Riya Hansen

2022-07-12

Linear approximation of quotient
$\frac{\left(2.01{\right)}^{2}}{\sqrt{.95}}$

Kaya Kemp

set $f\left(x\right)=\frac{\left(2+x{\right)}^{2}}{\sqrt{1-5x}}$ expand it then set $x=0.01$.
using linear approximation $f\left(x\right)=4+14x$ $f\left(0.01\right)=4.14$.

Savanah Boone

You may also do it using the linear approximation of $f\left(x,y\right)=\frac{{x}^{2}}{\sqrt{y}}$ at $\left({x}_{0},{y}_{0}\right)=\left(2,1\right)$ with $\mathrm{\Delta }x=0.01$ and $\mathrm{\Delta }y=0.05$:
$f\left(x,y\right)\stackrel{linear}{\approx }f\left({x}_{0},{y}_{0}\right)+\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\left({x}_{0},{y}_{0}\right)\mathrm{\Delta }x+\frac{\mathrm{\partial }f}{\mathrm{\partial }y}\left({x}_{0},{y}_{0}\right)\mathrm{\Delta }y$
With
$\frac{\mathrm{\partial }f}{\mathrm{\partial }x}=\frac{2x}{\sqrt{y}}$
$\frac{\mathrm{\partial }f}{\mathrm{\partial }y}=-\frac{{x}^{2}}{2\sqrt{{y}^{3}}}$
you get
$\frac{\left(2.01{\right)}^{2}}{\sqrt{.95}}\approx 4+4\cdot 0.01-2\cdot 0.05=3.94$

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