Problem 11 - Advanced Engineering Mathematics

Question

In Problems, find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=5 x^{3} y^{6} ;(-1,1), \theta=\pi / 6 $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{15\sqrt{3}}{2} - 15 \).

Step 1: Understanding the Directional Derivative

The directional derivative of a function at a given point measures the rate at which the function changes at that point as you move in a specific direction. It's a generalization of the notion of a partial derivative.

Step 2: Calculate the Gradient

The gradient of the function \( f(x, y) = 5x^3y^6 \) is given by the vector of partial derivatives: \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right). \] First, find \( \frac{\partial f}{\partial x} \):\[ \frac{\partial f}{\partial x} = 15x^2y^6. \]Next, find \( \frac{\partial f}{\partial y} \):\[ \frac{\partial f}{\partial y} = 30x^3y^5. \] So, \( abla f(x, y) = (15x^2y^6, 30x^3y^5). \)

Step 3: Evaluate the Gradient at the Given Point

Substitute \((-1, 1)\) into the gradient:\[ abla f(-1, 1) = (15(-1)^2(1)^6, 30(-1)^3(1)^5). \] Calculate each component:\[ 15(-1)^2(1)^6 = 15, \]\[ 30(-1)^3(1)^5 = -30. \]So, \( abla f(-1, 1) = (15, -30). \)

Step 4: Determine the Unit Vector in the Given Direction

The direction is given by \( \theta = \pi/6 \). The unit vector \( \mathbf{u} \) in this direction is \[ \mathbf{u} = (\cos(\pi/6), \sin(\pi/6)). \] Calculate each component:\[ \cos(\pi/6) = \frac{\sqrt{3}}{2}, \] \[ \sin(\pi/6) = \frac{1}{2}. \] Thus, \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right). \)

Step 5: Compute the Directional Derivative

The directional derivative \( D_\mathbf{u} f \) is the dot product of the gradient and the unit vector:\[ D_\mathbf{u} f = abla f(-1, 1) \cdot \mathbf{u}. \] Compute this dot product:\[ D_\mathbf{u} f = (15, -30) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \]\[ = 15\left( \frac{\sqrt{3}}{2} \right) + (-30)\left( \frac{1}{2} \right), \]\[ = \frac{15\sqrt{3}}{2} - 15. \] Hence, the directional derivative at the point in the given direction is \( \frac{15\sqrt{3}}{2} - 15 \).

Key Concepts

Gradient Vector

When dealing with functions of multiple variables, the gradient vector is your guiding tool. It points in the direction of the greatest rate of increase of the function. Think of it as a multi-dimensional generalization of the derivative that provides both direction and magnitude. The gradient vector is composed of partial derivatives for each variable involved.

For a function like \(f(x, y)\), the gradient is denoted as \( abla f(x, y) \), represented by \(\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\).
In simpler terms, *each component of the gradient is the rate of change of the function with respect to that variable.*

Here's how it connects to the original exercise:

• For \( f(x, y) = 5x^3y^6 \), the partial derivatives are calculated as \( 15x^2y^6 \) and \( 30x^3y^5 \) for \(x\) and \(y\) respectively.
• This gives the gradient vector as \( (15x^2y^6, 30x^3y^5) \).

The gradient forms the cornerstone for finding the directional derivative, which tells us how the function changes as we move in a specific direction.

Partial Derivatives

Partial derivatives are fundamental when working with functions of several variables. They measure how a function changes as only one variable is varied, while all other variables are held constant. They extend the basic idea of a derivative to functions with more than one input.

To calculate the partial derivative of a function with respect to, say, \(x\), you treat \(y\) as a constant, and vice versa.

Let's apply this to the given function:

• The partial derivative with respect to \(x\), \( \frac{\partial f}{\partial x} \), is derived by treating \(y\) as a constant. Hence, for \( f(x, y) = 5x^3y^6 \), it becomes \( 15x^2y^6 \).
• Similarly, the partial derivative with respect to \(y\), \( \frac{\partial f}{\partial y} \), treats \(x\) as a constant, resulting in \( 30x^3y^5 \).

Partial derivatives form each component of the gradient vector, making them crucial for understanding how changes in inputs affect the output of a function.

Unit Vector

To find the directional derivative, knowing the direction in which you are evaluating the derivative is critical. This is where unit vectors come into play. A unit vector has a magnitude of one and points in the specific direction you are interested in.

In the context of the exercise, the direction is given by an angle \( \theta \). The unit vector \( \mathbf{u} \) for this direction is found using trigonometric functions:

• The cosine of \( \theta \) provides the \(x\)-component, \( \cos(\pi/6) = \frac{\sqrt{3}}{2} \).
• The sine of \( \theta \) gives the \(y\)-component, \( \sin(\pi/6) = \frac{1}{2} \).

Combined, these form the unit vector \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \). The unit vector ensures that only the direction is considered without altering the magnitude when calculating the directional derivative.

Dot Product

Understanding the dot product is crucial for calculating the directional derivative. The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.

In our context, the dot product measures the projection of one vector onto another. When you take the dot product of the gradient vector and the unit vector, you determine how much the function's value increases in the direction of the unit vector.

Given the gradient \( abla f(-1, 1) = (15, -30) \) and the unit vector \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \), the dot product is calculated as follows:

• Multiply the \(x\)-components: \(15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2}\).
• Multiply the \(y\)-components: \(-30 \times \frac{1}{2} = -15 \).
• Now, add these products: \( \frac{15\sqrt{3}}{2} - 15 \).

The result of this operation gives the directional derivative, highlighting how the function changes as you move in the given direction.