 # Approximations

If your height is 5 ft and 10 inches, don’t you call yourself approximately 6 ft tall whenever asked? Similarly, if you scored 97.75% in your examinations, don’t you boast about it by chanting that you scored approximately 98%? Such is the intuitive nature of the topic of approximations that it doesn’t even need an explanation!

### Suggested Videos        However, the general notion of approximating values differs from its mathematical treatment. While 9.99 is still approximately equal to 10; here we will talk about approximating the values of quantities whose exact value is not known to us or cannot be calculated through any known means.

For example, try working out the value of the square root of 10 without using a calculator. You’ll fumble around for some time and then get stuck up on the answer that it is approximately equal to 3. You won’t be able to get more accurate than that. This is what the method of approximating certain quantities deals within mathematics.

## Approximations by Differentials

As the name suggests, this method relies on the derivatives of the functions whose values are to be determined at some points.

• Problem – Given a function y = f(x), determine its value at x = x′.
• Approach – We will use the definition of the derivative of a function y = f(x) with respect to x.

$${\frac{d}{dx} {(f(x))} = \text{change in y with respect to change in x as } {dx \rightarrow{0}}}$$

Thus if we can get to the value of x = x′ from a value of x near it, such that the difference in the two values i.e. dx is vanishingly small; we can obtain the change in the value of the function y = f(x) corresponding to the change dx in x. However, the notion of vanishingly small is not realizable in practical examples.

What we can achieve is a value of dx = Δx, which is very small as compared to the values of x on which we are working. This is what actually forms the ‘approximation’ part of our approach. Under these conditions, we can approximate the change in the value of the function by the value of the differential at that point i.e. dy.

### General Form

Notation-wise let us define y(x = x′) = y(x = x0) + Δy; implying that Δy is the change in the value of the function y when the change in x is given by Δx = x′ – x0. Then we proceed as follows:

• Find a point x0 near the point x′, at which the value of the function is known.
• Differentiate the function with respect to x.
$${\frac{dy}{dx} = \frac{d}{dx} {(f(x))}} \\ {dy = f'(x)dx}$$
• Use the approximations i.e. the value of the change in x i.e. dx = Δx = x′ – x0 and calculate the derivative at x = x0 to get dy, which is approximated as Δy:
$${\Delta{y} = f'(x_0)\Delta{x}} \\ {\Delta{y} = f'(x_0){(x’ – x_0)}}$$
• This would be the change in the value of the function y as x changes from x0 to x′. Thus, we have
$${f(x’) = f(x_0) + \Delta{y}} \\ {[f(x’) = f(x_0) + f'(x_0){(x’-x_0)}]}$$

The general form of writing the result obtained above is:
$${f(x+\Delta{x}) = f(x) + f'(x)\Delta{x}}$$
which enables one to get the value of the function at a point near x. In connection with this formula, look at the figure below: You should spot out that we have actually approximated the curve of the function by a straight line marked as AB. Using the formula for the slope of the straight line i.e. $${ m = tan\theta = tan(\angle{ABC}) = \frac{dy}{dx}}$$ ; we have arrived at the relation $${ \Delta y = dy}$$, which is approximately true, considering the small values of Δx that we are going to choose.

Thus, this form of approximations is also known as Linear Approximations! Confused on the implementation of this technique in mathematical problems? Just look at the solved example Question 1!

## Error Calculations

One can extend the notion of approximations of functions to approximating the errors in calculating certain functions as well, provided their exact dependence on the independent parameters is known. For example, let us have a function:

$${y = f(a,b,c…)}$$

Then the error in the calculation of y at the points a = a’, b = b’….., owing to the errors in the values of a, b, c…

$${\Delta{y} \approx{ dy} = [\frac{\partial{y}}{\partial{a}}]_{a = a’, b = b’, c = c’…..} \Delta{a} + [\frac{\partial{y}}{\partial{b}}]_{a = a’, b = b’, c = c’…..} \Delta{b} + ……}$$ where, $$\frac{\partial{y}}{\partial{a}}]_{a = a’}$$ is the derivative of y with respect to a (at a = a’), keeping other variables like b, c…. constant.

This formula follows directly as a consequence of our method of approximations explained above, and the logical assumption that the total error in the calculation of y is the sum of the different errors obtained in y due to the errors in the measurement of independent variables a, b, c….

## Solved Examples for You

Question 1: Approximate the value of √51 up to three decimal places.

Answer : Just by pure guesswork, you can work out the value of √51 to be approximately equal to 7; but as you can see: such approximations are not accurate enough for this problem. So let us resort to the method of differentials. Define a function

$${ f(x) = \sqrt{x}}$$

Can you see how this function would be helpful to us for solving this problem? What we seek is the value of this function at x = 51. We know the value of this function at x = 49. [The square root of 49 is 7]. We can use Δx = (51-49) = 2, and the derivative of f(x) at x = 49 in the formula of Δy and get our answer! Let us put the method into action!

$${ f(x) = \sqrt{x}} \\ { f'(x) = \frac{1}{2\sqrt{x}}} \\ \text{Now comes the approximations part:}\\ {\Delta{f} = f'(x = 49) \Delta{x}} \\ {\Delta{f} = \frac{1}{2\sqrt{49}} \times{(51 – 49)}} \\ {\Delta{f} = \frac{1}{14} \times{2}} \\ {\Delta{f} = \frac{1}{7} = 0.143 \text{(up to 3 decimal places)}}$$

Then the value of the function at x = 51 can be calculated as:
$${f(x+\Delta{x}) = f(x) + \Delta{f}} \\ { \sqrt{51} = 7 + 0.143} \\ { \sqrt{51} = 7.143}$$

The result obtained thus is very close to the actual value, which is 7.141. You must note that we will get a more precise value if we calculate the value of √50 through the same method. Why is that? Actually, in that case the value of Δx = 51 – 50 = 1, would be less that the Δx here. Hence, the resulting Δf would be closer to the actual change in the value of the function.

### Question 2

What is the error in calculating the area of an ellipse of major axis = 10 cm and the minor axis = 2 cm; if the error in measuring the major axis is 0.05 cm and the minor axis is 0.01 cm?

Answer : The area of the ellipse with a major axis a and a minor axis b is given as: $${ A = \Pi ab}$$

Here A is just a function of a and b. So let us calculate the error in the area using the above-mentioned technique:

$${\Delta{A} \approx{ dA} = [\frac{\partial{A}}{\partial{a}}]_{a = a’, b = b’} \Delta{a} + [\frac{\partial{y}}{\partial{b}}]_{a = a’, b = b’} \Delta{b}} \\ {\Delta{A} = (\Pi b)_{b = 2}\Delta{a} + (\Pi a)_{a = 10}\Delta{b}} \\ {\Delta{A} = (\Pi \times{2}\times{0.05} + \Pi\times{10}\times{0.01})} \\ {\Delta{A} = \Pi \times{(0.1 + 0.1)}} \\ {\Delta{A} = 0.628}$$

Question 3: Why do we use the approximation in Mathematics?

Answer: In Mathematics, approximation refers to using an easier process or model when the right model is tough to use. An approximate model can also be used to make the calculations easier. The approximations can also be used if the inadequate information prevents the use of exact demonstrations.

Question 4: How can we do successive approximation?

Answer: The technique for conducting the successive approximations is given below:

1. Assume an approximate value for the variable that will help in simplifying the equation.
2. Find a solution for that variable.
3. Use that answer as the 2nd approximate value and solve that equation one more time.
4. Finally, we will repeat this procedure until we obtain a constant value for the variable.

Question 5: Is there any difference between the approximation provided by a differential and the approximation provided by a linearization?

Answer: No, they both use a similar tangent line at the same time. Its 2 different ways of observing the same approximation.

Question 6: What is the linearization formula?

Answer: Any differentiable function ‘f’ can be approximated through its tangent line at the point a: ‘L(x) = f(a) + f (a)(x − a)2’. If ‘y = f(x)’, then the differentials will be defined through ‘dy = f (x)dx’.

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