I need help finding an equation for a tangent plane to the following graph at the point $(1,2,5)$: $$z=f(x,y)=x^2+2xy$$ For this question, I got $z=6x+2y-5$ as the tangent plane. Can someone verify this?
Also, I need help with linear approximation. I want to approximate $f(1.1,1.9)$. So for this I think that I should start with $f(1,2)$. However what do I do now?
$\endgroup$1 Answer
$\begingroup$The standard way is that $f(1,1.9) \approx. f(1,2)+ f_x(1, 2)(x-x_0)+ f_y (1,2)(y-y_0)=5+ 6(1-1.1)+ 2(2-1.9) = 5-0.6+ 0.2= 4.6$. i.e., you consider the change of the function along the tangent plane. The actual value at $(1.1, 1.9)$ is $5.39$, so the approximation is not that good.
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