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Given a function $z=f(x,y)$ the tangent plane at $(a,b,f(a,b))$ has equation $$0=z-f(a,b)-f_x(a,b)(x-a)-f_y(a,b)(y-b)$$It's clear, just by subbing in $(a,b)$ into this equation that the limit of $z=f(x,y)$ as $(x,y)\to(a,b)$ is equal to $f(a,b)$. That is, locally this plane looks like the function. My question is, why is this the best approximation. For example, $$0=z-f(a,b)-P(x-a)-Q(y-b)$$where $P,Q\not=0$ is also a plane containing $(a,b,f(a,b))$ which gets closer to $f(x,y)$ as $(x,y)$ approaches $(a,b)$. I can see via picture why the tangent plane is the best, but how would I show it mathematically. I guess by best I mean the following: For any $\varepsilon>0$ then exists $\delta>0$ such that

$\left|(x,y)-(a,b)\right|<\delta$.

implies that

$\left|f(x,y)-f(a,b)-f_x(a,b)-f_y(a,b)\right|<\varepsilon$.

Is it true that given the $\varepsilon>0$ the $\delta$ used in the tangent plane approximation is bigger than the one used for any other plane? Does this mean "better" approximation? How would I prove this?

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1 Answer 1

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You can explain it with asymptoic analysis: the equation of the tangent plane is the only linear approximation $L(x-a,y-b)$ such that the error in approximating: $\,f(x,y)-L(x-a,y-b)\,$ is $\,o\bigl(\lVert(x-a,y-b)\rVert\bigr)$.

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