Chapter 4: Problem 412
Find the directional derivative of \(f(x, y)=x^{2}+6 x y-y^{2}\) in the direction \(v=\mathbf{i}+4 \mathbf{j}\).
Short Answer
Expert verified
The directional derivative is \( \frac{1}{\sqrt{17}}(26x - 2y) \).
Step by step solution
01
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 \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2x + 6y, 6x - 2y). \]
02
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:\[ \mathbf{u} = \left( \frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right). \]
03
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 \(\mathbf{u} = \left( \frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right)\):\[ D_{\mathbf{u}}f(x, y) = abla f \cdot \mathbf{u} = (2x + 6y)\left(\frac{1}{\sqrt{17}}\right) + (6x - 2y)\left(\frac{4}{\sqrt{17}}\right). \]
04
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:\[ D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(2x + 6y + 24x - 8y) = \frac{1}{\sqrt{17}}(26x - 2y). \]
05
Write the Final Answer
Therefore, the directional derivative of \(f(x, y)\) in the direction of \(\mathbf{v} = \mathbf{i} + 4\mathbf{j}\) is:\[ D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(26x - 2y). \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( y \): \( \frac{\partial f}{\partial y} = 6x - 2y \)
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:
- \( \frac{\partial f}{\partial x} = 2x + 6y \)
- \( \frac{\partial f}{\partial y} = 6x - 2y \)
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: \( ||\textbf{v}|| = \sqrt{1^2 + 4^2} = \sqrt{17} \)
- Unit vector \( \textbf{u} = \left( \frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right) \)
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:
- \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)
- The dot product here gives \( D_{\mathbf{u}}f(x, y) = \frac{1}{\sqrt{17}}(2x + 6y) + \frac{4}{\sqrt{17}}(6x - 2y) \).