Tangent plane to a function $z=f(x,y)$

Question

I'm given with the following question:

Part A: prove that for the tangent plane to a function $z=f(x,y)$ at a point $(x_0,y_0,z_0)$ is given by $z=f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) +f_y(x_0,y_0)(y-y_0)$ (I did it)

Part B: prove that if $f$ is differentiable at $(x_0,y_0)$ then the tangent plane mentioned above contains the tangent line to every curve lying inside the surface $z=f(x,y)$ at the point $(x_0,y_0, f(x_0,y_0))$ .

I can't understand how to solve part B. Such a curve has a parameterization $\gamma(t) =(x(t), y(t) , f(x(t),y(t)) ) $ and by differentiating I get: $\gamma'(t) = (\gamma_x ,\gamma_y ,\gamma_z) \cdot (x'(t),y'(t), f'(x(t),y(t) ) ) $ and I can't understand why this should lie in the tangent plane I found in part A. I guess I should prove that $x'(t)= x-x_0,...$ but can't understand why this is true

Thanks in advance

Answer 1

I think it is easier to work in $\mathbb{R}^3$ rather that dealing with $f$ and $\gamma$ and stitching them together in some way.

Let $\phi: \mathbb{R}^3 \to \mathbb{R}$ be given by $\phi(x) = f(x_1,x_2)-x_3$. Let $C = \{ x | \phi(x) = 0 \}$.

The tangent plane to $C$ at the point $x_0$ is given by $T = \{ x | \langle \nabla \phi( x_0 ) , x-x_0 \rangle = 0 \} $.

Note that $\nabla \phi( x_0 ) = \binom{ \nabla f( [x_0]_1, [x_0]_2) }{-1}$.

Suppose $\gamma$ is a curve that lies in $C$ and passes through $x_0$. Without loss of generality, suppose $\gamma(0) = x_0$. Since the curve lies in $C$, we have $\phi(\gamma(t)) = 0$ for all appropriate $t$, in particular at $t=0$, we have $\phi(x_0) = 0$ and $\langle \nabla \phi( x_0 ), \gamma'(0) \rangle = 0$.

The tangent line to $\gamma$ at $t=0$ is given by $L = \{ \gamma(0) + s \gamma'(0) \}_{s \in \mathbb{R}}$.

The question asks to show that $L \subset T$, this is straightforward to verify given the definition of $L$ and the fact that $\langle \nabla \phi( x_0 ), \gamma'(0) \rangle = 0$.

Answer 2

Clearly, the curve (as a whole) does not necessarily lie in the tangent plane.

What DOES lie in the tangent plane is the tangent line of the curve at $(x_0,y_0,z_0)$, where $z_0=f(x_0,y_0)$.

First of all, let us be reminded that this tangent plane is the one passing from $(x_0,y_0,z_0)$ and perpendicular to $\big(\,f_x(x_0,y_0),\; f_y(x_0,y_0),\; -1\big)$.

Let's first see what is this tangent line: It is the straight line passing from $(x_0,y_0,z_0)$ with slope \begin{align*} \big(x'(0),y'(0),z'(0)\big)=&\left.\frac{d} {dt}\right|_{t=0}\bigg(x(t),\; y(t),\; f\big(x(t),y(t)\big)\bigg) \\ =&\bigg(x'(0),\; y'(0),\; f_x(x_0,y_0)x'(0)+f_y(x_0,y_0)y'(0)\bigg)\\ =&x'(0)\bigg(1,\; 0, \; f_x(x_0,y_0) +y'(0)\big(0,1,f_y(x_0,y_0)\bigg) \end{align*}

Now this line DOES lie of the tangent plane since both vectors $$ \big(1,\; 0,\; f_x(x_0,y_0)\big)\quad\text{and}\quad \big(0,\; 1,\; f_y(x_0,y_0)\big) $$ and perpendicular to $\big(\,f_x(x_0,y_0),\; f_y(x_0,y_0),\;-1\big)$.