Problem 263 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=x y, \quad P(0,-2), \quad \mathrm{v}=\frac{1}{2} \mathbf{i}+\frac{\sqrt{3}}{2} \mathbf{j}$$

Step-by-Step Solution

Answer

The directional derivative is -1.

Step 1: Find the Gradient of the Function

The gradient of a function \( f(x, y) \) is given by \( abla f (x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \). For the function \( f(x, y) = xy \), we first find the partial derivatives:\[ \frac{\partial f}{\partial x} = y \quad \text{and} \quad \frac{\partial f}{\partial y} = x. \ \text{Thus,} \quad abla f(x, y) = (y, x). \]

Step 2: Evaluate the Gradient at Point P

Substitute the values of point \( P(0, -2) \) into the gradient \( abla f(x, y) = (y, x) \).\[abla f(0, -2) = (-2, 0).\]

Step 3: Normalize the Direction Vector

The direction vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \) must be a unit vector. First, check its magnitude:\[\left| \mathbf{v} \right| = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1.\]As the magnitude is 1, \( \mathbf{v} \) is already a unit vector.

Step 4: Compute the Directional Derivative

The directional derivative of \( f \) at the point \( P \) in the direction \( \mathbf{v} \) is given by \( D_{\mathbf{v}}f = abla f \cdot \mathbf{u} \), where \( \mathbf{u} \) is the unit vector in the direction of \( \mathbf{v} \). We have \( abla f(0, -2) = (-2, 0) \) and \( \mathbf{u} = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \), hence:\[D_{\mathbf{v}}f = (-2, 0) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) = -2 \cdot \frac{1}{2} + 0 \cdot \frac{\sqrt{3}}{2} = -1.\]

Key Concepts

Gradient

The gradient is a vector that shows the direction of the greatest rate of increase of a function. For a function of two variables, like our example function \( f(x, y) = xy \), the gradient is represented as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Here, the components are partial derivatives, meaning they represent the rate of change of the function in relation to each individual variable.
In our exercise, finding the gradient involves computing the partial derivatives:

• \( \frac{\partial f}{\partial x} = y \)
• \( \frac{\partial f}{\partial y} = x \)

Thus, the gradient becomes \( (y, x) \). This gradient tells us how steeply the function is climbing or falling along each axis.
Once we evaluated the gradient at point \( P(0, -2) \), we found it to be \( (-2, 0) \), directing us towards the steepest descent at that point.

Partial Derivatives

Partial derivatives are a foundational concept in multivariable calculus that measure how a function changes as one of the variables changes, while keeping other variables constant. They're essential when dealing with functions of multiple variables and give insight into the behavior of the function along each axis.
To compute partial derivatives, treat the variable of interest as variable and others as constants.
In our example, for the function \( f(x, y) = xy \), we want to find how \( f \) changes with respect to each variable individually:

• With respect to \( x \): \( \frac{\partial f}{\partial x} = y \)
• With respect to \( y \): \( \frac{\partial f}{\partial y} = x \)

This approach lets us understand the influence of each variable on the function, which is crucial for forming the gradient.

Unit Vector

A unit vector is a vector with a magnitude (length) of 1. It's used to specify a direction without considering the magnitude. In directional derivative calculations, ensuring the directional vector is a unit vector is crucial, as it helps in determining the rate of change strictly along that direction.
In the provided exercise, the directional vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \) already has a magnitude of 1, which makes it a unit vector. We can confirm this by calculating its length:

• Magnitude: \( \left| \mathbf{v} \right| = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = 1 \)

This step is vital as using a non-unit vector could scale the directional derivative improperly.

Directional Vector

A directional vector defines the direction along which we want to find the rate of change of the function, i.e., the directional derivative. Typically, this vector must be a unit vector to give a true measure of the function's change rate in that specific direction.
For our example, the directional vector is initially given as \( \mathbf{v} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \). After confirming it's a unit vector, it becomes \( \mathbf{u} = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \).

• Direction vectors are useful to locate the maximal change in function value at a given point, like \( P(0, -2) \) in the exercise.

This reveals how the function behaves in specific directions, which can be critical in fields like physics for understanding force directions or in data science for multi-dimensional scaling.