Abstract Nonsense

Crushing one theorem at a time

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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Photo Credit:

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/tangent/tangent.html

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June 9, 2011 - Posted by | Analysis | , , , , ,

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