Differential Calculus and Approximations

Differential calculus and approximation is a sub-branch of calculus which is a part of mathematics. Integration, differentiation, limits, and functions are dealt with in calculus. Calculus has various science and technology applications and in economics, too, where algebra alone is insufficient to apply. The basic understanding of differential calculus approximations is to make smaller segments of something to study the rate of changes. Differentiation is nothing but finding the derivative of the function. A differential equation is an equation that has the derivative of a dependent variable concerning an independent variable.

The differential equation is given by f(x)=dy/dx

Where ‘y’ is the dependent variable and ‘x’ is the independent variable

Usually, the following are used to find out the derivatives.

• With respect to other variables, the rate of change of quantity can be determined.

• For certain quantities, approximate values can be determined.

• For the given function, intervals in which increase or decrease of function can be studied.

• Derivatives are used to find out the equation of tangent.

Consider the example, r=10cm find the rate of change of circle per second for the given radius.

It is known that Area of circle A is given by A = πr² 

So, the rate of change of area with respect to radius will be

dA/dr=(d/dr) πr²

dA/dr= 2πr

it is given r=10cm then

dA/dr=20 π

so, at 20 π cm2/s the area of the circle is changing.

Approximation of Differential Equation

Certain values are approximated using differentials. Consider a function f(x) is defined as f:D tends to R where D⊂R . Say y=f(x). An increment in x can be noted as ∆x. When the x is increased by ∆x, corresponding y has to increase by ∆y=f(x+∆x)-f(x). it is depicted in the below picture:

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The following conclusions are derived from the above-mentioned points.

A differential of the independent variable always equals the increment of the variable. On the other side, the differential of the dependent variable does not equal the increment of the variable.

In some cases, like when dx=∆x is too negligible a value to consider compared to x, ∆y is the best approximation of dy and dy≈∆y.

So now dx=∆x

dy=f ‘(x)dx=(dy/dx) ∆x

For better understanding let’s consider an example

Approximation Example

1. Using the differential approximate √36.5

Ans.   Say y=√x where x=36 and ∆x=0.5

∆y = \[\sqrt{x}\] + ∆x - \[\sqrt{x}\]

∆y = \[\sqrt{36}\] + 0.5 - \[\sqrt{36}\]

∆y =  \[\sqrt{36.5}\] - 6

∆y + 6 = \[\sqrt{36.5}\]

As dy≈∆y 

Now, dy=(dy/dx) ∆x

dy=½ \[\sqrt{x}\](0,5)=0.05

\[\sqrt{36.5}\] = 6 + 0.05 = 6.05 

Riemann Sum Example

Consider approximating the area under the graph of f(x) = \[\sqrt{x}\] between x=0.5 and y=3.5

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& when we consider doing it with the expression of Riemann right sum with 4 equal subdivisions then, 

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Assume A(i) represent the area of ith rectangle in this approximation

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A(1)+A(2)+A(3)+A(4)=i=1∑4​A(i)

Now we need to find the expression for A(i)

The width of the interval [0.5,3.5] is 3 units but we need 4 so ¾=0.75

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So now f(xi) = \[\sqrt{xi}\] = 0.5 + 0.75i​
A(i)​=width⋅height

= \[\sum_{i=1}^{4}\] A(i)

= \[\sum_{i=1}^{4}\] 0.75 ∗ \[\sqrt{0.5}\] + 0.75

Problems for Practice 

1. Determine the approximate for the \[\sqrt{25.5}\] using the differential.

2. Using differentials find the approximate value for (26)1/3.