 1996). The results of the tasks will help in giving advice to an engineering organization. The three tasks act as a test of logical and mathematical concepts in the field of engineering. Various laws are stated and explained. The general engineering study entails understanding challenges in different fields of study especially in the field of mathematics and other sciences like physics. Understanding these subjects helps an engineering student in tackling some of the challenges that they may face in future. Application of theoretical formulas learnt in class to get the required results as shown in this essay has been used to show that the students understands the theories learnt in class and can be able to apply them in solving various tasks. Below is a summary of the test areas in the Analysis Task 1 The instantaneous value of voltage in an a.c circuit is given by the equation: V=300sin (50πt-0.6) volts. Where 300, 50 and 0.6 are the constant parameters. The 300 in the equation indicated the maximum amount of voltage that can be reached by the circuit at different time intervals. Different voltages at various time intervals can be deduced from a sine graph with time drawn against voltage (Gong, 1996). The sine wave is used as the guiding factor in the analyses of the results. A single wave of the sine wave is interpreted as follows: The top most part of the wave (shows the maximum voltage, a line drawn a perpendicular to the x-axis cuts the wave at a point where the cut time yields the voltage indicated at the intercept with the curve). From the graph, it is deduced that when the t=0; V=-169v when t=10; V=246.2v and finally, voltage is at a maximum when the time elapsed is 13.9ms. Periodic time shows the amount of time taken by the current in the circuit to make one complete cycle (Gong, 1996). It is measured from the graph by calculating the difference between the last point and the start of the cycle. B Decay period of a capacitor is the duration a capacitor takes to become fully discharged. The capacitor in task 1 has an initial voltage of 200 volts, and after 6 sec the capacitor has zero volts. The defining equation for voltage at after a certain amount of time is given by; V=200 e –t/3. A curve of time is plotted to show the relationship between the two variables (Kreyszig, 1988). Drawing perpendicular lines falling to the x-axis, where the lines touch the curve shows values of voltage at the time indicated at the x-intercept of the line. The value of the voltage when t=0 and when t=10 is derived from the voltage decay diagram by drawing straight lines perpendicular to the x-axis line at x=0 and x=10. The value of v when t=0 is 200v, and after 3.4 seconds, the voltage is 65v. The decay of the voltage is exponential meaning that the voltage disappearance happens rapidly with as time elapses. Log graph The values for x(mm) and L (Kilo Newtons) are estimated using the graph by reading the values of x-axis against the corresponding values of y-axis. K is derived from the relationship; L-kxn=In(L) to find the value of k from the estimated values, the following general equation is applied; In(L) =In (k.xn). Making k the subject of the formula; k=L/xn the value of k for the function f L1 is; K=L1/Xn1. K is then from the table with the approximated results, n={In(L1)-In(L2)}/ln(x1)-ln(x2). L1 and x1 are the first values in the table where x=3 and L=1.7. K, therefore, becomes; L/X=1.7/30.93=0.612 Newton-Raphson method.Two results from graphical a graphical method and the The Newton-Raphson procedure is a scientific method of analysis that was discovered by Isaac Newton and Joseph Raphson. It is a procedure that is used in finding most appropriate approximations of roots of real-valued functions. It is commonly used in the analysis of thermodynamic functions. Given the first estimations of the values, the procedure is then used to find the rest of the estimate values of the data. The second (r) value of the data is derived from the first R-value by application of the procedure, and the answer 0.171 was found through the formula (x1-f(x1)/f’(r1)) then the answer rounded off to three decimal places. Newton-Raphson formula is compared for clarity. Calculation for A is given by the formula; f(x)=1/3(x3-6X2 + 9x -1 When A is assumed to be 0.1, the equation becomes 1/3{(0.1)3- 6(0.1)2 +9(0.1)-1}= -0.053. The values of x1 are substituted for X, subX2, and X3 …. Sub x1=-0.053. The function therefore becomes f’(x)=X2-4x + 3. X2=0.1—0.053/2.61=0.1203. The process continues until the value of X4 is derived and the results tabulated as shown below; X1 0.1 X2 0.120306513 X3 0.120614688 X4 0.120614758 Which are the cordinates of the for corners of the spar? Finding the area of the graph using the Simpsons' rule. Simpson’s rule is a mathematical formula that uses quadratic polynomials to estimate integrals of functions. Simpsons rule states that; The length fro (A to B) is given by subtracting b from A which gives; B-A=2.347-0.121=2.226m the interval is given by the division of length with the vertical values in the graph which is; length/yn=2.226/10 = 0.2226m. From the values of X and Y, the area is given by; The Simpsons rule can only be used correctly when applying an even number of intervals. From the first Simpsons calculation which applied ten strips, we can see that there is 0% error when compared with definite integration (Kreyszig, 1988). However when an odd number of intervals are used on but the Simpsons rule should be completely accurate for any quadratic up to the third derivative, therefore proving an odd number of intervals cannot be used to achieve an accurate estimation of the area. Definite integration Definite integration has a starting point and the last point under a quadratic curve. The start and end point are shown on the top and lower part of integral symbol respectively. For the area under the curve, the start point, in this case, is at 0.120614. The formula; which substituting the values for x in the equation gives area of M2 as approximately 1.4095 which is then converted into cm2 by multiplying by 10000cm2 to make it 14095.38931. Task 2 Torricelli law The Torricelli law states that the speed of a liquid flowing under a force of gravity out of a gap in a tank is proportional to the square root of the vertical distance (h) between the fluid surface and the center of the opening (Kreyszig, 1988). The equation can summarize it; V=2gh1/2. Acceleration due to gravity on the earth’s surface is approximated as 9.8 meters per second. The theory was discovered by Evangelisa Torricelli and is widely used in the field of physics to compute dynamics of fluids in machinery systems. When the water starts flowing, time is at zero. The speed of the water at this point is zero since substituting the value of time to the initial equation gives a result of zero. The outflow of water from the cylindrical tank The equation gives the volume of water outflow in a short time also known as instantaneous time; ∆v= Ar.∆t. The volume of water in the tank is, therefore, ∆v= -B.∆h. The symbol ∆ denotes a small change in the system (Kreyszig, 1988). Substituting -B.∆h for ∆v in the first equation gives; Ar.∆t.= -B.∆h. The equation is then re-arranged to find the ratio of a small change in height to a small change in time.The derivative of height versus the derivative of time hence give –A/B(0.600√2gh(t)) where g=981cm/sec2.. When the tank is full; When the value of h=0 the take is empty hence; Tank is empty when h=0 B Newton’s law of cooling Newton law of cooling indicates: “the rate of change of the temperature of an object is proportional to the dissimilarity between its temperature and the ambient temperature.” This means that the surrounding environment of an object directly influences the rate at which the object cools. The results are calculated using the formula; dT/dt= k(T-TA) integrating the equation gives; where k is found to be – 0.0558 at 6am approximately 6 hours down the process. T = 45+25e-0.055 = 60.99F Euler's method 2 Euler method is an explicit numerical method that is used to generate numerical solutions. Euler method is introduced by the functions y and y prime functions. These y values are responsible for the initial value problem, and they take the form of; y’=f(x,y); y(x0)=y0 The range of the area to be calculated is then determined and by creating stripes with heights of h. The process is terminated when the right end of the desired interval is reached. Below, the table shows the data that was used to compute A. X0 Y0 (Y1)0 1 1 1/3 2.5 2 1.2 0.83 2.288 3 1.4 1.406 3.22 4 1.6 2.05 3.52 5 1.8 2.754 3.78 6 2 3.51 4 When x=1 and y=1/3 the integral = 0.2 the range is x=1 is equivalent to x=2. Through the Euler’s method, line 2= 1.8; line 3=1.6;line 4=1.4 and so on. Spring system The operation of spring is described by hooks law which states that the force needed to extend or compress an elastic material such as a spring by some distance is proportional to the distance moved by the elastic material. The law is summarized by the formula; F= Restoring Force= -ky. Another law that applies to the Spring laws is the newton’s law (Stroud, 2013). The law states that; the mass of a body multiplied by the acceleration of that body gives the force of application to towards that body. The formula then calculates the result of the spring problem; F=k.X, making k the subject of the formula the equation becomes k= F/x. the sabin w (F) = 98nt and x = 1.09m then K becomes 89.91. Mass is defined as weight/acceleration due to gravity which is 10N and weight are 2.998Kg. f= w0/2π= 0.477Hz to find cycles per minute, a multiple of 0.477 by sixty sec that makes a minute is computed hence cycles per minute = 28.62. Task 2 E A second order differential equation can be defined as an equation that involves the unknown function y and its derivatives y’ and y” using the variable x Substituting the value of t with a zero gives; dy/dx=0. A particular integral and a particular solution is then computed using the following equetions; Conclusion The aim of the report was to test mathematical concepts in the field of engineering some laws have been defined and problems solved using the laws. Graphs show the relationship of the data, and they are used to compare answers derived from formulas. Mathematical concepts learned in class are put into test through calculation of sample exercises. Some of the concepts learned in school are as listed in the introduction summary. Their applications in calculating different sample exercises prove understanding of the concepts studied. References Feynman, R.P. and Weinberg, S., 1999. Elementary particles and the laws of physics: The 1986 Dirac memorial lectures. Cambridge University Press. Gong, W., Taylor, P.A. and Dörnbrack, A., 1996. Turbulent boundary-layer flow over fixed aerodynamically rough two-dimensional sinusoidal waves. Journal of Fluid Mechanics, 312, pp.1-37. Kreyszig, E., 1988. Engineering mathematics. Wiley. Stroud, K.A. and Booth, D.J., 2013. Engineering mathematics. palgrave macmillan. Read More