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Problem (translating from Spanish as exactly as I can, so please bear with me): Knowing that $f(x,y)$ is differentiable in $(5; -3)$ which is inside the function domain, and that $\frac{\partial f}{\partial x}(5;-3)=2$ and $\frac{\partial f}{\partial \vec u}(5;-3)=4$ for unit vector $\vec u=<\frac{\sqrt 2}{2}; \frac{\sqrt 2}{2}>$. Use the differential to approximate $f(4.9;-2.8)$.

I'm completely lost.

I know that the total differential ($df$) can be expressed as $f_x(a,b).(x-a)+f_y(a,b).(y-b)$ and that $\Delta z$ is $f(a+(x-a), b+(y-b))-f(a,b)$. However, I don't know how to use them in this problem. I'm not even sure what I'm being asked to do.

I know the geometrical interpretation of a linear approximation, where you use a tangent plane to approximate the value of a surface in 3D and this plane will have a similar value to the surface while you're near the tangency point while it will differ more and more as you move far away from the point (error). But I don't know how to apply that to solve this problem.

Floella
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1 Answers1

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Instead of breaking down the problem coordinate by coordinate, try looking at it in terms of entire vectors. Let’s review a bit. The best linear approximation to a differentiable function $f:\mathbb R^2\to\mathbb R$ near $\mathbf v$ is via its differential: $$f(\mathbf v+\Delta\mathbf v)=f(\mathbf v)+\operatorname{d}f_{\mathbf v}[\Delta\mathbf v]$$ plus a small error. In terms of the tangent plane at $\mathbf v$ this says, as you know, that to find the approximate change in $f$ for some distance and direction given by $\Delta\mathbf v$, we can move that distance and direction along the tangent plane at $\mathbf v$. The inclination of this plane is encoded in the differential $\operatorname{d}f_{\mathbf v}$, which is a linear map that takes a vector and returns a scalar.

Since $\operatorname{d}f_{\mathbf v}$ is linear, $$\operatorname{d}f_{\mathbf v}[a\mathbf w_1+b\mathbf w_2]=a\operatorname{d}f_{\mathbf v}[\mathbf w_1]+b\operatorname{d}f_{\mathbf v}[\mathbf w_2].\tag{1}$$ If $\mathbf u$ is a unit vector, then $\operatorname{d}f_{\mathbf v}[\mathbf u]$ is just the directional derivative in the direction of $\mathbf u$, which measures the slope of the tangent plane in that direction. Relative to the standard basis, then, we have the familiar $$\operatorname{d}f_{\mathbf v}[\Delta\mathbf v]=\Delta x\operatorname{d}f_{\mathbf v}[\mathbf e_x]+\Delta y\operatorname{d}f_{\mathbf v}[\mathbf e_y] = f_x(\mathbf v)\Delta x+f_y(\mathbf v)\Delta y,\tag{2}$$ since the partial derivatives $f_x$ and $f_y$ are just the directional derivatives in the directions of the coordinate axes. We can use a different pair of unit vectors $\mathbf u_1$ and $\mathbf u_2$, though. Referring to (1) we see that in that case $\Delta x$ and $\Delta y$ in (2) are replaced by the components of $\Delta\mathbf v$ in those directions and the partial derivatives are replaced by the respective directional derivatives.

That’s exactly what you’ve been given in this problem: a pair of linearly independent unit vectors and the directional derivatives in those directions. All that remains for you to do is to express the displacement $\Delta\mathbf v$ in terms of these two vectors and then put it all together.

amd
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  • Thanks for the lengthy explanation :D . Let me see if I know how to approach this: I should compute $df$ using $\vec u$, so I have: $df(5;-3)=f_x(5;-3).(x-\frac{\sqrt 2}{2}) + f_{\vec u}(5;-3).(y-\frac{\sqrt 2}{2})$ which, using the data I was given, would be $2x+\sqrt 2 y-3$. Now I'm tempted to evaluate this at $(4.9, -2.8)$, which would be roughly $0.57$, but since that is way far from $f(5;-3)=4$ I know I did something wrong... – Floella Dec 09 '16 at 00:13
  • @Anarelle Not quite. Among other things, it look like you might be confusing the point at which $\operatorname{d}f$ is computed with the argument to this map. You need to find $a$ and $b$ such that $(4.9; -2.8)-(5;-3)=a(1,0)+b\vec u$, then $\operatorname{d}f_{(5;-3)}[(4.9;-2.8)-(5;-3)]=af_x(5,-3)+bf_{\vec u}(5,-3)$. – amd Dec 09 '16 at 04:12
  • @Anarelle Alternatively, you can recover $f_y$ from the given directional derivatives. Since $\vec u=\frac1{\sqrt2}(\vec e_x+\vec e_y)$, equation (1) tells us that $f_{\vec u}=\frac1{\sqrt2}(f_x+f_y)$. – amd Dec 09 '16 at 04:56