# Intersection of a function f(x,y) with a plane

• Poetria
In summary, the solution to the equation 2(x-a)+(y-b)+0z=0 in 3D space is a plane that contains the line 2(x-a)+(y-b)=0 at the xy-plane but also contains all the parallel lines of the points that have positive or negative z.f

#### Poetria

Homework Statement
Which of the following expressions best approximates the slope of the function created by intersecting the graph of the function z=f(x,y) with a plane 2(x-a)+(y-b)=0 at the point (a,b)
Relevant Equations
2(x-a)+(y-b)=0 is a line
y=2a+b-2x
at the point (a,b) b=2a+b-2a
Since z=0, the only variable that counts is x.
So the solution would be:

$$\frac {f \left(a + \Delta\ x, b \right) - f(a,b)} {\left( \Delta\ x\right)}$$

2(x-a)+(y-b)=0 is a line in 2D space but in 3D space it becomes a plane. You can write it as $$2(x-a)+(y-b)+0z=0$$ to see it mathematically as the general equation of a plane in 3Dimensional space is ##Ax+By+Cz+D=0##.

To see it intuitively this plane contains the line 2(x-a)+(y-b)=0 at the xy-plane (where z=0) but it also contains all the parallel lines of the points that have positive or negative z (all the lines that are above and below the basic line 2(x-a)+(y-b)=0).

The function that is "created " by the above intersection is the function $$g(x)=f(x,2a+b-2x)$$, or $$h(y)=f(\frac{b+2a-y}{2},y)$$ so you should look for options that give the slope at ##g(a)## or ##h(b)##.

Poetria
Sorry I had a typo in post #2 in the expression of ##h(y)## I think now it is fixed.

Poetria
Actually I might be wrong in what I say in post #2 about which are the functions that are created by the intersection. Is this question from the theory of directional derivatives?

I haven't learned anything about directional derivatives yet. It is the question about geometry of partial derivatives. But I think you are right. It does make sense. The option with g(x)

$$\frac {f \left(a + x - a, b - 2 x + 2 a \right) - f \left( a, b\right)} {x - a}$$

$$\frac {f \left(a + \Delta\ x, b - 2 \Delta x \right) - f \left( a, b\right)} {\Delta x}$$

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Delta2
I will try to do the same with h(y).

Delta2
$$\frac {f\left(\frac {2 a + b - \left( y - b\right)} {2}, b + y - b \right) - f\left(a, b \right)} {\left(y - b\right)}$$

Of course, y-b is ##\Delta y##

This is a multi-choice exercise and there is no option with ##\Delta y## as the one above. I was wondering why the option with ##\Delta x## in the denominator was supposed to be correct and not e.g. ##\Delta x \Delta y##. Now I understand it. :)
By the way ##\Delta x \Delta y## would mean the second derivative if I understand it correctly.
Tricky stuff.

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Delta2
$$\frac {\frac {2 a + b - \left( y - b\right)} {2} - f\left(a, b \right)} {\left(y - b\right)}$$

Of course, y-b is ##\Delta y##

This is a multi-choice exercise and there is no option with ##\Delta y## as the one above. I was wondering why the option with ##\Delta x## in the denominator was supposed to be correct and not e.g. ##\Delta x \Delta y##. Now I understand it. :)
By the way ##\Delta x \Delta y## would mean the second derivative if I understand it correctly.
Tricky stuff.
not sure what you saying at the top, your expression is incomplete/has some typos i think, but yes if you were given an expression with ##\Delta x \Delta y## in the denominator that would come from the approximation expression for the second mixed partial derivative ##\frac{\partial^2 f}{\partial x\partial y}## or ##\frac{\partial ^2 f}{\partial y\partial x}## (those 2 are equal if f is continuous btw).

Poetria
I meant the slope at h(b) you have suggested to find. Have I made a mistake?

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ehm I think you mean $$f(\frac{2a+b-\Delta y}{2},b+\Delta y)$$ there instead of just $$\frac{2a+b-\Delta y}{2}$$

Poetria
Silly me. :(

Delta2