Differential Calculus

Differential Calculus

Name

Institution

Tutor

Date

Differentiation is about determining the rate of change of a certain quantity compared to another one. Differentiation is useful when the rate of change is unvarying. Economists and business analysts use it to select rational course of action, forecast the effects of diverse business policies and to describe what is happening in the business world. They also use it to find optimal solutions. For example to minimize the costs of production, or to minimize the amount of material used in packaging. Other than in business world, calculus is applied in the analysis of crash data and it is also used in motor industry to improve the safety of motor vehicles. Also, differential calculus is significant in civil engineering. Engineers in designing dams, they do this by first determining the force of water from the surface area of the reservoirs. And to assess the suspension system of dump trucks (trucks are used to clear unused material from the construction sites) (Lebedev, 2012).

One significant application of derivatives is optimization. That is finding the situation in which a minimum or maximum will occur. Optimisation is important in engineering (minimum cost, maximum strength) and business (profit increase, cost reduction).

In using mathematical functions to model the things that happen in the physical world, the physical quantity under study if often referred to as the variable. An engineer is only interested in the turning point of a function, minimum and maximum values, and its ups and downs. They then draw the graph of the function using either a computer or graphical calculator. However, to locate the exact position of the points, differential calculus is used.

Differentiation is used to find the displacement of dumper trucks as well as its displacement. It is always recommended that the suspension systems of the dump trucks are assessed especially when large masses of waste material are loaded on it from a high point. This is because the truck bed undergoes velocity, displacement as well as acceleration.

To model this velocity, displacement and acceleration, a mathematical function is used. The function is used to get the acceleration and maximum displacement of the truck at a certain point.

This is the equation: Y=0.26066E-εωt sin (ωdt)

Where; ε is about 5 radians per second, Ω is about 0.18 and Ω is about 4.85 radians per second

To get the maximum displacement of the dumper truck; we first get the derivative of y with respect to time t.

That is QUOTE = QUOTE .v + QUOTE .u

And to get the first derivative, that is QUOTE , product rule is used. In product rule, the first function in the equation is multiplied by the derivative of the second function plus twice the derivative of the first function (Narayan 2005).

For example, given the function g(x) = f(x). H(x)

The first derivative g’(x) = f(x). h’(x) + h’(x). f’(x).

U= 0.26066 e-εωt

QUOTE = -εωt 0.26066e-εωt

v=sin (wdt)

QUOTE = wd cos (wdt)

To get the minimum or maximum velocity (v)

QUOTE = 0.261 e-εωt [-εωsin (wdt) +ωd cos (ωdt)] =0

Which implies that ωd cos (ωdt) – -εωsin (ωdt)

Therefore, εωsin (ωdt) = ωd cos (ωdt)

Εωtsn (ωdt) =ωd

Thus 0.9 tan (4.85t) = 4.85

Thus t= 0.3 seconds

Therefore when time t is 0.3 second, the displacement is approximately 0.2 meters. This is obtained by substituting time t with 0.3 seconds.

Acceleration of the track is determined by getting the second derivative of the equation.

From equation one, QUOTE = v= 0.27e-εωt [ωd cos (wdt) –εωsin (ωdt)]

Therefore acceleration a = QUOTE = QUOTE .v = QUOTE .u

Thus u= 0.27e-εωt

QUOTE =-εω0.3e-εωt

V=-εωsin (ωdt) +ωd cos (ωdt)

QUOTE =-ω2dsin(ωdt) -εωωd cos (ωdt)

Thus QUOTE = -εω 0.27e-εωt (-εωsin (ωdt) + ωd cos (ωdt)) + 0.27e-εωt (-εωωd cos (ωdt) – ω2d sin (ωdt))

0.27e-εωt [ε2ω2 sin (ωdt) -εωωd cos (ωdt) –εωωd cos (ωdt) – ω2d sin (ωdt)]

0.27 e-εωt [ε2ω2sin (ωdt) –εωωd cos (ωdt) –εωωd cos (ωdt) – ω2d sin (ωdt)]

0.27 e-εωt [(ε2ω2 – ω2d) sin (ωdt) – 2εωωd cos (ωdt)]

0.27 e -0.88t [ – 22.7 sin ( 4.9 t) – 8.5 cos(4.9t)

Since t=0.3, QUOTE = -4.9 metres per square second.

Thus after the maximum displacement at 0.3 seconds, the acceleration of the dump truck is about 5 meters per square second downwards.

For trigonometric functions; cos bx differentiates to –bsin bx and Sin bx differentiates to bcos bx.

Differentiation is used to get the stationary points namely minimum, maximum and the point of inflection. That is when the first derivative is equal to zero. The second derivative is used to classify the stationary points (Narayan 2005).

In graphical analysis, the derivative gives information on the slope or the gradient of a graph of a given function. Additionally, the point where the gradient is zero in a graph can be located using differentiation. This is done by first determining the points where the vertical axis crosses the horizontal axis. Also, find out what happens when x tends to infinity on both sides (Ayres 2000).

References

Ayres, F., Hademenos, G. J., Ayres, F., & Mendelson, E. (2000). Calculus: Based on Schaum’s

outline of differential and integral calculus. New York: McGraw-Hill.

Lebedev, L. P., Cloud, M. J., & Eremeyev, V. A. (2012). Advanced engineering analysis: The

calculus of variations and functional analysis with applications in mechanics. Singapore: World Scientific Pub. Co.

Narayan, S. (2005). Differential calculus. S.l.: Dk Agencies Ltd.