Problem 263 - Calculus 3

Question

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

Step-by-Step Solution

Answer

The directional derivative is -1.

Step 1: Understand the Problem

We need to find the directional derivative of the function \( f(x, y) = xy \) at the point \( P(0, -2) \) in the direction of the vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \). To do this, we will calculate the gradient of \( f \) and then take the dot product with the unit direction vector \( \mathbf{v} \).

Step 2: Compute the Gradient

The gradient of the function \( f(x, y) = xy \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate the partial derivatives:1. \( \frac{\partial f}{\partial x} = y \) 2. \( \frac{\partial f}{\partial y} = x \)Thus, \( abla f = (y, x) \).

Step 3: Evaluate the Gradient at Point P

Substitute the point \( P(0, -2) \) into the gradient:- At \( P(0, -2) \), \( abla f = (-2, 0) \).

Step 4: Confirm the Direction Vector is a Unit Vector

The vector \( \mathbf{v} \) is given as \( \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \) which is a unit vector because its magnitude is 1. Verify:\[ \text{Magnitude} = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 } = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

Step 5: Calculate the Directional Derivative

The directional derivative \( D_{\mathbf{v}} f \) is calculated using the dot product of the gradient \( abla f \) and the unit vector \( \mathbf{v} \):\[ D_{\mathbf{v}} f = abla f \cdot \mathbf{v} = (-2, 0) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \]Calculate the dot product:\[ D_{\mathbf{v}} f = (-2) \cdot \frac{1}{2} + 0 \cdot \frac{\sqrt{3}}{2} = -1 \]

Step 6: Conclusion

The directional derivative of the function \( f(x, y) = xy \) at point \( P(0, -2) \) in the direction of vector \( \mathbf{v} \) is \(-1\).

Key Concepts

Gradient

To understand the concept of gradient, envision a surface in a 3D space that represents a function of two variables, like our function \(f(x, y) = xy\). The gradient helps us find the slope of the surface at any point.
The gradient is a vector that consists of partial derivatives with respect to each variable. In formal terms, for the function \(f\), the gradient is written as \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\).
In our example, the gradient at point \(P(0, -2)\) is calculated from the partial derivatives:

• \(\frac{\partial f}{\partial x} = y\), which becomes \(-2\) at \(P\).
• \(\frac{\partial f}{\partial y} = x\), which becomes \(0\) at \(P\).

This results in the gradient vector \((y, x) = (-2, 0)\) at the point \(P\). The gradient's direction gives us the direction of the steepest ascent of the function, similar to following the path uphill.

Partial Derivative

Partial derivatives are the building blocks of the gradient vector. They measure how a function changes as we vary one variable while holding the other constant.
For the function \(f(x, y) = xy\), we differentiate with respect to \(x\) and \(y\) individually to obtain the partial derivatives.

• To find \(\frac{\partial f}{\partial x}\), treat \(y\) as a constant, resulting in \(y\). For example, for \(f(x, y) = xy\), \(\frac{\partial f}{\partial x} = y\).
• To find \(\frac{\partial f}{\partial y}\), treat \(x\) as a constant, resulting in \(x\). So, \(\frac{\partial f}{\partial y} = x\).

Each partial derivative captures the rate of change of the function along one axis, making them essential for constructing the gradient.

Dot Product

The dot product is an operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number. It is widely used to find the angle between vectors or to project one vector onto another.
In the context of directional derivatives, the dot product of the gradient vector and a unit vector gives us the directional derivative. It measures the rate at which the function changes in the direction given by the unit vector.
In our example, to find the directional derivative, we calculate the dot product \(abla f \cdot \mathbf{v}\), where \(abla f\) is the gradient \((-2, 0)\), and \(\mathbf{v}\) is the direction vector \(\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\).
The dot product computation is:

• \((-2) \times \frac{1}{2} = -1\)
• \(0 \times \frac{\sqrt{3}}{2} = 0\)

Thus, the result is \(-1\). This indicates how much and in what direction the function increases as we move along \(\mathbf{v}\).

Unit Vector

A unit vector is a vector with a magnitude of 1. It is used to specify a direction but not a magnitude, making it a perfect tool when studying directional derivatives.
In any practical problem, when someone refers to a "direction" in terms of a vector, they're often talking about a unit vector.
To confirm that \(\mathbf{v}\) is a unit vector, compute its magnitude, which should equal 1. For \(\mathbf{v} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j}\), calculate:

• Magnitude \(= \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 } = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\)

This confirms that \(\mathbf{v}\) is indeed a unit vector. Using a unit vector is crucial as it ensures that we're only focusing on direction, not scaling the derivative by any length.