Question

Prove that the angle of inclination \(\theta\) of the tangent plane to the surface \(z=f(x, y)\) at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) is given by \(\cos \theta=\frac{1}{\sqrt{\left[f_{x}\left(x_{0}, y_{0}\right)\right]^{2}+\left[f_{y}\left(x_{0}, y_{0}\right)\right]^{2}+1}}\).

Short answer

As shown in the steps above, by applying the definitions of the normal vector to the tangent plane and the angle of inclination, and then using the cosine formula, we find that \(\cos \theta=\frac{1}{\sqrt{\left[f_{x}\left(x_{0}, y_{0}\right)\right]^{2}+\left[f_{y}\left(x_{0}, y_{0}\right)\right]^{2}+1}}\).

Step-by-step solution

Calculation of partial derivatives

First, find the partial derivatives \(f_{x}\) and \(f_{y}\) of the function \(f(x, y)\) at the point \((x_{0}, y_{0})\). These values represent the slopes of the function in the x and y directions, respectively.

Define the normal vector

The normal vector to the tangent plane at the point \(P(x_{0}, y_{0}, z_{0})\) is given by the gradient of \(f\) at the point \((x_{0}, y_{0})\), that is \(-f_{x}(x_{0}, y_{0})\hat{i} -f_{y}(x_{0}, y_{0})\hat{j} + \hat{k}\). The negative signs are due to the direction of the normal.

Define the inclination angle

The inclination angle \(\theta\) is the angle between the positive z-axis and the normal vector to the tangent plane.

Applying cosine rule

By geometry, we know the cosine of the angle between two vectors is given by the dot product of the vectors divided by the product of their magnitudes. Applying this, we have \(\cos \theta = \frac{\hat{k} \cdot \left(-f_{x}(x_{0}, y_{0})\hat{i} -f_{y}(x_{0}, y_{0})\hat{j} + \hat{k}\right)}{||\hat{k}|| ||-f_{x}(x_{0}, y_{0})\hat{i} -f_{y}(x_{0}, y_{0})\hat{j} + \hat{k}||}\).

Simplifying the expression

The magnitude of the unit vector \(\hat{k}\) along the z axis is 1 and the dot product \(\hat{k} \cdot \left(-f_{x}(x_{0}, y_{0})\hat{i} -f_{y}(x_{0}, y_{0})\hat{j} + \hat{k}\right)\) simplifies to 1. The magnitude of the normal vector is \(\sqrt{f_{x}^{2}(x_{0}, y_{0}) + f_{y}^{2}(x_{0}, y_{0}) + 1}\). Substituting these gives \(\cos \theta=\frac{1}{\sqrt{\left[f_{x}\left(x_{0}, y_{0}\right)\right]^{2}+\left[f_{y}\left(x_{0}, y_{0}\right)\right]^{2}+1}}\).

Key concepts

Partial Derivatives

Understanding the concept of partial derivatives is crucial when dealing with functions of multiple variables. In the context of the mentioned exercise, partial derivatives represent how the function's output changes as we vary each input slightly, holding other inputs constant.

Consider a function \(z=f(x, y)\) of two variables, \(x\) and \(y\). The partial derivative with respect to \(x\), denoted as \(f_x\), measures the rate at which \(z\) changes with a small increase in \(x\) while \(y\) remains the same. Similarly, \(f_y\) looks at how \(z\) changes as \(y\) is varied slightly with \(x\) held fixed. These derivatives are typically found using methods of differentiation akin to those used for functions of a single variable but require one to consider each variable in isolation.

Normal Vector

The normal vector plays a pivotal role when it comes to understanding geometric properties of surfaces. In our case, the normal vector to the tangent plane is essential to finding the inclination angle.

The normal vector is perpendicular to the tangent plane at a given point. It can be found by taking the gradient of function \(f\), which is composed of the partial derivatives \(f_x\) and \(f_y\), and the unit vector in the \(z\)-direction, \(\hat{k}\). Intuitively, the direction of this normal vector encapsulates the steepest ascent from the point on the surface. Thereby, knowing the normal vector helps us find the exact orientation of the tangent plane. It's necessary for the exercise to correctly define the normal vector, as mistakes here will propagate through subsequent calculations.

Cosine Rule in Vector Geometry

The cosine rule, or the law of cosines, is a principle in vector geometry that can help us understand angles and relationships in various geometric configurations. When applied to vectors, the cosine of the angle between two vectors is equal to the dot product of the vectors normalized by the product of their magnitudes.

In our problem, we're using the cosine rule to calculate the angle between the normal vector to the tangent plane and the vertical, represented by the \(z\)-axis. By defining the normal vector and calculating its dot product with the unit vector \(\hat{k}\), we can use the cosine rule to find the desired inclination angle \(\theta\). The elegance of this rule lies in not having to visually measure angles; instead, we use algebraic operations on vectors to deduce the angle accurately.