# Statement(s) > [!NOTE] Theorem (Normal to Level Curve) > Let $F:\mathbb{R}^{3}\to \mathbb{R}$ be a [[Real-Valued Function on Real n-Space (Multivariable Function)|multivariable function]]. Then the [[Fréchet Differentiation|gradient]] of $F$ is [[Orthonormal Set of Real Vectors|orthogonal]] to the [[Algebraic Surface|surface]] $F(x,y,z)=k,$ where $k\in \mathbb{R}$ is constant. > > [!NOTE] Theorem (Normal to Level Surface) > Let $F:\mathbb{R}^{2}\to \mathbb{R}$ be a [[Real-Valued Function on Real n-Space (Multivariable Function)|multivariable function]]. Then the [[Fréchet Differentiation|gradient]] of $F$ is [[Orthonormal Set of Real Vectors|orthogonal]] to the [[Algebraic Curve|curve]] $F(x,y)=k,$ where $k\in \mathbb{R}$ is constant. > # Proof(s) ###### Proof of 1: Consider a point $(x_{0},y_{0})$ on the curve $F(x,y)= k.$ Clearly, $k=F(x_{0},y_{0}).$ Suppose we use the variable $t$ to parametrise this curve so that $F(x(t),y(t))=F(x_{0},y_{0})$and let $(x_{0},y_{0})$ correspond to the point where $t=0$. Differentiating the above equation with respect to $t$ and using the Chain rule, we find that $\frac{ \partial F }{ \partial x } \frac{dx}{dt} + \frac{ \partial F }{ \partial y } \frac{dy}{dt} = 0 $Evaluating this expression at $(x_{0},y_{0})$ gives $\left. \begin{pmatrix} \frac{ \partial F }{ \partial x } \\ \frac{ \partial F }{ \partial y } \end{pmatrix} \right |_{(x_{0},y_{0})} \cdot \begin{pmatrix}x'(0) \\ y'(0) \end{pmatrix} = 0$Which shows that $\nabla F$ is perpendicular to the tangent vector of the curve at $(x_{0},y_{0})$. ###### Proof of 2: Consider any parametrised curve on the surface $F (x,y,z)=k$, passing through $(x_{0},y_{0},z_{0})$ where $t = 0$. Proceeding as above, we come to the conclusion that $\nabla F$ is perpendicular to all possible tangent vectors at $(x_{0},y_{0},z_{0})$. All such tangent vectors lie on the tangent plane at $(x_{0}, y_{0}, z_{0})$, hence we conclude that $\nabla F$ is normal to the surface