Please help me understand the solution to this three dimensional graph

Question

I have a function $$f(x,y)=-x^2 -y^2 +4$$

Text:

The figure shows a plane that is parallell to the xz-plane, and goes through the point $(1,1(f,1,1))$

The straight line that is drawn in this plane touches the graph $f$ in the point $(1,1,f(1,1))$

Question: What is the slope of the straight line?

Solution:

Can someone please help me understand?

Answer 1

The line is tangential to the curve in the $xz$-plane that is the intersection between the surface $z=-x^2-y^2+4$ and the vertical plane $y=1$, which can obtained by eliminating the variable $y$, i.e.

$$z=-x^2+3$$

Its slope is then given by $z’=-2x$, which is -2 at $x=1$.

Answer 2

The idea is to reduce the problem of multivariate derivatives to singlevariable derivatives. We use a plane to cut through the 3d function. The intersection of the plane and the function is a singlevariable function (see sketch of the parabols).

Now we need to calculate the derivative of this parabola, that is -2x, and put in the x and y values and you get your result. By reducing the problem to singlevariable functions we can interpret y as a constant and the derivative of a constant is 0.

The interpretation of your result is like this: At the point x,y = 1 the slope in the x direction is -2.