{"id":5581,"date":"2021-09-21T18:34:20","date_gmt":"2021-09-21T13:04:20","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5581"},"modified":"2021-11-15T00:14:07","modified_gmt":"2021-11-14T18:44:07","slug":"differentials-errors-and-approximations","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentials-errors-and-approximations\/","title":{"rendered":"Differentials Errors and Approximations"},"content":{"rendered":"
Here you will learn what is differentials errors and approximations with examples.<\/p>\n
Let’s begin –<\/p>\n
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,<\/p>\n
\nIn general dy = f'(x)dx or df(x) = f'(x)dx<\/p>\n
Or \\(\\Delta y\\) = \\(dy\\over dx\\) \\(\\Delta x\\)\u00a0<\/p>\n
Because [ f'(x) = \\(dy\\over dx\\) ]<\/p>\n<\/blockquote>\n
Absolute Error\u00a0<\/strong><\/h4>\n
\nThe error \\(\\Delta x\\) in x is called the absolute error in x.<\/p>\n<\/blockquote>\n
Relative Error\u00a0<\/strong><\/h4>\n
\nIf \\(\\Delta x\\) is an error in x, then \\(\\Delta x\\over x\\) is called the relative error in x<\/p>\n<\/blockquote>\n
Percentage Error\u00a0<\/strong><\/h4>\n
\nIf \\(\\Delta x\\) is an error in x, then \\({\\Delta x\\over x} \\times 100\\) is called the percentage error in x.<\/p>\n<\/blockquote>\n
Remark<\/strong> : 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,<\/p>\n
y + \\(\\Delta y\\) = \\(f(x + \\Delta x)\\)<\/p>\n
But, \\(\\Delta y\\) = \\(dy\\over dx\\) \\(\\Delta x\\) = f'(x) \\(\\Delta x\\) , approximately<\/p>\n
\\(\\therefore\\) \\(f(x + \\Delta x)\\) = y + \\(\\Delta y\\)<\/p>\n
\\(\\implies\\) \\(f(x + \\Delta x)\\) = y + f'(x) \\(\\Delta x\\) , approximately<\/p>\n
\\(\\implies\\) \\(f(x + \\Delta x)\\) = y + \\(dy\\over dx\\) \\(\\Delta x\\) , approximately<\/p>\n
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.<\/p>\n
Algorithm :<\/strong><\/p>\n
\n1). Choose the initial value of the independent variable as x and the changed value as x + \\(\\Delta x\\).<\/p>\n
2). find \\(\\Delta x\\) and assume that dx = \\(\\Delta x\\).<\/p>\n
3). find \\(dy\\over dx\\) from the given relation y = f(x).<\/p>\n
4). find the value of \\(dy\\over dx\\) at (x, y).<\/p>\n
5). find dy by using the relation dy = \\(dy\\over dx\\)dx.<\/p>\n
6). Put \\(\\Delta y\\) = dy to obtain an approximate change in y.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : If y = \\(x^4\\) – 10\u00a0 and x changes from 2 to 1.99, what is the approximate change in y ? Also, find the changed value of y.<\/p>\n
Solution<\/strong><\/span> : Let x\u00a0 = 2, x + \\(\\Delta x\\) = 1.99. Then, \\(\\Delta x\\) = 1.99 – 2 = -0.01<\/p>\n
Let dx = \\(\\Delta x\\)\u00a0 = -0.01<\/p>\n
We have,\u00a0<\/p>\n
y = \\(x^4\\) – 10<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = \\(4x^3\\) \\(\\implies\\) \\(({dy\\over dx})_{x=2}\\) = \\(4(2)^3\\) = 32<\/p>\n
\\(\\therefore\\) dy = \\(dy\\over dx\\) dx<\/p>\n
\\(\\implies\\) dy = 32(-0.01) = -0.32<\/p>\n
\\(\\implies\\) \\(\\Delta y\\) = -0.32 approximately<\/p>\n
So, approximate change in y = -0.32<\/p>\n
When, x = 2, we have<\/p>\n
y = \\(2^4\\) – 10 = 6<\/p>\n
So, changed value of y = y + \\(\\Delta y\\) = 6 + (-0.32) = 5.68<\/p>\n
\nRelated Questions<\/h3>\n
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.<\/a><\/p>\n
Find the approximate value of f(3.02), where f(x) = \\(3x^2 + 5x + 3\\).<\/a><\/p>\n\n\n