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My book states without proof that the directional derivative at any point is orthogonal to the tangent to the level set at the same point.

I don't even know where to get started.

All I contribute is that :

Assume $f : R^n \rightarrow R$ (I can make this assumption as per question)

  • $D_v f(a_1,a_2,\cdots,a_n) = ||\nabla{f(x,y)}||_{a_1,a_2,\cdots,a_n} \cdot \dfrac{v}{||v||}$

  • I need to show that 2 vectors are orthogonal and thus, I feel that there is a point where I'd need to show that the inner product of the above directional derivative vector with the tangent vector is 0. (But what is the tangent vector?)

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Just a curiosity: what is your book? –  Siminore Jul 6 '12 at 11:33
    
The scalar function $f$ is constant on all curves located on the level set going through the point $x$, hence the directional derivative $D_v f=\nabla f\cdot v$ of $f$ with respect to any direction vector $v$ parallel to the tangent hyperplane is zero, and hence $\nabla f\in \langle v\rangle^\perp$. –  anon Jul 6 '12 at 11:35
    
@anon tangent hyperplane? I follow that the "value" of f won't change as I move along the level set (by definition of a level set) but couldn't follow beyond that. –  Real_Analysis Jul 6 '12 at 11:39
    
The tangent hyperplane is just like a tangent plane: it is the subspace of vectors tangent the the level set at a certain point, translated through space so that the (it's called a hyperplane, because it has dimension one less than the full ambient space) hyperplane contains said point. –  anon Jul 6 '12 at 11:42
    
@anon " the directional derivative Dvf=∇f⋅v of f with respect to any direction vector v parallel to the tangent hyperplane is zero" How did you arrive at that? –  Real_Analysis Jul 6 '12 at 11:43
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2 Answers 2

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A differentiable function $f:\ {\mathbb R}^n\to {\mathbb R}$ has at each point ${\bf p}$ of its domain a gradient $$\nabla f({\bf p})=\Bigl({\partial f\over\partial x_1},{\partial f\over\partial x_2},\ldots,{\partial f\over\partial x_n} \Bigr)_{\bf p}\ .$$ On the other hand, the directional derivative of $f$ at ${\bf p}$ in direction ${\bf v}$ is given by $$D_{\bf v}f({\bf p}):=\lim_{t\to0}{f({\bf p}+ t{\bf v})-f({\bf p})\over t} =\nabla f({\bf p})\cdot{\bf v}\ .$$ Now the level surface: If $\nabla f({\bf p})\ne{\bf 0}$ then the level set of $f$ through the point ${\bf p}$ is locally a smooth surface $S$. Consider a curve $\gamma:\ t\mapsto {\bf x}(t)$ drawn on $S$ with ${\bf x}(0)={\bf p}$. Then the function $$\phi(t):=f\bigl({\bf x}(t)\bigr)$$ is constant, namely $\equiv f({\bf p})$. It follows that $\phi'(t)\equiv0$. In particular we have by the chain rule $$0=\phi'(0)=\nabla f\bigl({\bf x}(0)\bigr)\cdot {\bf x}'(0)=\nabla f({\bf p})\cdot {\bf x}'(0)\ ,$$ which says that $\nabla f({\bf p})$ is orthogonal to the tangent vector ${\bf x}'(0)$ to $\gamma$ at ${\bf p}$. Since this is true for any level curve $\gamma$ through ${\bf p}$ it follows that the vector $\nabla f({\bf p})$ is orthogonal to the level surface $S$.

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The answer is a bit involved. You can read it on page 335 of this link. The idea is to parametrize level curves and to differentiate implicitly the equation $f(x,y)=c$.

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I had gone through this link. But the Theorem 5 (The answer) calls a result from Theorem 4 which is stated without proof. Also, the book I'm using isn't a mainstream one. Just a bunch of notes my professor put together. It's based on Apostol though. –  Real_Analysis Jul 6 '12 at 11:42
    
Actually, you must define tangent vectors to the level curves. In a general setting, the implicit function theorem is the best tool. –  Siminore Jul 6 '12 at 12:39
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