tg

tg
tgt

Sunday, 3 August 2014

CHAPTER 19 - Errors and Approximations

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable. The theory behind it is quite simple: From the chapter on differentiation, we know that
\mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\Delta y}}{{\Delta x}} = \dfrac{{dy}}{{dx}} = f'\left( x \right)
For small \Delta x, we can therefore approximate \Delta y\,{\rm{as}}\,f'\left( x \right)\Delta x. This is all there is to it!
Suppose we have to calculate {(4.016)^2}.
We let
y = {x^2}.{x_0} = 4\,\,\,{\rm{and}}\,\,{y_0} = 16
y' = 2x,\,\,\Delta x = 0.016
 \Rightarrow  \,\,\,\, \Delta y = f'\left( x \right) \cdot \Delta x
 = {\left. {2x} \right|_{{x_0} = 4}} \times 0.016
8 \times 0.016
=0.128
 \Rightarrow  \,\,\,\, y = {y_0} + \Delta y = 16.128
     Example: 1      

Find the value of {\left( {8.01} \right)^{4/3}} + {\left( {8.01} \right)^2}
Solution: 1

Let y = f\left( x \right) = {x^{4/3}} + {x^2}
Let {x_0} = 8\,\,\,{\rm{so}}\,\,{\rm{that}}\,\,{y_0} = 16 + 64 = 80
\Delta x = 0.01
 \Rightarrow  \,\,\,\, \Delta y = {\left. {f'\left( x \right)} \right|_{x = {x_0}}} \times \Delta x
{\left. { = \left( {\dfrac{4}{3}{x^{1/3}} + 2x} \right)} \right|_{{x_0} = 8}} \times \Delta x
 = \left( {\dfrac{8}{3} + 16} \right) \times 0.01
 = \dfrac{{0.56}}{3}
 = 0.1867
 \Rightarrow  \,\,\,\, {y_0} = {y_0} + \Delta y
 = 80.1867
Post a Comment