I am in calculus 3 and I have a question on gradient versus tangent. I know that this has been answered several times before:

Difference between a Gradient and Tangent

But I still have much trouble understanding them.

Assume $z = x^2 + y^2$ (a sphere)

In the topic of directional derivatives and gradients, gradient is the steepest slope (or, more formally, the direction to move so that z increases most rapidly). AT THE SAME TIME, it is a normal to the level curve (to my understanding, it means when z is fixed at a particular value, a vector that is perpendicular to the tangent of a curve at a particular point. The gradient points to the concave side of the level curve).

Combining the two ideas above, I assume it is correct that a normal to the level curve is the same as the direction in which z will increase most rapidly (although I am unsure why it is true).

Somehow, then I thought that gradient points to the direction where a tangent line to the 3D object is, because gradient points to the direction of the steepest slope.

And then in the following topic of tangent planes, now it tells me that gradient is actually a normal line instead of a tangent line to a 3D object? What a surprise!

In the case of 2D, gradient is the slope of a line. therefore, I naturally think that in the case of 3D, gradient is also the slope (or, the vector that points to the steepest slope for z). It is not the case? What is wrong with my thinking?

• You're trying to apply gradient to a surface; this does not make sense. The gradient is an operator on a scalar field. $z = x^2 + y^2$ does not define a scalar field. It could perhaps be the level surface of some scalar field. – Muphrid Nov 14 '15 at 5:01
• 1. Gradient for a line is not the same as gradient for a surface? 2. What is a scalar field? – Joseph Nov 14 '15 at 5:05
• I found the following from another thread: "Gradient is a mathematical term that is equal to the slope in one dimension. In more than one dimension it is no longer possible to talk about a general slope - but he gradient will give you a vector value essentially describing the nn slopes needed to describe the general nn-dimensional case." – Joseph Nov 14 '15 at 5:13
• Okay, let's back up. When you say $z = x^2 + y^2$, you mean $z(x,y)$ is a function on a 2d region? You don't mean $z$ is a coordinate in a 3d space? – Muphrid Nov 14 '15 at 5:15
• z is the same as f(x,y) here, I think? So, z is a function of x and y, and it is the 'vertical' axis in a 3D graph? Oh, the equation should be f(x,y) = sqrt(4 - x^2 - y^2)? – Joseph Nov 14 '15 at 5:20

It sounds like you still think of functions like $z$ here as though they are "plotted" along additional axes. In my opinion, this is great for visualization but can be confusing: such a viewpoint adds structure to what you're picturing, structure that isn't actually there.
Instead, let $f(x,y)$ merely associate a number with any particular $(x,y)$ point on the 2d plane. This is what is meant by a "scalar field"--a scalar value is assigned to every point in a region (here, the entire 2d plane). In mathematics, this is sometimes simply called a function.
When you use this intrinsic viewpoint, the level curves of the function $f$ are literally that: curves on the 2d plane. Each curve has well-defined tangent and normal vectors. The normal vector, called the gradient, does indeed point toward the direction of greatest increase. It must: the tangent sure doesn't. The tangent points along the level curve, which is not increasing or decreasing at all.
It seems to me you're getting confused precisely because you're still imagining the 2d function plotted with its values on a fictitious third axis. When we're talking about tangent planes, we're moving to the case of functions on 3d space: a function $g(x,y,z)$. The level curves for this function in 3d space are no longer curves: they are surfaces, and these surfaces have tangent planes. The gradient still points perpendicular to these tangent planes. And here, the idea of plotting this function $g$ with an additional axis breaks down because you can't imagine (easily) a 4d space.