Question

Suppose \(\mathbf{n}\) is a vector normal to the tangent plane of the surface \(F(x, y, z)=0\) at a point. How is \(\mathbf{n}\) related to the gradient of \(F\) at that point?

Short answer

Answer: At a specific point on the surface, the normal vector to the tangent plane is equal to the gradient of the function defining the surface. Thus, $\mathbf{n} = \nabla F$.

Step-by-step solution

Recall the definition of the gradient vector

The gradient of a function \(F(x, y, z)\) is a vector that points in the direction of the maximum rate of change of the function at a given point, and its magnitude represents the rate of change in that direction. The gradient of \(F\) is defined as: $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right).$$

Compute the tangent plane at a point on the surface

To compute the tangent plane, consider a point \((x_0, y_0, z_0)\) that lies on the surface. The tangent plane at this point can be written as a linear equation: $$\Delta F = \frac{\partial F}{\partial x}(x - x_0) + \frac{\partial F}{\partial y}(y - y_0) + \frac{\partial F}{\partial z}(z - z_0) = 0.$$

Derive the normal vector

The normal vector \(\mathbf{n}\) is a vector that is perpendicular to the tangent plane, which can be derived from the tangent plane equation: $$\mathbf{n} = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right).$$

Relate the gradient to the normal vector

We observe that \(\mathbf{n}\) is equal to the gradient vector \(\nabla F\). This means that at a given point on the surface \(F(x, y, z) = 0\), the normal vector \(\mathbf{n}\) to the tangent plane is equal to the gradient of \(F\) at that point: $$\mathbf{n} = \nabla F.$$