Problem 11 - A History of Mathematics: An Introduction

Question

Show that the plane \(z=\alpha y-\beta x+\gamma\) is perpendicular to the surface \(z=f(x, y)\) if \(\beta \frac{\partial z}{\partial x}-\alpha \frac{\partial z}{\partial y}=1\). (Show that the plane contains a normal line to the surface.)

Step-by-Step Solution

Answer

Question: Prove that the plane \(z = \alpha y - \beta x + \gamma\) is perpendicular to the surface \(z = f(x, y)\) if \(\beta \frac{\partial z}{\partial x} - \alpha \frac{\partial z}{\partial y} = 1\). Answer: Since the dot product of the normal vectors of the plane and the surface is equal to zero (\(\mathbf{n}_p \cdot \mathbf{n}_s = 0\)), the plane is perpendicular to the surface.

Step 1: Find the normal vector of the plane

The normal vector of the plane \(\mathbf{n}_p\) is the gradient of the plane function \(z = \alpha y - \beta x + \gamma\). To find \(\mathbf{n}_p\), take the partial derivatives with respect to \(x\) and \(y\) to obtain: $$\frac{\partial z}{\partial x} = -\beta,$$ $$\frac{\partial z}{\partial y} = \alpha.$$ Thus, the normal vector of the plane \(\mathbf{n}_p\) is: $$\mathbf{n}_p = \langle -\beta, \alpha, 1 \rangle.$$

Step 2: Find the normal vector of the surface at a point

The normal vector of the surface \(\mathbf{n}_s\) is the gradient of the surface function \(z = f(x, y)\). To find \(\mathbf{n}_s\), take the partial derivatives with respect to \(x\) and \(y\) to obtain: $$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x},$$ $$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial y}.$$ Thus, the normal vector of the surface \(\mathbf{n}_s\) is: $$\mathbf{n}_s = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \rangle.$$

Step 3: Show that the dot product of normal vectors is zero.

In order for the plane to be perpendicular to the surface, the dot product of the plane's normal vector and the surface's normal vector must be equal to zero. The dot product of \(\mathbf{n}_p\) and \(\mathbf{n}_s\) is: $$\mathbf{n}_p \cdot \mathbf{n}_s = \langle -\beta, \alpha, 1 \rangle \cdot \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \rangle=-\beta \frac{\partial f}{\partial x} + \alpha \frac{\partial f}{\partial y} + 1.$$ From the given condition, $$\beta \frac{\partial z}{\partial x}-\alpha \frac{\partial z}{\partial y}=1,$$ we know that: $$-\beta \frac{\partial f}{\partial x} + \alpha \frac{\partial f}{\partial y} + 1 = 0.$$ Thus, the dot product of normal vectors is zero: $$\mathbf{n}_p \cdot \mathbf{n}_s = 0.$$

Step 4: Conclusion

Since the dot product of the normal vectors of the plane and the surface is equal to zero, the plane is perpendicular to the surface. The plane contains a normal line to the surface, as required.

Key Concepts

Partial Derivatives

Partial derivatives are a core concept in multivariable calculus and differential geometry. They measure how a function changes as its variables change, focusing on one variable at a time while treating others as constants. Consider a function of two variables, say \( f(x, y) \). If we want to see how \( f \) changes as \( x \) changes while keeping \( y \) constant, we use the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \). Similarly, \( \frac{\partial f}{\partial y} \) looks at changes in \( f \) as \( y \) changes.

In the context of surfaces in differential geometry, like the plane given by \( z = \alpha y - \beta x + \gamma \), partial derivatives help us construct gradients and understand the surface's orientation in space. For the plane mentioned, partial derivatives help us find the normal vector, which is crucial for understanding how the surface interacts with other objects.

Normal Vector

A normal vector is a vector perpendicular to a surface at a given point. It is an important concept in geometry and physics, helping to describe the orientation of surfaces. To find the normal vector to a plane, such as \( z = \alpha y - \beta x + \gamma \), we use the gradient of the plane's equation. This is because the gradient points in the direction of the greatest rate of change, which is perpendicular to the level curves (or surfaces).

In the exercise you are working with, the normal vector to the plane \( \mathbf{n}_p = \langle -\beta, \alpha, 1 \rangle \) was determined using partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). This vector is fundamental when assessing how this plane is positioned in relation to a given surface, allowing us to examine perpendicularity conditions through geometric means.

Dot Product

The dot product is an operation that takes two equal-length sequences of numbers (usually, coordinate vectors) and returns a single number. In mathematical terms, if we have two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

The result is a scalar, not a vector. The dot product is particularly useful in determining angles between vectors. When the dot product of two vectors is zero, the vectors are perpendicular to each other.

In this exercise, the dot product checks if the normal vector of the plane \( \mathbf{n}_p = \langle -\beta, \alpha, 1 \rangle \) is perpendicular to the normal vector of the surface \( \mathbf{n}_s = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \rangle \). By computing \( \mathbf{n}_p \cdot \mathbf{n}_s = -\beta \frac{\partial f}{\partial x} + \alpha \frac{\partial f}{\partial y} + 1 \) and showing it equals zero, we confirm perpendicularity, hence showing that the plane contains a normal line to the surface.