Question

(a) Given \(\phi=x^{2}-y^{2} \varepsilon\), find \(\nabla \phi\) at \((1,1,1)\). (b) Find the directional derivative of \(\phi\) at \((1,1,1)\) in the direction \(i-2 j+k\). (c) Find the equations of the normal line to the surface \(x^{2}-y^{2} z=0\) at \((1,1,1)\),

Short answer

Gradient: (2, -2, -1). Directional derivative: \backslash \frac{5}{\backslash \sqrt{6}}. Parametric equations: x=1+2t, y=1-2t, z=1-t.

Step-by-step solution

Compute the Gradient of \(\backslash \phi\)

To find \(\backslash \abla \phi\), first calculate the partial derivatives of \(\backslash \phi (x, y, z) = x^{2} - y^{2} z \). \(\frac{\backslash \partial \phi}{\backslash \partial x} = 2x\), \(\backslash \frac{\backslash \partial \phi}{\backslash \partial y} = -2yz\), and \(\backslash \frac{\backslash \partial \phi}{\backslash \partial z} = -y^{2}\).

Evaluate the Gradient at (1,1,1)

Substitute x=1, y=1, z=1 into the partial derivatives. \(\backslash \abla \phi (1,1,1) = (2(1), -2(1)(1), -1^{2}) = (2, -2, -1)\).

Find the Directional Derivative

The directional derivative of \(\backslash \phi\) in the direction of \(\vec{v} = i - 2j + k\) is given by \(\backslash \abla \phi (1,1,1) \backslash \cdot \backslash \hat{v} = (2,-2,-1) \backslash \cdot \frac{1}{\backslash \sqrt{6}}(1, -2, 1)= \backslash \frac{2 + 4 - 1}{\backslash \sqrt{6}} = \backslash \frac{5}{\backslash \sqrt{6}}\).

Find the Normal Line Equations

The normal line to the surface \(x^{2} - y^{2} z = 0\) at (1,1,1) has direction vector \((2, -2, -1)\). The parametric equations are: \(x=1+2t\), \(y=1-2t\), and \(z=1-t\).

Key concepts

Partial Derivatives

Partial derivatives are a cornerstone of multivariable calculus. They measure how a function changes as only one of the input variables changes while keeping the others constant. For a function \(f(x, y, z)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\backslash \partial f}{\backslash \partial x}\). In our exercise, we started with \(\backslash \phi(x, y, z) = x^{2} - y^{2} z \). We computed:

• \(\frac{\backslash \partial \phi}{\backslash \partial x} = 2x\)
• \(\frac{\backslash \partial \phi}{\backslash \partial y} = -2yz\)
• \(\frac{\backslash \partial \phi}{\backslash \partial z} = -y^{2}\). Exploring partial derivatives helps to understand the behavior of functions in a multidimensional space. By focusing on one variable at a time, we can simplify complex functions. This makes partial derivatives incredibly useful for analyzing and solving real-world problems like optimization and in this case, computing gradients.

Directional Derivative

The directional derivative extends the concept of a derivative into multidirectional spaces. It measures how a function changes as we move in any particular direction. To find the directional derivative of \(\backslash \phi(x, y, z) = x^{2} - y^{2} z \) at \( (1,1,1) \), in the direction of \(\backslash \vec{v} = i - 2j + k\), we used the gradient vector \(\backslash \backslash \abla \phi(1,1,1) = (2, -2, -1)\).

First, normalize \(\backslash \vec{v}\):
• \(\backslash \hat{v} = \frac{1}{\backslash \sqrt{6}} (1, -2, 1)\)
Then, we compute the dot product:
• \(\backslash \backslash \abla \phi(1,1,1) \backslash \cdot \backslash \hat{v}\) = \(\frac{5}{\backslash \sqrt{6}}\)

The directional derivative helps us understand how the function behaves as we move in different specific directions, which is crucial in fields like physics and engineering.

Normal Line

In vector calculus, the normal line to a surface at a given point is a line that is perpendicular to the tangent plane of the surface at that point. This concept is essential in differential geometry and helps us understand surface orientation. For the surface \(x^{2}-y^{2}z=0\) at \((1,1,1) \), the gradient at this point \(\backslash \abla \phi(1,1,1) = (2, -2, -1)\) provides the direction of the normal line.
The parametric equations of the normal line are then:

• \(x=1+2t\)
• \(y=1-2t \)
• \(z=1-t)\). These equations describe how to move along the normal line originating from the point \( (1, 1, 1)\) in 3D space.

Normal lines are practical for many applications, including finding distances from points to surfaces, analyzing light reflection, or even optimizing geometric structures.

Vector Calculus

Vector calculus is an extension of calculus into multiple dimensions, which deals with vector fields and the operations on these fields. In the context of our example, several concepts from vector calculus were utilized:

• Gradients: Represent the rate and direction of fastest increase of a scalar field.
• Partial Derivatives: Measure the function’s rate of change with respect to each variable independently.
• Directional Derivatives: Extend partial derivatives to show changes in any specified direction.
• Normal Lines: Apply the concept of gradients to establish perpendicular lines to surfaces. Understanding these concepts equips students with the tools necessary to navigate and solve complex problems involving multidimensional functions, optimizing real-world scenarios, and analyzing physical phenomena.