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# Applications of Calculus

Calculus (differentiation and integration) has a wide application not only for the classical optimization techniques but also for other areas of research including inventory models, nonlinear programming etc.

Today, in real life problems too, Calculus is applied in a major way.

Applications of Differential Calculus:
Differentiation has many applications, which can be used to find out the rate of change of quantities with respect to time or some other variables, to find the equation of the normal and tangent to a curve at a point, to find out the moving points on the graph of the function which helps to find out the points which has largest or the smallest value throughout the graph, to find out the intervals for which the provided function is increasing or decreasing and of course, to find out the approximate value of certain quantities.

Applications of Integral Calculus:
Integration is widely used to find the length of curves, or the area of the region bounded by a curve and a line, or area bounded by two or more curves.

Solved Examples:
Example 1: Find the intervals in which the function f given by f x) = 4x3– 6x2– 72x + 30 is strictly increasing and strictly decreasing?
Solution: Here, f x) = 4x3– 6x2 – 72x + 30
Therefore, f’(x) = 12x2 – 12x - 72 = 12 (x2– x – 6) = 12(x - 3) (x + 2).
Now, f′(x) = 0 gives x = – 2, 3.
The points x = – 2 and x = 3 divides the real number line into three disjoint intervals, namely, (– infinity, – 2), (– 2, 3) and (3, infinity).
In the intervals (– infinity, – 2) and (3, infinity), f′(x) is positive while in the interval (– 2, 3), f′(x) is negative.
Therefore, the function f is increasing in interval (– infinity, – 2) and (3, infinity) while the function is decreasing in the interval (– 2, 3).

Example 2: Find the slope of the tangent to the curve y = x3– x at x = 2.
Solution:  The slope of the tangent at x = 2 is given by $\frac{dy}{dx}$ at x = 2

$\frac{dy}{dx}$ = 3x2 – 1, at x = 2, which is equal to 11.

Example 3: Find the area enclosed by the curve xy2 = a2 (a - x) and y axis?
Solution: This curve is symmetrical about x axis and it intersects x axis at (a, 0) and its asymptote is x = 0, i.e., y-axis.
Now, x y2 = a2 (a - x) implies, $x =$$\frac{a^{3}}{a^{2}+y^{2}}. Thus, required area = 2\int_{0}^{\infty }xdy = 2a^{3}\int_{0}^{\infty }$$\frac{1}{a^{2}+y^{2}}$$dy = 2a^{3} \times$$\frac{1}{a}$ $\tan^{-1}$$\frac{y}{a}$$|_0^\infty$ = $2a^{2} \times$$\frac{\pi }{2}$ = $\pi a^{2}$.

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