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This is my question,

A function of 2 variable is given by,

$f(x,y) = e^{2x-3y}$

How to find tangent approximation to $f(0.244, 1.273)$ near $(0,0)?$

I need some guidance for this question. Am i suppose to do the linear approximation or quadratic approximation?

Need some explanation for the formula. Thanks

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I've never heard of the word tangent used outside of lines, planes, or any "flat" surfaces – Hawk Sep 25 '12 at 6:27
up vote 2 down vote accepted

Linear(Tangent) approximation of $f(x,y)$ about $(a,b)$ is given by,

$$f(x,y)\approx f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$$

In your problem $f(x,y)=e^{2x-3y}$

$f_x(x,y)=2e^{2x-3y}$ and $f_y(x,y)=-3e^{2x-3y}$

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So, (a,b) refers to near point which is (0,0) while fx(x,y) and fy(x,y) is where f(0.244,1.273) will be applied? – David Sep 25 '12 at 6:37
$(a,b)$ is $(0,0)$ while $(x,y)$ is $(0.244,1.273)$, so $f_x,f_y$ are also to be computed at $(0,0)$ not $(0.244,1.273)$ – Aang Sep 25 '12 at 6:40
1 + 0.244(2) + 1.273(-3) = -2.331 – David Sep 25 '12 at 6:51
@David: exactly :) – Aang Sep 25 '12 at 6:51

More precise approximation we obtain if represent $f(x,\,y)$ as $$f(x,\,y)=e^{2x-3y}=e^3e^{2x-3y-3}=e^3e^{2x-3(y-1)}.$$ Then apply formula for tangent approximation to function $g(x,\,y)=e^{2x-3(y-1)}$ with $a=0; \,b=1.$

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