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Can someone please help me understand how to start this problem? I have posted this up before but have not received any help. I can obviously see that the gradient is 4, that the line passes through (0,0) and possibly (-2,2), when z=-3 and z=9, respectively, but beyond this, nothing comes to me naturally. I have no idea how I am supposed to come up with a function given this information.
From this picture, you can visually estimate the partial derivatives with respect to y and to x. With the partials, you can find the equation of a plane that satisfies the initial condition g(0,0) = -3
Use the following equation for a plane:
$g-g_0=\frac{dg}{dx}(x-x_0)+\frac{dg}{dy}(y-y_0)$
When $y=x$, you have $g = -3$. So the form of the linear function is $g(x, y) = a(y-x)-3$. Now you also have noted $g(-2, 2)= 9$. Solve for $a$.
We see that as $g$ changes from $-3$ to $-7$, $x$ changes from $0$ to roughly $1.25$. So we have $\frac{\partial g}{\partial x}= \frac{-7-(-3)}{1.25}=-\frac{4}{1.25}=3.2$, and equivalently for the change in $y$: $\frac{\partial g}{\partial y}= \frac{1-(-3)}{1.25}=\frac4{1.25}=3.2$
Using the equation of a plane with point $(0,0)$: $$g-(-3)=-3.2(x-0)+3.2(y-0) \longrightarrow g=-3.2x+3.2y-3$$
When $(x,y)=(0,0)$ this gives $-3$, but $(x,y)=(-2,2)$ does not give 9, so refining gets us $$g=-3x+3y-3$$
which is the correct answer and so: $\partial x = \partial y = \frac43$.
