# Directional Derivative and Level Sets Orthogonality

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

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 $S$.
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 $f(x,y)=c$.