Question

How do you estimate the quantity using the Linear Approximation and find the error using a calculator `(15.8)^(1/4)`?

Answer 1

The linear approximation (about `x=16`) is:

`(15.8)^(1/4) ~~ 1.975`

This is within `1%` of the actual value.

Explanation:

Linear Approximation
The linear approximation formula for `f(x)` about `x=a` is;

`f(x) ~~ f(a) + f'(a)(x-a)`

Let `f(x) = x^(1/4) => f'(x)=1/4x^(-1/4)`

And so the Linear approximation for `f(x)` is:

`f(x) ~~ f(a) + (x-a)/(4a^(1/4))`

If we choose `a=16` (close to `15.8`); then

`f(x) ~~ 16^(1/4) + (x-16)/(4*16^(1/4))`
`\ \ \ \ \ \ \ = 2 + (x-16)/(4*2)`
`\ \ \ \ \ \ \ = 2 + (x-16)/8`

So a linear approximation of `15.8` is given by;

`f(15.8) ~~ 2 + (15.8-16)/8`
`\ \ \ \ \ \ \ \ \ \ \ = 2 - 0.2/8`
`\ \ \ \ \ \ \ \ \ \ \ = 2 - 0.025`
`\ \ \ \ \ \ \ \ \ \ \ = 1.975`

Calculator Result
Using a calculator we find:

`f(15.8) = 1.99372048 ...`

Error Analysis
The %age error is calculated using:

`%"age error" = |("estimate-actual")/"actual "* 100|`
`" " = | (1.975-1.99372048 ...)/(1.99372048 ... ) * 100|`
`" " = 0.9389725 ...`

So our linear approximation estimate was within `1%` of the actual value.