Why are these tangent vectors?

Question

I have a differentiable function $f:\mathbb{R}^2\to\mathbb{R}$. Let the graph of $f$ be the surface $\Sigma = \{(x,y,z):z=f(x,y)\}$.

My question is why are the vectors $v=(1,0,\frac{\partial f}{\partial x})$ and $w=(0,1,\frac{\partial f}{\partial y})$ tangent vectors to the surface $\Sigma$?

I know that $\frac{\partial f}{\partial x}$ is the instantaneous change in $f$ in the $x$-direction given $y$ is fixed. So graphically it makes sense that the tangent vector $v$ has $0$ in its y-coordinate. But why is the x-coordinate 1? I'm still shaky and not fully confident. Any intuition or clarification would help.

For context, I'm learning about the method of characteristics in solving semi-linear PDEs and so I need to construct a normal to the solution surface using two tangent vectors to the surface.

Maybe I don't understand what the partial derivative means.

Answer 1

In general, a tangent vector to this surface will have the form $ \Big ( a , \; b , \; a \frac { \partial f } { \partial x } + b \frac { \partial f } { \partial y } \Big ) $. This is because, if you move along the surface for a little distance $ a $ in the $ x $-direction, then you'll move about $ a \frac { \partial f } { \partial x } $ in the $ z $-direction; and similarly, if you move along the surface for a little distance $ b $ in the $ y $-direction, then you'll move about $ b \frac { \partial f } { \partial y } $ in the $ z $-direction. Once you know that, you can pick pretty much whatever you want for $ a $ and $ b $, as long as you pick two linearly independent vectors; but $ ( a = 1 , \; b = 0 ) $ and $ ( a = 0 , \; b = 1 ) $ are two particularly convenient choices.

Answer 2

At a given point $(a,b)$, the quantity $\frac{\partial f}{\partial x}(a,b)$ represents the slope of a tangent line to the graph above $(a,b)$, facing in the $x$-axis direction. In other words, if I start at $(a,b)$ and go $1$ unit in the $x$ direction and $0$ units in the $y$ direction, I will move up this tangent line by $\frac{\partial f}{\partial x}(a,b)$ units in the $z$ direction, by the fact that this is the slope of the tangent line in the $x$-axis direction. The vector that represents this movement up the tangent line is $(1,0,\frac{\partial f}{\partial x}(a,b))$, or just $(1,0,\frac{\partial f}{\partial x})$ if we drop the inclusion of the point $(a,b)$ from our notation. Because the tangent vector in the $x$ direction should describe unit movement of the linear approximation at $(a,b)$, this is in fact the tangent vector in the $x$ direction. This point about unit movement is the same reasoning behind choosing the slope to represent the change in $y$ after a unit change in $x$ is made.

To contrast, we could also start at $(a,b)$ and move $2$ units in the $x$ direction and $0$ units in the $y$ direction, which would take us up by $2\frac{\partial f}{\partial x}(a,b)$ in the $z$ direction if we follow the $x$-direction tangent line. This movement gives us the vector $(2,0,2\frac{\partial f}{\partial x}(a,b))$. This would not be the tangent vector in the $x$ direction, as it is not the result of unit movement along the linear approximation in the $x$ direction at the point $(a,b)$.

Answer 3

One of the abstract definitions of a tangent vector at a point $x \in M$ on manifolds is the velocity at time zero of a "short curve" $\dot{\gamma}(0)$ that is at the point $x$ at time zero $\gamma(0) = x$. A consequence of this definition is that the tangent vectors of a manifold defined as the level set of a function $F=c$ are the kernel of the derivative of $F$. This is by the chain rule since

$$0 = \left.\frac{d}{dt}\right|_{t=0}F(\gamma(t)) = DF \cdot \dot{\gamma}(0)$$

The specific case of a graph of a function $z = f(x,y)$ is the zero level set of $F = z-f$ and you can calculate the derivative matrix to get the answer of Toby Bartels by solving $DF \cdot v=0$.