Problem 13 - Advanced Engineering Mathematics

Question

Find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=\tan ^{-1} \frac{y}{x} ;(2,-2), \mathbf{i}-3 \mathbf{j} $$

Step-by-Step Solution

Answer

The directional derivative is \(\frac{-1}{2\sqrt{10}}\).

Step 1: Calculate the Gradient of the Function

To find the directional derivative, we first need the gradient of the function \( f(x, y) = \tan^{-1} \left( \frac{y}{x} \right) \). The gradient vector, \( abla f(x, y) \), is given by the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Calculate \( \frac{\partial f}{\partial x} \):Using the chain rule, \( \frac{\partial}{\partial x} \tan^{-1} \left( \frac{y}{x} \right) = \frac{-y}{x^2+y^2} \).Calculate \( \frac{\partial f}{\partial y} \):Using the chain rule, \( \frac{\partial}{\partial y} \tan^{-1} \left( \frac{y}{x} \right) = \frac{x}{x^2+y^2} \).Thus, \( abla f(x, y) = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right) \).

Step 2: Evaluate the Gradient at the Given Point

Substitute \((x, y) = (2, -2)\) into the gradient \( abla f(x, y) \) to find its value at the point:\( abla f(2, -2) = \left( \frac{-(-2)}{2^2 + (-2)^2}, \frac{2}{2^2 + (-2)^2} \right) = \left( \frac{2}{8}, \frac{2}{8} \right) = \left( \frac{1}{4}, \frac{1}{4} \right) \).

Step 3: Create the Unit Vector for the Direction

The given direction vector is \( \mathbf{i} - 3\mathbf{j} \), which corresponds to \( \langle 1, -3 \rangle \). First, find the magnitude of this vector:\( ||\langle 1, -3 \rangle|| = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \).The unit vector \( \mathbf{u} \) in this direction is:\( \mathbf{u} = \frac{1}{\sqrt{10}} \langle 1, -3 \rangle = \langle \frac{1}{\sqrt{10}}, \frac{-3}{\sqrt{10}} \rangle \).

Step 4: Calculate the Directional Derivative

The directional derivative \( D_{\mathbf{u}} f \) of the function at the point \((2, -2)\) in the direction of the unit vector \( \mathbf{u} \) is the dot product \( abla f(2, -2) \cdot \mathbf{u} \):\( D_{\mathbf{u}} f = \left\langle \frac{1}{4}, \frac{1}{4} \right\rangle \cdot \left\langle \frac{1}{\sqrt{10}}, \frac{-3}{\sqrt{10}} \right\rangle \).Calculate the dot product:\( D_{\mathbf{u}} f = \frac{1}{4} \cdot \frac{1}{\sqrt{10}} + \frac{1}{4} \cdot \frac{-3}{\sqrt{10}} \).\( D_{\mathbf{u}} f = \frac{1}{4\sqrt{10}} - \frac{3}{4\sqrt{10}} = \frac{-2}{4\sqrt{10}} = \frac{-1}{2\sqrt{10}} \).

Key Concepts

Gradient Vector

The gradient vector is a crucial concept in multivariable calculus. It is denoted by \( abla f(x, y) \) and provides the direction and rate of the steepest ascent of a function. For a function of two variables, \( f(x, y) \), the gradient is composed of its partial derivatives:

• \( \frac{\partial f}{\partial x} \): This represents the rate of change of the function along the x-axis.
• \( \frac{\partial f}{\partial y} \): This shows the rate of change along the y-axis.

By combining these, the gradient vector is \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
The direction in which the gradient vector points is where the function increases most rapidly. And its magnitude gives the rate of increase in that direction.
This vector has extensive applications in optimization problems and is essential for understanding directional derivatives.

Partial Derivatives

Partial derivatives express how a function changes as each input variable is varied, holding the others constant. In the context of the directional derivative problem, you'll frequently encounter chains of partial derivatives for functions like \( \tan^{-1} \left( \frac{y}{x} \right) \).
For such a function, we calculate:

• \( \frac{\partial f}{\partial x} \): This uses the chain rule to differentiate arctangent with respect to \( x \), resulting in \( \frac{-y}{x^2 + y^2} \).
• \( \frac{\partial f}{\partial y} \): Similarly, this yields \( \frac{x}{x^2 + y^2} \) using the chain rule.

Partial derivatives enable the creation of the gradient vector, bridging the gap between single-variable calculus and multivariable scenarios. They play a pivotal role in understanding and solving problems pertaining to surfaces and curves in three-dimensional space.

Unit Vector

Unit vectors are vectors that have a magnitude of 1 and are used to specify a direction without any scaling. When working on directional derivatives, they are critical for ensuring that you're defining direction accurately.
For any vector, like \( \langle 1, -3 \rangle \), you need to convert it to a unit vector before using it to determine direction. The process involves:

• Calculating the magnitude of the vector: \( ||\langle 1, -3 \rangle|| = \sqrt{1^2 + (-3)^2} = \sqrt{10} \).
• Normalizing the vector: \( \mathbf{u} = \frac{1}{\sqrt{10}} \langle 1, -3 \rangle = \langle \frac{1}{\sqrt{10}}, \frac{-3}{\sqrt{10}} \rangle \).

This unit vector \( \mathbf{u} \) ensures that the resulting directional derivative reflects the pure directional component without being distorted by any inherent scaling factor in the original vector. In practice, aligning vectors to create a unit vector is essential in various fields, such as physics and computer graphics.

Dot Product

The dot product is an operation that takes two vectors and returns a scalar. It often represents the magnitude of one vector projected onto another, which is exceptionally useful when finding directional derivatives. For vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is defined as:

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

In the context of the directional derivative, you'd calculate the dot product between the gradient vector and the unit vector as follows:

• For example, \( \left\langle \frac{1}{4}, \frac{1}{4} \right\rangle \cdot \left\langle \frac{1}{\sqrt{10}}, \frac{-3}{\sqrt{10}} \right\rangle = \frac{1}{4} \cdot \frac{1}{\sqrt{10}} + \frac{1}{4} \cdot \frac{-3}{\sqrt{10}} \).

The result denotes how quickly the function is changing in the specified direction defined by the unit vector. Understanding the dot product's role can deepen comprehension of vector projections in any dimensional space.