Did each point on the curve has exactly one tangent plane? They seem to have different derivatives in different directions.

silentQ

@ruoyanz That is not necessarily a problem, as I understand it. A point can have one derivative in the x-direction, a different derivative in the y-direction, and a derivative in every direction in between, with those all being parallel to the tangent plane. If a function isn't differentiable, then it can actually be said to have multiple tangent planes. The one in this slide looks pretty differentiable, though.

keenan

@ruoyuanz Right, @silentQ has the right idea. It's fine if the directional derivative is different in different directions, as long as all the directional derivatives can be expressed as a linear function of the direction $u$. In particular, it must be the case that

$$ D_u f = \langle u, v \rangle $$

for some fixed vector $v$, i.e., a constant vector that does not depend on $u$. If such a vector exists, we call it the gradient vector, and usually denote it by

$$v = \nabla f.$$

This is where we get the definition of the gradient seen on this slide, i.e.,

$$ \langle \nabla f, u \rangle = D_u f\ \forall u. $$

What do you use to generate these nice function visualizations?

See the bottom of this page.

Did each point on the curve has exactly one tangent plane? They seem to have different derivatives in different directions.

@ruoyanz That is not necessarily a problem, as I understand it. A point can have one derivative in the x-direction, a different derivative in the y-direction, and a derivative in every direction in between, with those all being parallel to the tangent plane. If a function isn't differentiable, then it can actually be said to have multiple tangent planes. The one in this slide looks pretty differentiable, though.

@ruoyuanz Right, @silentQ has the right idea. It's fine if the directional derivative is different in different directions, as long as

allthe directional derivatives can be expressed as alinearfunction of the direction $u$. In particular, it must be the case that$$ D_u f = \langle u, v \rangle $$

for some fixed vector $v$, i.e., a constant vector that does not depend on $u$. If such a vector exists, we call it the

gradientvector, and usually denote it by$$v = \nabla f.$$

This is where we get the definition of the gradient seen on this slide, i.e.,

$$ \langle \nabla f, u \rangle = D_u f\ \forall u. $$