# Abstract Nonsense

## The Geometry of the Derivative for Real Valued Mappings (Pt. I)

Point of Post: In this post I’d like to discuss some of the geometric aspects of what the total and partial derivatives mean including the idea of approximating lines and tangent planes.

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Motivation

As usual in math it’s helpful to have a picture to backup the ideas. In this post we discuss what it geometrically looks like when a mapping $f:\mathbb{R}^n\to\mathbb{R}$ is differentiable at a point in terms of tangent planes. This of course generalize the notion that a mapping $\mathbb{R}\to\mathbb{R}$ is differentiable at a point if it has a tangent line there.

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Tangent Planes

Recall from basic geometry that a hyperplane in $\mathbb{R}^n$ is a set of vectors which are mutually orthogonal to some fixed vector, call the normal, which pass through some specified point. In particular, if $a,b\in\mathbb{R}^n$ with $b\ne\bold{0}$ we define the hyperplane through $a$ with normal $b$, denoted $\Pi_b(a)$, to be the set $\left\{x\in\mathbb{R}^n:(x-a)\cdot b=0\right\}$. If $n=3$ we call a hyperplane just a plane.  Of course a hyperplane is just an $n-1$-dimensional subspace of $\mathbb{R}^n$ which is shifted by a certain vector. Indeed, if one defines the linear functional $f:\mathbb{R}^n\to\mathbb{R}$ by $x\mapsto x\cdot b$ then we know that $\ker f$ is a $n-1$-dimensional subspace of $\mathbb{R}^n$ and it’s easy to see that $\Pi_b(a)=\ker f+a\cdot b$. Said in more common language we see that $\Pi_b(a)=\{b\}^\perp+a\cdot b$. Thus, we can alternatively define $\Pi_b(a)$ as $\left\{(x_1,\cdots,x_n)\in\mathbb{R}^n:x_1b_1+\cdots+x_nb_n=a_1b_1+\cdots+a_nb_n\right\}$ if one likes more explicit formulae.

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Let $S\subseteq\mathbb{R}^n$, we say that the hyperplane $\Pi_b(a)$ is tangent to $S$ at $c$ if $c\in\Pi_b(a)$ and $\displaystyle \lim \frac{c-c_n}{\|c-c_n\|}\cdot b=0$ for any sequence in $S-\{c\}$ with $\lim c_n=c$. Note that if we use the common definition of angle in $\mathbb{R}^n$ given by $x\cdot y=\|x\|\|y\|\cos\left(\angle(x,y)\right)$ then tangency is equivalent to saying that $\displaystyle \angle (c-c_n,b)\to \frac{\pi}{2}$ for every $c_n\to c$ which clearly satisfies our intuitive notion of tangency.

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What we now show is that every differentiable function possesses a tangent plane. But first, we recall the definition of the graph $\Gamma_f$ of a function $f:X\to Y$ is defined by $\left\{(x,f(x)):x\in X\right\}\subseteq X\times Y$. In what follows we shall think of  the graph of a mapping $f:\mathbb{R}^n\to\mathbb{R}$ as sitting inside $\mathbb{R}^{n+1}$ instead of $\mathbb{R}^n\times\mathbb{R}$ in the obvious way.

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Theorem: Let $f:U\to\mathbb{R}$, with $U\subseteq\mathbb{R}^n$ open, be differentiable at $a\in U$. Then the hyperplane $\Pi\overset{\text{def.}}{=}\Pi_{(\nabla f(a),-1)}((a,f(a))$ is tangent to $\Gamma_f$ at $a$ where $\nabla f(a)$ is the gradient of $f$ (where we, as stated before this problem, freely identify $(\nabla f(a),-1)$ with $(D_1f(a),\cdots,D_nf(a),-1)$ and $(a,f(a))$ with $(a_1,\cdots,a_n,f(a))$).

Proof: Let $c_m$ be a sequence in $\Gamma_f-\{(a,f(a))\}$, then by definition $c_m=(a_m,f(a_m))$ with $a_m\to a$ (this is enough since we know that $f$ must be continuous) Thus, we get that

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\displaystyle \begin{aligned}\left|\frac{(a_k,f(a_k))-(a,f(a))}{\|(a_k,f(a_k))-(a,f(a))\|}\cdot (\nabla f(a),-1)\right| &=\left|\frac{1}{\|(a_k-a,f(a_k)-f(a))\|}\left(\nabla f(a)\cdot (a-a_k)-(f(a_k)-f(a))\right)\right|\\ &\leqslant -\frac{\left|f(a)-f(a_k)-\nabla f(a)\cdot(a_k-a)\right|}{\|a_k-a\|}\end{aligned}

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but this last term goes to zero by definition of the derivative. Since the sequence was arbitrary the conclusion follows. $\blacksquare$

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For the sake of pedagogy let us do an actual example. Take $f$ to be the hyperboloid $x^2-y^2$. Clearly then $f$ is smooth  on all of $\mathbb{R}^2$. Let us then compute the tangent plane to $f$ at $(0,2)$. In particular since $(0,2,f(0,2))=(0,2,-4)$ and $(\nabla f(0,2),-1)=(0,4,-1)$ and so the tangent plane $\Pi$ is equal to $\left\{(x_1,x_2,x_3)\in\mathbb{R}^2:(x_1,x_2-2,x_3+4)\cdot(0,4,-1)=0\right\}$ or when one writes it out it is given by the equation $z=4y-12$. Graphing this gives us the following picture:

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Evidently from the above theorem we recover the formula taught in every multivariable calculus course. Namely, if $f:U\to\mathbb{R}$, with $U\subseteq\mathbb{R}^n$ open, is differentiable at $(x_0,y_0)\in U$ then the tangent plane $\Pi$ to $\Gamma_f$ at $(x_0,y_0)$ is given by the equation

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$z=f(x_0,y_0)+D_1f(x_0,y_0)(x-x_0)+D_2f(x_0,y_0)(y-y_0)$

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References:

1.  Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.

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