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
Related Questions
Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\).
