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