How is the gradient of a curve or function its normal

Question

I see in several books that the gradient of a function (curve), f(x,y) is defined as the normal to the curve without the proof. Can someone prove mathematically that this is true?

Answer 1

Gradient of a function, i.e., $\vec{\nabla} f(x)$, is not normal to the curve $f(x)$. It is normal to the contour curve, and tangent to the curve $f(x)$.

Consider a function $z=f(x,y)$ then rewrite it as $f(x(t),y(t)) = c$ with some constant scalar $c$ for each level surface (i.e. a surface with constant value of $f(x,y)$). Then let $\vec{r}(t) = \begin{bmatrix}x(t) & y(t)\end{bmatrix}'$ be a parametric curve on the level surface . Also note that $x=x(t), y=y(t)$ for the general case.

Now we take the derivative

$\dfrac{df}{dt} = \dfrac{\partial f}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial f}{\partial y}\dfrac{dy}{dt} = 0$

$\dfrac{df}{dt} = \begin{bmatrix}\dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y}\end{bmatrix} \begin{bmatrix}\dfrac{dx}{dt} & \dfrac{dy}{dt}\end{bmatrix}' =0$

$\vec{\nabla} f(x,y) \cdot \vec{r}(t) = 0$

Because the inner product of the above two vectors is 0, it implies that $\vec{\nabla} f(x,y) \perp \vec{r}(t)$. Hence the gradient of $f(x,y)$ is perpendicular to the level surface (i.e. the contour line) of $f(x,y)$.

The same proof can be used for the function $f$ in higher dimension.