Problem 278 - Calculus 3

Question

For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\) $$ f(x, y)=x \arctan (y), \quad \theta=\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{x}{1+y^2} \).

Step 1: Understand the Directional Derivative

The directional derivative of a function at a point in the direction of a vector \(\mathbf{u}\) is given by the dot product of the gradient of the function with the unit vector \(\mathbf{u}\). Mathematically, \( D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \).

Step 2: Find the Gradient of the Function

The given function is \( f(x, y) = x \arctan(y) \). The gradient \( abla f \) is the vector of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 3: Calculate Partial Derivatives

First, calculate \( \frac{\partial f}{\partial x} \), which is \( \arctan(y) \), since the derivative of \( x \) with respect to \( x \) is 1. Then, calculate \( \frac{\partial f}{\partial y} \), which is \( \frac{x}{1+y^2} \) by using the derivative of \( \arctan(y) \), which is \( \frac{1}{1+y^2} \).

Step 4: Form the Gradient Vector

The gradient vector is \( abla f = (\arctan(y), \frac{x}{1+y^2} ) \).

Step 5: Define the Unit Vector

The unit vector in the direction is given by \( \mathbf{u} = \cos \frac{\pi}{2} \mathbf{i} + \sin \frac{\pi}{2} \mathbf{j} \). Since \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \), the unit vector is \( \mathbf{u} = \mathbf{j} = (0, 1) \).

Step 6: Compute the Dot Product

Compute the dot product \( abla f \cdot \mathbf{u} \) which is \( (\arctan(y), \frac{x}{1+y^2}) \cdot (0, 1) = \arctan(y) \cdot 0 + \frac{x}{1+y^2} \cdot 1 = \frac{x}{1+y^2} \).

Step 7: Conclusion for Directional Derivative

The directional derivative of the function \( f(x, y) = x \arctan(y) \) in the direction of the vector \( \mathbf{u} \) where \( \theta = \frac{\pi}{2} \) is \( \frac{x}{1+y^2} \).

Key Concepts

Gradient

The gradient is a vector that points in the direction of the greatest rate of increase of a function at a given point. It is like the compass for the function, telling us where to go uphill the fastest. If you imagine the height of a hill represented by a surface, the gradient gives you the steepest path upwards.

The gradient of a function with two variables, such as \(f(x, y) = x \arctan(y)\), is represented as a vector of partial derivatives. This means we compute the derivative with respect to each variable separately to form a vector. In our example, the gradient \(abla f\) is calculated as:

• \(\frac{\partial f}{\partial x} = \arctan(y)\)
• \(\frac{\partial f}{\partial y} = \frac{x}{1+y^2}\)

So, our gradient vector becomes \((\arctan(y), \frac{x}{1+y^2})\). This vector gives both magnitude and direction of the steepest ascent.

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one of its variables, while keeping other variables constant. It's like looking at slices of a multivariable function and checking how each slice changes when we tweak one ingredient at a time.

For a function \(f(x, y) = x \arctan(y)\), finding the partial derivatives means determining how \(f\) changes as \(x\) or \(y\) changes independently:

• To find \(\frac{\partial f}{\partial x}\), vary \(x\) while holding \(y\) constant, resulting in \(\arctan(y)\).
• To determine \(\frac{\partial f}{\partial y}\), vary \(y\) while \(x\) remains static, using the chain rule or known derivative \(\frac{d}{dy}[\arctan(y)] = \frac{1}{1+y^2}\), giving \(\frac{x}{1+y^2}\).

These derivatives allow us to understand the influence of each individual variable on the overall functional behaviour.

Vector Calculus

Vector calculus is a key mathematical tool for dealing with vector fields and multivariable functions. It includes operations like differentiation and integration over vector domains. In directional derivatives, vector calculus helps us gauge how a scalar function's value changes as we move in a specific direction.

The main operation involved in this context is obtaining the directional derivative, which combines the gradient of the function and the unit vector of the direction we are interested in. The general formula for the directional derivative of a function \( f \) in the direction of a vector \( \mathbf{u} \) is:

• \( D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \)

The dot product here evaluates the component of the gradient in the direction of the unit vector, revealing the rate of change of the function in that direction. This operation reflects how vector calculus ties together spatial directions and functional rates of change.

Unit Vector

A unit vector is a vector that has a length or magnitude of 1. It’s like a pointer telling us direction, but without any inherent 'size' or 'push'. It's like a wind vane pointing to where the wind is blowing without indicating how strong the wind is.

When dealing with directional derivatives, the unit vector specifies the direction in which we wish to measure the rate of change. In an exercise where \( \theta = \frac{\pi}{2} \), the corresponding unit vector is derived as:

• \( \mathbf{u} = \cos \frac{\pi}{2} \mathbf{i} + \sin \frac{\pi}{2} \mathbf{j} \)
• Because \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \), our unit vector becomes \( \mathbf{u} = (0, 1) \)

This unit vector points out purely in the \(y\)-direction, indicating that we are interested in how the function changes with vertical movement while holding horizontal position constant. It simplifies the calculation of directional derivatives by focusing on a one-dimensional direction of analysis.