Evaluation of Testing Theories for Global Manufacturing Assignment Sample

Introduction: Testing Theories for Global Manufacturing - A Detailed Evaluation

In this assignment, the engineering mathematical operations have been done for different problems given in the tasks. The calculus equations are used to complete these tasks. Mathematical operations are at the basic level for these given tasks. There are a total of four tasks; some of the tasks have two parts consisting of different mathematical questions. In some parts of the assignment, simple integration and some of the parts complex integration processes have been done where the formulas are used for these particular problems. Calculating these problems has been at a great pace as the formulas have helped in solving the problems. In some of the tasks of this assignment simple derivation and some parts of complex derivation are used as there are problems that have a mixture of these kinds of problems. The calculations are done carefully to avoid any mistakes. The calculation of the circuit is done based on the idea of physics, applying mathematical formulas. The calculations are done using trigonometric calculus, which is essential to determine the result. The minima and maxima concept is done by using the integration of the values given by evaluating the graph. The graphical representation of the formula and examples are used for solving the problems. The problems are related to the whole trigonometric part. The assignment has completed all the tasks step by step, and the description of the solved problems has been elaborated to create a complete idea about the method of the solving process.

Discussion on Key Mathematical Techniques: Differentiation, Integration, and Derivatives

Task 1: Calculation of Rate of Change Using Differentiation Techniques

The given task in this section involves the differentiation of the given questions. The differentiation is done using the proper formula. After that, the rate of change is calculated where the time t is given 2 seconds.

For the first calculation, the velocity,v, is calculated with the rate of change using the value of t, which is 2 seconds. The solved value has come out as 10 m/s^2, considering the value of time as 2 seconds (Torelli et al., 2020). The solving of the given problems has been done by using the power rule, chain rule, product rule, etc. The given equations have been derive, and the value of time is substituted in the equation (Pieroni et al., 2020). The rate of change is the derivation of the given equation by giving the given value for the equation.

• dv/dt = (20 + 30)/5 = 10 m/s^2, considering the value of t = 2s, given in the problem
• di/dt = (6t + 5 - 3)/3 = 7 A/s, considering the value of t = 2s, given in the problem
• dv/dt = (21t2)/e^(7t^3) = 42 m/s^2, considering the value of t = 2s, given in the problem
• di/dt = 12e^(6t-2) = 144 A/s, considering the value of t = 2s, given in the problem
• dv/dt = -6sin(4t + 14) = -6sin(26) = -5.416 m/s^2, considering the value of t = 2s, given in the problem
• di/dt = 3cos(t)e^sin(t) = 3cos(2)e^sin(2) = 2.818 A/s, considering the value of t = 2s, given in the problem

The different problems are calculated by doing proper derivation, and the results are estimated correctly for the given problems.

Task 2: Mastering Circuit Analysis: Calculating Current in Capacitors and Inductors Using Integration

Part 1: Calculating the Current Stored in a Capacitor Using Integration

In this part, the current stored in the capacitor is calculated. The circuit is closed. The calculation is done when the switch is off. The values of the power source, resistor, and capacitor are given in the problem. The values are substituted in the equation given for the flow of current through the circuit. Integration of the equation for the current answers the amount of current stored in the capacitor. The amount of current stored is the value of the capacitance of the capacitor in the circuit.

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• i = (E/R)e^(-t/RC)
• q = ∫ idt = (E/R)∫ e^(-t/RC) dt
• = (E/R)(-RC)e^(-t/RC)]|0^2
• = (E/R)RC[1 - e^(-2/RC)]
• Let R = 22 ohm, C = 47 μF = 47 *10^-6 F, E = 12 V
• Then q = (12/22)(22*47 *10^-6)[1 - e^(-2/22*47*10^-4)]
• = 2256[1 - e^(-2/22*47*10^-4)]
• = 9.99 F
• ≈ 10 F

The current stored in the closed circuit when the switch is off is 10 Faraday, which is calculated by integrating the equation of current flowing through the circuit.

Part 2: Calculating Inductor Current Using Integration for a Given Time Interval

In this calculation, integration is done to calculate the inductor current for this given problem. The calculation is done by substituting the value of L, which is 250 mH concerning the time (t). The integration of sin(100t) for the given time gap, which is between 0 seconds to 1.5 seconds. In the final step, the value of L is substituted to determine the value of il, and the value of the inductor current is calculated.

The equation is calculated as per the given value of time, and the integration is done by isolating the value of 1/L, which is then multiplied by the value of the integration of sin(100t) dt, which at last gave the proper value of the inductor current.

• il = (1/L)∫sin(100t) dt
• = -(1/100L)cos(100t)]|0^1.5
• = (1/250)[cos(0) - cos(150)] = 0.245 A

For the given equation, the induction current is calculated for the given time gap is 0.245 Ampere. The integration of the given equation is done to get the value of the induction current.

Task 3: Integrating Exponential Curves: Growth and Decay Explained

Part 1: Calculating the Area Under an Exponential Growth Curve Using Integration

In this section, the function y is integrated concerning the time gap given between 2 seconds and 4 seconds. The calculation for this given problem gives the result as the area comes under t 2 seconds and tt4 seconds. The exponential growth curve is calculated by substituting the proper values and creating the proper area within this curve.

The curve could be seen within the given time gap. The calculative part was done by integrating the function y c,onsidering the time gap in the integration from 2 seconds to 4 seconds (Waheed and Zhang, 2022).

• ∫ (5 - 2e^-3t ^^4 dt, considering the value of time between 2 seconds and 4 seconds
• = 64/3 - 32e^-6 + 16e^-12 - 64/9
• = 17.48

The area of the exponential growth curve is 17.4,8, which is calculated by considering the time gap given for the equation.

Part 2: Calculating the Area Under an Exponential Decay Curve Using Integration

The calculation in this part is done in the same way as part 1 of task 3. In this part, the exponential decay curve was calculated for the given equation (Nielsen et al., 2020). The function y was calculated by integrating the equation for the time g,ap which is provided between 0 seconds to 5 seconds.

The area under the curve was exponentially decay,ing which can be only calculated by applying integration.

• ∫ E e^-7t dt for the time gap given for the given value between 0 seconds to 5 seconds
• = -1/7 e^-7t ]|0^5
• = 1/7 (1 - e^-35)
• = 0.0465

The area of the exponential decay curve of the given equation calculated is 0.0465 by considering the time gap given for the equation.

Task 4: Understanding Maxima, Minima, and Turning Points Through Derivatives

Part 1: Determining Turning Points, Maxima, and Minima Using Derivatives

In this part of task 4, thecoordinatess of the turning point for the given function are to be determined along with determining whether the state of the equation is either maxima or minima.

In this part, the first derivative has been taken under consideration; the value of this first derivation is taken as 0, and the point (4/3, 8/3) has been shown by evaluating the sign of the derivative, which is done after the first derivative (Saidu and Aifuwa, 2020).

The location of the turning points is crucial to determine the value of maxima and minima. The maxima and minima are determined by seeing the value of the calculated result (Ahmed et al., 2020). If the value of the result is negative, then the value is considered as maxima, and if the value of the result is positive,e then the value is considered as minima.

• dy/dx = -6x + 8
• Let dy/dx = 0 calculation gives x = 4/3

The calculated value is a maxima as the second derivation is d2y/dx2 = -6, as the value is negative.

The maxima and minima determination is done by deriving the equation two times. The coordinates are also determined by calculating this equation.

Part 2: Determining Maxima and Minima Using First and Second Derivatives

In this part, the value of the first derivative is considered as zero. To determine the value, whether it is maxima or minima, the second derivative is calculated (Mardani et al., 2020). The derivatives are taken to determine the value and the sign.

To determine the state of the equation, which falls under maxima or minima, the first and second derivations of the equation are mandatory.

• dy/dx = 3x^2 + 4x - 5
• Setting dy/dx = 0 gives x = -1, 1
• d2y/dx2 = 6x + 4 > 0 at x = -1 -> minima
• d2y/dx2 = 6x + 4 < 0 at x = 1 -> maxima
• Minima has been determined at x = -1, y = -2
• Maxima has been determined at x = 1, y = 3

The maxima and minima are determined by calculating the equation. The calculation is done by deriving the given equation two times.

Conclusion

In the conclusion, the inference is determined as the calculations of the given problems are done thoroughly. I works as a test engineer for a global manufacturing company in the stream of developing electrical and mechanical components and systems. The calculation of the problems has been done by applying the formulas of derivation and integration of calculus. The theory of theory, hypotheses, and principles have been taken under consideration for solving these problems. The attended problems have the given values depending on which calculations have been done. The area of the curve is determined by the calculation of the given function. The functions are derived to calculate the maxima and minima. The coordinates of the turning point of the curve are determined from the equation given. The derivations are done concerning the given time. For all the problems, there are specific values of time and time gaps which are used to calculate the equations for the given problems. For the circuit, the given values of voltage, capacitor, and resistors are taken into consideration and substituted in the equation. The circuit is closed, and the calculator is done based on the given equation of the current flowing through the circuit. The calculation of the current stored in the capacitor is made by integrating the equation given for the current flowing in the circuit. After that, the inductor current is calculated for the given time, which consists of the current flow in the inductor. The exponential growth curve is also determined by the integration of the given y function. The given time gap is taken intor consideration for this calculation. In the same way, the exponential decay curve is calculated for the y function. The integration is done to get the coordinates of the curve. Here, the time gap is also taken into consideration to calculate the values and solve the problems. At last, it can be seen that the calculations are done successfully by interpreting the problems using the derivation and integration formulas, which has helped in getting the correct solutions.

References

Journals

• Torelli, R., Balluchi, F. and Furlotti, K., 2020. The materiality assessment and stakeholder engagement: A content analysis of sustainability reports. Corporate Social Responsibility and Environmental Management, 27(2), pp.470-484.
• Pieroni, M.P., McAloone, T.C. and Pigosso, D.C., 2020. From theory to practice: systematising and testing business model archetypes for circular economy. Resources, conservation and recycling, 162, p.105029.
• Patwa, N., Sivarajah, U., Seetharaman, A., Sarkar, S., Maiti, K. and Hingorani, K., 2021. Towards a circular economy: An emerging economies context. Journal of business research, 122, pp.725-735.
• Mardani, A., Kannan, D., Hooker, R.E., Ozkul, S., Alrasheedi, M. and Tirkolaee, E.B., 2020. Evaluation of green and sustainable supply chain management using structural equation modelling: A systematic review of the state of the art literature and recommendations for future research. Journal of cleaner production, 249, p.119383.
• Nielsen, B.B., Welch, C., Chidlow, A., Miller, S.R., Aguzzoli, R., Gardner, E., Karafyllia, M. and Pegoraro, D., 2020. Fifty years of methodological trends in JIBS: Why future IB research needs more triangulation. Journal of International Business Studies, 51(9), pp.1478-1499.
• Saidu, M. and Aifuwa, H.O., 2020. Board characteristics and audit quality: The moderating role of gender diversity. International Journal of Business & Law Research, 8(1), pp.144-155.
• Ahmed, W., Ashraf, M.S., Khan, S.A., Kusi-Sarpong, S., Arhin, F.K., Kusi-Sarpong, H. and Najmi, A., 2020. Analyzing the impact of environmental collaboration among supply chain stakeholders on a firm's sustainable performance. Operations Management Research, 13, pp.4-21.
• Waheed, A. and Zhang, Q., 2022. Effect of CSR and ethical practices on sustainable competitive performance: A case of emerging markets from stakeholder theory perspective. Journal of Business Ethics, 175(4), pp.837-855.