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 thelevel surface corresponding to setting the function w equal to a constant. ∇𝑤 ⊥ 𝑙𝑒𝑣𝑒𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑤 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 { } If we draw a contour plot of my function, then we will have a two-variable functionwith level curves representing the gradient at each point. If we then draw thegradient vector at each point on that contour plot, it will end up being perpendicularto the level curve at that point. If we have a function of three variables, then we can try to draw its contour plot. Ofcourse, 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 bea vector in space. That vector is perpendicular to the level surfaces. So let's try tosee that in a couple of examples. So let's do a first example: 𝑤 = 𝑎 1 𝑥 + 𝑎 2 𝑦 + 𝑎 3 𝑧 Let's consider the following linear function of x, y, and z. We will write w equal a1x +a2y + a3z. What is the gradient of this function? . ∇𝑤 = 〈𝑎 1 , 𝑎 2 , 𝑎 3 〉 𝑎 1 = ∂𝑤 ∂𝑥 The first component will be a1.That's the partial w, partial x.Then a2, that's the partialw, partial y;and a3, partial w, partial z. If we set w equal to some constant c, we find that the point (x, y, z) must satisfy theequation ax + by + cz = c. 𝑎 1 𝑥 + 𝑎 2 𝑦 + 𝑎 3 𝑧 = 𝑐 What kind of surface is that? It's a plane.To 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 thelevel surfaces by the tangent planes. You'll end up in this situation, but maybe that'snot very convincing.