Practical examples of Calculus Differential calculus is a branch of calculus that studies the rate at which quantities change. The other branch is integral calculus. There are several practical applications of differential calculus the most common being plotting graphs of some formulae and functions. The directions in which a graph will take at a specific point or the shape of the graph at a particular point as well as its dimensions can be determined using differential calculus. Moreover, a number of disciplines including physics, medicine, economics, engineering and statistics use differential calculus as it is useful in creating models that can help us obtain optimal solutions.

1.1 The Disciplines. (a) Uses of Calculus in Engineering In the field of engineering, calculus is essential in the day to day practice of an engineer. Over the years, there has been development of softwares based on calculus simplify engineering designs for optimization. (i) Civil Engineering Calculus is needed to obtain the basic fluid mechanics equations. Calculus numerical methods are used to obtain the results in all numerical analysis programs which aid in the design of storm drain and open channel systems.

Volume is evaluated as the area under the curve of a plot of flow versus time using calculus; this is inhydrology. (ii) Structural Engineering Forces in complex configurations of structural elements use calculus. For example, calculus is used to calculate bearing capacity and shear strength of the soil in a soil structure context. (iii) Mechanical Engineering. In computing the surface area of complex objects to determine frictional forces, designing a pump according to flow rate and head and calculating the power provided by a battery system, calculus is used. (i) Aerospace Engineering Thrust over time calculated using the ideal rocket equation, analysis of rockets that function in stages and gravitational modeling over time and space are some of the applications of calculus in aerospace engineering.

(b) Uses in Biology To find out the rate of muscle contraction, dissolution of drugs into the bloodstream and growth of bacteria, derivates in calculus are used. When a drug is administered to the body, it is dissolved into the bloodstream. The rate at which this occurs is known as dissolution. By using this rate, differential equations are used to relate the concentrations of drugs at different times. This can be modeled using the Noyes-Whitney equation, which is as follows- dW/dt is the rate of dissolution.

A is the surface area of the solid. C is the concentration of the solid in the bulk dissolution medium. Cs is the concentration of the solid in the diffusion layer surrounding the solid. D is the diffusion coefficient. L is the diffusion layer thickness.

Solving By using the Noyes-Whitney equation, the rate at which the drug absorbs can be found as dW/dT. In many cases the rate of elimination of a drug can be described as being dependent on or proportional to the amount of drug remaining to be eliminated. Therefore dX/dt = -K * X, -K would be the rate constant and X is the amount remaining to be eliminated, thus making dX/dt the rate of elimination. Solving pt 2 dX/dt = -K * X = rate of elimination after integrating this to find an original equation, the equation X = X0*e^-kt can be used.

Differentiating this equation, using the given components can solve for the rate of elimination. This, in accordance with the rate of dissolution can be used to find a drugs path in the body when administered Application of Integration Now that we can solve the rate of dissolution of a drug into the bloodstream, we can use this to find how much has been absorbed. Say a drug absorbs into the bloodstream at the equation modeled by R(M) = 3M2/(M3+1) Where M represents the dosage. Solving By integrating the equation, R(M) = 3M2/(M3+1) we can find the total amount of the drug absorbed into the bloodstream over a given interval. By using the rate found previously, the second interval can be found.

Say this drug was prescribed as a 4 miligram dosage. Solution R(M) = (0-4)? 3M^2/(M^3+1)dx ? 3M^2/M^3+1 U=(M^3+1) ? 3M^2/U Du=3M^2 dM ? 1/U Du LnU Ln(M^3+1) l(0-4) Ln((4)^3+1) – Ln((0)^2+1) Ln(65)-Ln(1) 4.17438727-0 = 4.17438727 total mg absorbed Application of Chain Rule When cancer cells form a tumor, they grow in many different shapes. Some tumors, such as an adenoma (a benign tumor originating in glandular tissue) tend to be spherical. Assuming the radius grows at a constant rate of k, we can find the rate of growth of the volume of tumor at a given radius. For this situation we decided to make the radius equal to 5 millimeters.

Solving Volume of a circle = 4/3-r^3 Now differentiate with respect to R dV/dR = (4p/3)(3R²) = 4pR² Now plug in the R 4-(5^2)=100- Solution The volume of the sphere is changing at 100- millimeters per second when the radius of the tumor is equal to 5 millimeters. (c) Uses in Business and Economics In Business