Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\gamma=\gamma(x, y, f(x, y))\) be the acute angle between the \(z\)-axis and a normal vector to the surface \(z=f(x, y)\) at the point \((x, y, f(x, y))\) on the surface. Show that sec \(\gamma=\sqrt{f_{x}^{2}+f_{y}^{2}+1}\). (Note that this gives another formula for surface area: \(A(G)=\) \(\left.\iint_{S} \sec \gamma d A .\right)\)

Step-by-Step Solution

Answer

\( \sec \gamma = \sqrt{f_x^2 + f_y^2 + 1} \) at point \((x, y, f(x, y))\).

Step 1: Understand the Geometry

To find the angle \(\gamma\) between the \(z\)-axis and a normal vector to the surface, we first identify that the normal vector to the surface \(z=f(x, y)\) at point \((x, y, f(x, y))\) can be given by the gradient vector \(abla g = (-f_x, -f_y, 1)\), where \( g(x, y, z) = z - f(x, y) = 0\). This is because the gradient vector is perpendicular to the level surface.

Step 2: Determine the Components

The normal vector to the surface is \( \mathbf{N} = (-f_x, -f_y, 1) \). Meanwhile, the \(z\)-axis is represented by the vector \(\mathbf{k} = (0, 0, 1)\). The angle \(\gamma\) is the angle between these two vectors.

Step 3: Use the Dot Product to Find Cosine of the Angle

The dot product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \). For our vectors \( \mathbf{N} \cdot \mathbf{k} = 0 \cdot (-f_x) + 0 \cdot (-f_y) + 1 \cdot 1 = 1\). To find \( \cos \gamma \), use the formula: \( \cos \gamma = \frac{\mathbf{N} \cdot \mathbf{k}}{||\mathbf{N}|| \cdot ||\mathbf{k}||} \).

Step 4: Find the Magnitudes of Vectors

Magnitude of \( \mathbf{N} \) is \( ||\mathbf{N}|| = \sqrt{(-f_x)^2 + (-f_y)^2 + 1^2} = \sqrt{f_x^2 + f_y^2 + 1} \). And \(||\mathbf{k}|| = \sqrt{0^2+0^2+1^2} = 1 \).

Step 5: Calculate Cosine and Secant of the Angle

Using the result from Step 3 \( \cos \gamma = \frac{1}{||\mathbf{N}||} \). Since \( \sec \gamma = \frac{1}{\cos \gamma} \), therefore \( \sec \gamma = ||\mathbf{N}|| = \sqrt{f_x^2 + f_y^2 + 1} \), completing the proof.

Key Concepts

Gradient Vector

When examining surfaces defined by functions like \(z = f(x, y)\), the gradient vector plays a crucial role in understanding their properties. The gradient vector is typically denoted as \(abla f\) and represents the rate and direction of the fastest increase of a function.

This vector is tangent to level surfaces (or contour lines) of the function. For a surface \(z = f(x, y)\), the 'level surface' is given by the equation \(g(x, y, z) = z - f(x, y) = 0\). Here, the gradient \(abla g = (-f_x, -f_y, 1)\) becomes the normal vector \(\mathbf{N}\), which is perpendicular to the surface.

In this context, the components of the gradient vector are derived from the partial derivatives \(f_x\) and \(f_y\), indicating how the function \(f\) changes with respect to \(x\) and \(y\), respectively. Hence,

• The gradient vector captures the steepest ascent's direction.
• It becomes a foundation for finding the normal vector necessary to evaluate angles with standard axes.

Surface Area

The concept of surface area is vital when dealing with surfaces in calculus. For surfaces defined by \(z = f(x, y)\), surface area can be computed using integrals. More specifically, the surface area \(A(G)\) over a region \(S\) can be expressed using the formula: \(A(G) = \iint_{S} \sec \gamma dA\).

In this formula, \(\gamma\) represents the angle between the z-axis and the normal vector. The geometric interpretation is that the surface area involves the planar region and is 'stretched' over the surface based on the orientation of \(\gamma\).

Here's how it works:

• When \(\gamma\) is small, meaning the surface is nearly vertical, \(\sec \gamma\) approaches infinity, increasing the contribution to the surface area.
• This approach utilizes calculus to measure the 'real' area of a 3D surface projected onto 2D space.

Dot Product

The dot product is a fundamental operation in vector calculus and is essential for determining angles between vectors. If you have vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), their dot product is given by: \(\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\).

In this specific problem, to find the angle \(\gamma\) between the normal vector \(\mathbf{N} = (-f_x, -f_y, 1)\) and the z-axis vector \(\mathbf{k} = (0, 0, 1)\), we leverage the dot product:

• The dot product \(\mathbf{N} \cdot \mathbf{k} = 1\) reveals how \(\mathbf{N}\) is aligned with \(\mathbf{k}\).
• Using the formula \(\cos \gamma = \frac{\mathbf{N} \cdot \mathbf{k}}{||\mathbf{N}|| \cdot ||\mathbf{k}||}\), the calculation simplifies due to the known dot product.

This approach provides a geometric perspective on how vectors interact and the angle between them via a simple scalar measurement.

Normal Vector

In the context of a surface \(z = f(x, y)\), a normal vector is essential to understanding the surface's orientation in space. The normal vector is a vector perpendicular to the surface at a given point. The normal vector is critical in calculating surface properties like angles and area.

For our surface, the normal vector \(\mathbf{N}\) can be found from the gradient \(abla g = (-f_x, -f_y, 1)\). It's derived such that it points out perpendicularly from the surface \(z = f(x, y)\) at the point \((x, y, f(x, y))\).

• This normal vector is useful in comparing angles, such as the acute angle \(\gamma\) with the z-axis.
• The magnitude of this vector is \(||\mathbf{N}|| = \sqrt{f_x^2 + f_y^2 + 1}\), which ties directly into the formula for calculating the angle \(\gamma\).

This magnitude helps in converting the geometric description of the surface into useful algebraic forms like finding \(\sec \gamma\). Understanding how the normal vector functions reveal the elegance of calculus in interacting with 3D objects through a 2D slice.