Question

Bestimmen Sie die Richtungsableitung der Funktion \(f(x, y)=\frac{x^{2}}{y^{2}}\) am Punkt \(\mathbf{x}=(1,1)^{T}\) in Richtung von \(\mathbf{a}=(1,2)^{T}\).

Short answer

The directional derivative is \(-\frac{2}{\sqrt{5}}\).

Step-by-step solution

Calculate the Gradient of the Function

To determine the directional derivative, we first calculate the gradient of the function \( f(x, y) = \frac{x^2}{y^2} \). The gradient is given by the vector of partial derivatives. First, calculate the partial derivative with respect to \( x \):\[ \frac{\partial f}{\partial x} = \frac{2x}{y^2}. \]Next, calculate the partial derivative with respect to \( y \):\[ \frac{\partial f}{\partial y} = -\frac{2x^2}{y^3}. \]Thus, the gradient is:\[ abla f(x, y) = \left( \frac{2x}{y^2}, -\frac{2x^2}{y^3} \right). \]

Evaluate the Gradient at the Given Point

Substitute the point \( \mathbf{x} = (1,1) \) into the gradient \( abla f(x, y) = \left( \frac{2x}{y^2}, -\frac{2x^2}{y^3} \right) \):\[ abla f(1, 1) = \left( \frac{2 \cdot 1}{1^2}, -\frac{2 \cdot 1^2}{1^3} \right) = (2, -2). \]

Normalize the Direction Vector

The direction vector given is \( \mathbf{a} = (1, 2) \). To normalize this vector, divide it by its magnitude:\[ \| \mathbf{a} \| = \sqrt{1^2 + 2^2} = \sqrt{5}. \]Thus, the normalized direction vector is:\[ \mathbf{u} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right). \]

Calculate the Directional Derivative

The directional derivative of \( f \) in the direction of \( \mathbf{u} \) at the point \( (1, 1) \) can be calculated using the dot product of \( abla f(1, 1) \) and the normalized direction vector \( \mathbf{u} \):\[ D_{\mathbf{u}} f(1, 1) = abla f(1, 1) \cdot \mathbf{u} = (2, -2) \cdot \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right). \]Calculate the dot product:\[ D_{\mathbf{u}} f(1, 1) = 2 \cdot \frac{1}{\sqrt{5}} + (-2) \cdot \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}} - \frac{4}{\sqrt{5}} = -\frac{2}{\sqrt{5}}. \]

Key concepts

Gradient

The gradient is a fundamental concept in vector calculus. It provides a way to discern how a function changes as one moves through space. When dealing with a function of two variables, for example, \( f(x, y) \), the gradient is a vector with two components: the partial derivative of \( f \) with respect to \( x \) and the partial derivative of \( f \) with respect to \( y \).

• The gradient vector is represented as \( abla f(x, y) \).
• It points in the direction of the greatest rate of increase of the function.
• The magnitude of the gradient gives the rate of increase in that direction.

To calculate the gradient of the function \( f(x, y) = \frac{x^2}{y^2} \), we use its partial derivatives:

• \( \frac{\partial f}{\partial x} = \frac{2x}{y^2} \)
• \( \frac{\partial f}{\partial y} = -\frac{2x^2}{y^3} \)

The gradient at any point \( (x, y) \) is hence expressed by vector \( \left( \frac{2x}{y^2}, -\frac{2x^2}{y^3} \right) \). Evaluating this at point \( (1, 1) \), we conclude the gradient is \( (2, -2) \). This vector indicates how \( f \) increases most steeply from the point \( (1, 1) \).

Partial Derivative

Partial derivatives are essential for understanding how a multivariable function changes when one variable changes, keeping the others fixed.

• For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), tells us how \( f \) changes as \( x \) varies, while \( y \) is held constant.
• Similarly, \( \frac{\partial f}{\partial y} \) reveals how the function shifts as \( y \) changes, with \( x \) fixed.

In our exercise, we have \( f(x, y) = \frac{x^2}{y^2} \). Let's break down its partial derivatives:

• Computing \( \frac{\partial f}{\partial x} \) by treating \( y \) as a constant yields \( \frac{2x}{y^2} \).
• The partial derivative \( \frac{\partial f}{\partial y} \) is more involved; it requires applying the chain rule, resulting in \( -\frac{2x^2}{y^3} \).

These results form the components of the gradient vector, helping us determine the rate and direction of change of \( f \) with respect to its variables.

Normalization of Vector

Normalization is a process of adjusting a vector to make its length or magnitude equal to one, without changing its direction. This is crucial when we need a direction vector in calculations, such as finding a directional derivative.

• The original direction vector in the step-by-step solution is \( \mathbf{a} = (1, 2) \).
• To normalize it, find the magnitude of \( \mathbf{a} \) using the formula \( \| \mathbf{a} \| = \sqrt{1^2 + 2^2} = \sqrt{5} \).

Divide each component of \( \mathbf{a} \) by this magnitude to create a new vector \( \mathbf{u} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \).

• \( \mathbf{u} \) now has a magnitude of 1 and retains the direction of \( \mathbf{a} \).

This normalization is important when calculating the directional derivative, ensuring that the effect of direction does not increase the derivative's magnitude beyond what the direction inherently suggests. Simply put, normalization helps differentiate between influence due to direction and that due to the vector's length.