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Let \(f(x, y)=x^{2}-x y+y^{2}-y .\) Find the directions \(\mathbf{u}\) and the values of \(D_{\mathbf{n}}, f(1,-1)\) for which a. \(D_{\mathrm{u}} f(1,-1)\) is largest b. \(D_{\mathrm{u}} f(1,-1)\) is smallest c. \(D_{\mathrm{u}} f(1,-1)=0\) d. \(D_{\mathrm{u}} f(1,-1)=4\) e. \(D_{\mathrm{u}} f(1,-1)=-3\)

Short Answer

Expert verified
a. 5 in direction (3/5, -4/5); b. -5 in direction (-3/5, 4/5); c. (4/5, 3/5); d. (12/25, -16/25); e. (-9/25, 12/25).

Step by step solution

01

Find Gradient of the Function

The gradient vector of a function \( f(x, y) \) is computed by taking the partial derivatives with respect to \( x \) and \( y \). For \( f(x, y) = x^2 - xy + y^2 - y \), \[ f_x = \frac{\partial}{\partial x}(x^2 - xy + y^2 - y) = 2x - y, \]\[ f_y = \frac{\partial}{\partial y}(x^2 - xy + y^2 - y) = -x + 2y - 1. \]Thus, the gradient vector is \( abla f(x, y) = (2x - y, -x + 2y - 1) \).
02

Evaluate Gradient at Point (1, -1)

Substitute \( x = 1 \) and \( y = -1 \) into the gradient vector:\[ abla f(1, -1) = (2(1) - (-1), -(1) + 2(-1) - 1) = (3, -4). \]
03

Determine Unit Vector Direction

In general, if \( D_{\mathrm{u}} f(1, -1) \) achieves its largest value, the direction \( \mathbf{u} \) should be in the direction of the gradient vector itself. The unit vector \( \mathbf{u} \) is \[ \mathbf{u} = \frac{1}{\|abla f(1, -1)\|} abla f(1, -1), \]where \( \|abla f(1, -1)\| = \sqrt{3^2 + (-4)^2} = 5 \). Therefore, \[ \mathbf{u} = \left( \frac{3}{5}, -\frac{4}{5} \right). \]
04

Largest and Smallest Directional Derivative

The largest directional derivative is the magnitude of the gradient: \[ D_{\mathrm{u}}f(1, -1) = \|abla f(1, -1)\| = 5. \]The smallest directional derivative, occurring in the exact opposite direction of \( abla f \), is:\[ D_{\mathrm{-u}}f(1, -1) = -5. \]
05

Directional Derivative Equal to 0

For \( D_{\mathrm{u}} f(1, -1) = 0 \), \( \mathbf{u} \) must be orthogonal to \( abla f(1, -1) \). Thus, \( 3u_1 - 4u_2 = 0 \), a direction vector could be \( (4, 3) \). The unit vector is then:\[ \mathbf{u} = \left( \frac{4}{5}, \frac{3}{5} \right). \]
06

Directional Derivative Equal to 4

We need \( \mathbf{u} \cdot abla f(1, -1) = 5k = 4 \), so \( k = \frac{4}{5} \). The direction \( \mathbf{u} = \left(\frac{3}{5} \cdot \frac{4}{5}, -\frac{4}{5} \cdot \frac{4}{5}\right) = \left( \frac{12}{25}, -\frac{16}{25} \right). \)
07

Directional Derivative Equal to -3

For \( D_{\mathrm{u}} f(1, -1) = -3 \), we solve \( 5k = -3 \), obtaining \( k = -\frac{3}{5} \). The direction vector becomes \( \mathbf{u} = \left(\frac{3}{5} \cdot -\frac{3}{5}, -\frac{4}{5} \cdot -\frac{3}{5}\right) = \left( -\frac{9}{25}, \frac{12}{25} \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient Vector
The gradient vector is a powerful tool used in multivariable calculus. It's essentially a vector that stores all the partial derivative information of a function.

For a function of two variables, like our example with function \( f(x, y) = x^2 - xy + y^2 - y \), the gradient vector \( abla f(x, y) \) is expressed as \( (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \). Each component of this vector represents the rate of change of the function in the respective dimension.

When you evaluate this gradient at a specific point, such as (1, -1) in our problem, you find the vector \( abla f(1, -1) \). This specific vector indicates the direction of steepest ascent of the function at that particular point.
Partial Derivatives
Partial derivatives are used to understand how a function changes as one of its input variables varies, while the others are held constant.

In our function \( f(x, y) \), the partial derivative with respect to \( x \) is found by treating \( y \) as a constant and differentiating the equation just like you would a single-variable function. This gives us \( f_x = 2x - y \).

Similarly, to find the partial derivative with respect to \( y \), we treat \( x \) as a constant. This yields \( f_y = -x + 2y - 1 \).

The partial derivatives help us construct the gradient vector, indicating how the function behaves in various directions.
Unit Vector
A unit vector is a vector that has a magnitude of one and points in the same direction as the original vector, essentially "normalizing" it.

To find a unit vector, you divide the vector by its magnitude. In our case, with the gradient vector at (1, -1) given by \( (3, -4) \), its magnitude is calculated as \( \sqrt{3^2 + (-4)^2} = 5 \).

The unit vector \( \mathbf{u} \) is calculated as \( \left( \frac{3}{5}, -\frac{4}{5} \right) \). This unit vector retains the direction of the gradient vector but scales it down to a length of one.
Directional Derivative Calculation
The directional derivative represents the rate at which the function changes as one moves in a specified direction, quantified by a unit vector.

It is computed as the dot product of the gradient vector and the unit vector: \( D_{\mathbf{u}} f(x, y) = abla f(x, y) \cdot \mathbf{u} \).

At the point (1, -1), the maximum directional derivative is equal to the magnitude of the gradient vector, which is 5. This happens when \( \mathbf{u} \) points in the same direction as \( abla f(1, -1) \). Conversely, the smallest (most negative) directional derivative, -5, occurs in the opposite direction.

To achieve specific directional derivative values like 4 or -3, the vector \( \mathbf{u} \) must be scaled to maintain orthogonality, ensuring the desired directional sensitivity when projected onto the gradient vector.

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