Differentials Errors and Approximations

Here you will learn what is differentials errors and approximations with examples.

Let’s begin –

Differentials Errors and Approximations

In order to calculate the approximate value of a function, differentials may be used where the differential of a function is equal to its derivative multiplied by the differential of the independent variable,

In general dy = f'(x)dx or df(x) = f'(x)dx

Or \(\Delta y\) = \(dy\over dx\) \(\Delta x\)

Because [ f'(x) = \(dy\over dx\) ]

Absolute Error

The error \(\Delta x\) in x is called the absolute error in x.

Relative Error

If \(\Delta x\) is an error in x, then \(\Delta x\over x\) is called the relative error in x

Percentage Error

If \(\Delta x\) is an error in x, then \({\Delta x\over x} \times 100\) is called the percentage error in x.

Remark : Let y = f(x) be a function of x, and let \(\Delta x\) be a small change in x. Let the corresponding change in y be \(\Delta y\). Then,

y + \(\Delta y\) = \(f(x + \Delta x)\)

But, \(\Delta y\) = \(dy\over dx\) \(\Delta x\) = f'(x) \(\Delta x\) , approximately

\(\therefore\) \(f(x + \Delta x)\) = y + \(\Delta y\)

\(\implies\) \(f(x + \Delta x)\) = y + f'(x) \(\Delta x\) , approximately

\(\implies\) \(f(x + \Delta x)\) = y + \(dy\over dx\) \(\Delta x\) , approximately

Let x be the independent variable and y be the dependent variable connected by the relation y = f(x). We use the following algorithm to find an approximate change \(\Delta y\) in y due to small change \(\Delta x\) in x.

Algorithm :

1). Choose the initial value of the independent variable as x and the changed value as x + \(\Delta x\).

2). find \(\Delta x\) and assume that dx = \(\Delta x\).

3). find \(dy\over dx\) from the given relation y = f(x).

4). find the value of \(dy\over dx\) at (x, y).

5). find dy by using the relation dy = \(dy\over dx\)dx.

6). Put \(\Delta y\) = dy to obtain an approximate change in y.

Example : If y = \(x^4\) – 10 and x changes from 2 to 1.99, what is the approximate change in y ? Also, find the changed value of y.

Solution : Let x = 2, x + \(\Delta x\) = 1.99. Then, \(\Delta x\) = 1.99 – 2 = -0.01

Let dx = \(\Delta x\) = -0.01

We have,

y = \(x^4\) – 10

\(\implies\) \(dy\over dx\) = \(4x^3\) \(\implies\) \(({dy\over dx})_{x=2}\) = \(4(2)^3\) = 32

\(\therefore\) dy = \(dy\over dx\) dx

\(\implies\) dy = 32(-0.01) = -0.32

\(\implies\) \(\Delta y\) = -0.32 approximately

So, approximate change in y = -0.32

When, x = 2, we have

y = \(2^4\) – 10 = 6

So, changed value of y = y + \(\Delta y\) = 6 + (-0.32) = 5.68

Differentials Errors and Approximations