Problem 412 - Calculus Volume 3

Question

Find the directional derivative of \(f(x, y)=x^{2}+6 x y-y^{2}\) in the direction \(v=\mathbf{i}+4 \mathbf{j}\).

Step-by-Step Solution

Answer

The directional derivative is \( \frac{1}{\sqrt{17}}(26x - 2y) \).

Step 1: Compute the Gradient

The gradient of the function, \(f(x, y) = x^2 + 6xy - y^2\), is found by taking the partial derivatives of \(f\) with respect to \(x\) and \(y\). So, we have \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2x + 6y, 6x - 2y). \]

Step 2: Normalize the Direction Vector

The direction vector given is \(\mathbf{v} = \mathbf{i} + 4\mathbf{j}\). First, we need to normalize this vector to ensure it has a unit length. Calculate the magnitude of \(\mathbf{v}\):\[ ||\mathbf{v}|| = \sqrt{1^2 + 4^2} = \sqrt{17}. \]The unit vector \(\mathbf{u}\) in the direction of \(\mathbf{v}\) is:\[ \mathbf{u} = \left( \frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right). \]

Step 3: Compute the Dot Product

To find the directional derivative, compute the dot product of the gradient \(abla f(x, y) = (2x + 6y, 6x - 2y)\) and the unit vector \(\mathbf{u} = \left( \frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right)\):\[ D_{\mathbf{u}}f(x, y) = abla f \cdot \mathbf{u} = (2x + 6y)\left(\frac{1}{\sqrt{17}}\right) + (6x - 2y)\left(\frac{4}{\sqrt{17}}\right). \]

Step 4: Simplify the Expression

Simplify the expression for the directional derivative:\[ D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(2x + 6y) + \frac{4}{\sqrt{17}}(6x - 2y). \]Combine the terms to get:\[ D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(2x + 6y + 24x - 8y) = \frac{1}{\sqrt{17}}(26x - 2y). \]

Step 5: Write the Final Answer

Therefore, the directional derivative of \(f(x, y)\) in the direction of \(\mathbf{v} = \mathbf{i} + 4\mathbf{j}\) is:\[ D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(26x - 2y). \]

Key Concepts

Gradient

A gradient is a vector summarizing the rate of change of a function. It points in the direction of the steepest ascent. For the function \( f(x, y) = x^2 + 6xy - y^2 \), the gradient is found by calculating the partial derivatives with respect to each variable. This gives us:

• Partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 2x + 6y \)
• Partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = 6x - 2y \)

Together, these form the gradient vector \( abla f(x, y) = (2x + 6y, 6x - 2y) \). This combination tells us how the function changes as \( x \) and \( y \) change. In real-world terms, it helps in understanding depictions like the slope of a terrain.

Partial Derivatives

Partial derivatives are like regular derivatives but focus on one variable at a time while treating others as constants. They represent how a function changes as just one variable is altered. For the function \( f(x,y) = x^2 + 6xy - y^2 \), we have two partial derivatives:

• \( \frac{\partial f}{\partial x} = 2x + 6y \)
• \( \frac{\partial f}{\partial y} = 6x - 2y \)

Each partial derivative shows the rate of change of \( f \) with respect to \( x \) and \( y \) respectively. When solving problems, partial derivatives become crucial in constructing the gradient and analyzing directions of change.

Normalization

Normalization is the process of adjusting a vector so that its magnitude becomes 1 while maintaining its direction. This is essential to work with unit vectors, which simplify mathematical operations in calculus. For a vector \( \textbf{v} = \textbf{i} + 4 \textbf{j} \), we determine its magnitude first:

• Magnitude: \( ||\textbf{v}|| = \sqrt{1^2 + 4^2} = \sqrt{17} \)

Then, create a unit vector by dividing each component by the magnitude:

• Unit vector \( \textbf{u} = \left( \frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right) \)

Normalization helps in ensuring computations like the dot product result in meaningful values, such as finding directional derivatives.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers and returns a single number. This operation is crucial for finding directional derivatives. For vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is:

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

In the context of directional derivatives, you need the dot product of the gradient vector \( abla f(x, y) = (2x + 6y, 6x - 2y) \) and the normalized direction vector \( \mathbf{u} \) to find how rapidly the function changes in a specified direction:

• The dot product here gives \( D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(2x + 6y) + \frac{4}{\sqrt{17}}(6x - 2y) \).

The result is the scalar value representing the directional derivative, which indicates the rate at which the function \( f \) changes at a point \( (x, y) \) in the direction \( \mathbf{u} \).