Linear approximation problem I don't know how to approach

[patr1c1a]

This exercise is driving me crazy:

Knowing that \(\displaystyle f(x,y)\) is differentiable in (5; -3) which is inside the function domain, and that \(\displaystyle \frac{\partial f}{\partial x}(5;-3)=2\) and \(\displaystyle \frac{\partial f}{\partial \vec u}(5;-3)=4\) for unit vector \(\displaystyle \vec u=<\frac{\sqrt 2}{2}; \frac{\sqrt 2}{2}>\). Use the differential to approximate \(\displaystyle f(4.9;-2.8)\).


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, and it will differ more and more as you move far away from the point (error).

I also know that the total differential (df) can be computed as:
\(\displaystyle \displaystyle d_f (x,y)=f_x(a,b).(x-a)+f_y(a,b).(y-b) \)

And that the increment (\(\displaystyle \Delta f\)) can be computed as:
\(\displaystyle \displaystyle \Delta f = f(a+(x-a), b+(y-b))-f(a,b)\)

But I don't know how to apply all of this to solve this problem.

I'm just learning about linear and quadratic approximations and I'm quite a math dufus, so please keep it simple (just so I don't go even crazier)

 

[HallsofIvy]

You can't do that without knowing what f(5, -3) is. Aren't you given that information? Also the unit vector is not \(\displaystyle \langle\sqrt{2}/2, \sqrt{2}/2\rangle\). That is the unit vector at 45 degrees to the horizontal and the vector from (5, -3) to (4.9, -2.8) is \(\displaystyle \left<4.9- 5, -2.8- (-3)\right>= \left<-0.1, 0.2\right>\).The function value of f(4.9, -2.8) is f(5, -3) plus the vector having the partial derivatives as components dot product that vector. (Not the unit vector).