14. Directional Derivatives and Gradients

e. Tangent and Normal, Lines and Planes to Level Sets

3. Normal Lines to Level Sets

The gradient is normal to each level set. So not only can it be used to find the tangent line to a level curve in \(\mathbb R^2\) and the tangent plane to a level surface in \(\mathbb R^3\), it can also be used to find a normal line to a level set in \(\mathbb R^2\) or \(\mathbb R^3\).

The parametric equation of a line is \(X=P+t\vec v\) where \(P\) is a particular point, \(\vec v\) is a direction (or tangent) vector and \(X\) is a general point, either \((x,y)\) or \((x,y,z)\). For the normal line to a level set at a point \(P\), the direction vector is the normal to the level set, which in turn is the gradient vector at \(P\): \[\vec v=\vec N=\left.\vec\nabla F\right|_P\]

Find the line perpendicular to the ellipse \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=2\) at the point \((3,2)\).

The ellipse is a level set of the function \(F(x,y)=\dfrac{x^2}{9}+\dfrac{y^2}{4}\) with value \(2\). Its gradient is: \[ \vec\nabla F=\left(\dfrac{2x}{9},\dfrac{2y}{4}\right) \] So the direction vector of the normal line is the normal vector at \((3,2)\): \[ \vec N=\left.\vec\nabla F\right|_{(3,2)} =\left(\dfrac{6}{9},\dfrac{4}{4}\right) =\left(\dfrac{2}{3},1\right) \] Thus the equation of the line is \[ X=P+t\vec N \qquad \text{or} \qquad (x,y)=(3,2)+t\left(\dfrac{2}{3},1\right) \] or \[ x=3+\dfrac{2}{3}t \qquad \text{and} \qquad y=2+t \]