Partial Derivatives and the Gradient. Gradient and Theorem

Partial derivatives and the gradient. Gradient and theorem

Here's the first cool property of a gradient: The gradient vector is perpendicular to the
level surface corresponding to setting the function w equal to a constant.
∇w I level surface w ={constant}
If we draw a contour plot of my function, then we will have a two-variable function
with level curves representing the gradient at each point. If we then draw the
gradient vector at each point on that contour plot, it will end up being perpendicular
to the level curve at that point.

[Graph]
If we have a function of three variables, then we can try to draw its contour plot. Of
course, we can't really do it, because the control plot would be living in space with x,
y, and z. But it would be a bunch of level surfaces. And the gradient vector would be
a vector in space. That vector is perpendicular to the level surfaces.

So let's try to
see that in a couple of examples.
SO lets do a first example: w=a x_1+ay_2+az^3
Let's consider the following linear function of x, y, and z. We will write w equal alx +
a2y + a3z. What is the gradient of this function'
∇w={a_1,a_2,a_3}
a_1=δw/δx
The first component will be a1. That's the partial W, partial x. Then a2, that's the partial
w, partial y;and a3, partial w, partial z.

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If we set w equal to some constant c. we find that the point (x, y, z) must satisfy the
equation ax + + by + cz = c.
a_1x+a_2y+a_3z=c
What kind of surface is that? It's a plane. Io find the normal vector to such a surface,
we only need to look at the coefficients.
{a_1,a_2,a_3}
And in fact, this is the only case you need to check because of linear approximation.
If you replace a function by its linear approximation, that means you will replace the
level surfaces by the tangent planes. You'll end up in this situation, but maybe that's
not very convincing.