Question

If I have a function F(x,y,z) and a plane. How can I find the point P where the plane is tangent to a level surface of F?

Answer 1

Do you have a particular function or plane?

Answer 2

For \(z=f(x,y)\) at the point \(P(x_0,y_0,z_0)\) the tangent plane is:
\[\large z-z_0 = f_x(x_0,y_0)\cdot (x-x_0)+f_y(x_0,y_0)\cdot (y-y_0)\]

Answer 3

But in general, two functions are going to be tangent when they intersect and their rates of change (partial derivatives) intersect. I hope that helps.

Answer 4

Right, I got it about the tangent plane, but not knowing where they should intersect confuses me.
This is what I have: \[F(x,y,z) = \frac{1}{4} . x^{2} + \frac{1}{9} . y^{2} + \frac{1}{2}. z^{2}\]

And the plane: \[z = 3x - 2y + z = 74\]

Answer 5

The gradient of F I calculated as: \[Grad(F)(x,y,z) = (\frac{1}{2}x,\frac{2}{9}y,z)\]

That direction should be parallel to the normal of the plane at the point P, right?

Answer 6

yes that is correct
so the normal vector of plane is <3,-2,1>
since a vector can be multiplied by a scalar, it becomes --> <3c, -2c , c>
Now set these components equal to gradient for point (x,y,z)

3c = x/2 --> x = 6c
-2c = 2y/9 --> y = -9c
c = z --> z = c

Now plug into plane equation to solve for c:
3(6c) -2(-9c) +c = 74
37c = 74
c = 2

Thus:
x = 12
y = -18
z = 2

Answer 7

Awesome! I followed a similar path at one point but couldn't get a result. It was much more simpler than I thought.

Thanks a lot! :)