Problem 13 - Advanced Engineering Mathematics

Question

In Problems, 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 \(-\frac{\sqrt{10}}{20}\).

Step 1: Calculate the Gradient of the Function

The gradient of a function \( f(x, y) \) is the vector \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For the function \( f(x, y) = \tan^{-1} \left( \frac{y}{x} \right) \), we need to find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

Step 2: Compute Partial Derivatives

For \( \frac{\partial f}{\partial x} \), use the derivative of \( \tan^{-1} u \), where \( u = \frac{y}{x} \). Using the chain rule, \( \frac{\partial f}{\partial x} = \frac{-y}{x^2 + y^2} \). For \( \frac{\partial f}{\partial y} \), similarly, we find \( \frac{x}{x^2 + y^2} \). Hence, the gradient is \( abla f = \left( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right) \).

Step 3: Evaluate the Gradient at the Given Point

Substitute the point \((2, -2)\) into \( abla f(x, y) \). This gives \( abla f(2, -2) = \left( \frac{-(-2)}{4 + 4}, \frac{2}{4 + 4} \right) = \left( \frac{2}{8}, \frac{2}{8} \right) = \left( \frac{1}{4}, \frac{1}{4} \right) \).

Step 4: Normalize the Direction Vector

The direction vector is \( \mathbf{i} - 3\mathbf{j} = (1, -3) \). Normalize it to get a unit vector: \( \|\mathbf{v}\| = \sqrt{1^2 + (-3)^2} = \sqrt{10} \). Thus, the unit direction vector is \( \frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \).

Step 5: Calculate the Directional Derivative

The directional derivative of \( f \) in the direction of the unit vector \( \mathbf{u} \) at point \( \mathbf{a} \) is given by \( D_{\mathbf{u}} f(\mathbf{a}) = abla f(\mathbf{a}) \cdot \mathbf{u} \). Compute: \[ D_{\mathbf{u}} f = \left( \frac{1}{4}, \frac{1}{4} \right) \cdot \left( \frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \right) = \frac{1}{4} \frac{1}{\sqrt{10}} + \frac{1}{4}\left(-\frac{3}{\sqrt{10}}\right) = \frac{1}{4\sqrt{10}} - \frac{3}{4\sqrt{10}} = -\frac{2}{4\sqrt{10}} = -\frac{1}{2\sqrt{10}} \].

Step 6: Simplify the Final Result

Simplify \( -\frac{1}{2\sqrt{10}} \) to form: \( -\frac{\sqrt{10}}{20} \) (by rationalizing the denominator).

Key Concepts

Gradient of a Function

The gradient of a function is a fundamental concept in calculus. It gives us a vector that points in the direction of the greatest rate of change of the function. For a function of two variables, such as \( f(x, y) \), the gradient \( abla f(x, y) \) is composed of two partial derivatives: \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

The gradient is often written as a vector:

• \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)

In our example, the function given is \( f(x, y) = \tan^{-1} \left( \frac{y}{x} \right) \). The gradient helps us understand how the function changes around any given point. Calculating it involves finding these partial derivatives and checking their values at specific points of interest.

Partial Derivatives

Partial derivatives are like regular derivatives but applied to functions with more than one variable. When we take a partial derivative, we are "holding other variables constant" and looking at how the function changes as we change just one variable.

For our function \( f(x,y) = \tan^{-1} \left( \frac{y}{x} \right) \), we need to find:

• The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \)
• The partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \)

Using the chain rule, these become:

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

These partial derivatives are the building blocks of our gradient, showing how the function changes in each direction across the \(x\) and \(y\) axes.

Chain Rule

The chain rule is a crucial technique in calculus, especially when dealing with composite functions. It allows us to compute the derivative of a function that depends on another function. It may seem a bit abstract, but it's essentially about tracing how changes in one variable lead to changes in another, ultimately affecting the outcome we care about.

In the context of our function \( f(x, y) = \tan^{-1} \left( \frac{y}{x} \right) \), the chain rule helps break down the derivatives into manageable parts. By understanding how \( \tan^{-1} \) behaves as a function of \( u = \frac{y}{x} \), and how \( u \) itself changes with respect to \( x \) and \( y \), we can systematically find each partial derivative. The basic idea is:

• Take the derivative of the outer function (\( u \) here) with respect to its inside part.
• Multiply it by the derivative of that inside part with respect to the variable of interest.

These steps combined help calculate the required partial derivatives effectively.

Normalizing a Vector

Normalizing a vector makes it a unit vector, meaning it has a length of 1. This process is important in directional derivatives because all we care about is direction, not magnitude. If a vector is pointing the way, its size should not influence our calculation.

For any vector, say \( \mathbf{v} = (a, b) \), its magnitude (or length) is given by \( \sqrt{a^2 + b^2} \). To normalize this vector:

• Divide each component of the vector by its magnitude.
• This gives us \( \mathbf{u} = \left( \frac{a}{\|\mathbf{v}\|}, \frac{b}{\|\mathbf{v}\|} \right) \)

In our problem, the direction vector is \( \mathbf{i} - 3\mathbf{j} = (1, -3) \). Its magnitude is \( \sqrt{1^2 + (-3)^2} = \sqrt{10} \).

Therefore, the normalized, or unit vector, becomes \( \mathbf{u} = \left( \frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \right) \). This vector maintains direction but has a standard length, perfect for calculating the directional derivative.