Problem 15 - Advanced Engineering Mathematics

Question

Find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=(x y+1)^{2} ;(3,2) \text {, in the direction of }(5,3) $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{266}{\sqrt{34}} \).

Step 1: Differentiate Partial Derivatives

First, we need to find the gradient of the function \( f(x, y) = (xy + 1)^2 \). This requires computing the partial derivatives of \( f \) with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is obtained using the chain rule: \[ f_x = \frac{\partial}{\partial x}((xy + 1)^2) = 2(xy + 1) \cdot y. \]Similarly, the partial derivative with respect to \( y \) is:\[ f_y = \frac{\partial}{\partial y}((xy + 1)^2) = 2(xy + 1) \cdot x. \]

Step 2: Evaluate Gradient at Given Point

Next, evaluate the gradient \( abla f = (f_x, f_y) \) at the point \((3, 2)\). Substitute \( x = 3 \) and \( y = 2 \) into the partial derivatives:\[ f_x(3, 2) = 2((3)(2) + 1) \cdot 2 = 28, \]\[ f_y(3, 2) = 2((3)(2) + 1) \cdot 3 = 42. \]The gradient at the point is \( abla f(3, 2) = (28, 42) \).

Step 3: Normalize the Direction Vector

The direction vector given is \((5, 3)\). To use it in finding the directional derivative, it must be a unit vector. Find the magnitude of the vector:\[ |(5, 3)| = \sqrt{5^2 + 3^2} = \sqrt{34}. \]Normalize the vector by dividing each component by the magnitude:\[ \mathbf{u} = \left( \frac{5}{\sqrt{34}}, \frac{3}{\sqrt{34}} \right). \]

Step 4: Compute the Directional Derivative

Finally, compute the directional derivative using the dot product of the gradient and the unit direction vector:\[ D_{\mathbf{u}}f(3, 2) = abla f(3, 2) \cdot \mathbf{u}. \]Perform the dot product calculation:\[ D_{\mathbf{u}}f(3, 2) = (28, 42) \cdot \left( \frac{5}{\sqrt{34}}, \frac{3}{\sqrt{34}} \right) = \frac{28 \cdot 5 + 42 \cdot 3}{\sqrt{34}} = \frac{140 + 126}{\sqrt{34}} = \frac{266}{\sqrt{34}}. \]

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, such as \( f(x, y) = (xy + 1)^2 \), partial derivatives allow us to see how the function changes as each variable changes on its own. It's like zooming in to see the effect of one variable while holding the others constant.

For example, to find the partial derivative with respect to \( x \), denoted \( f_x \), you only focus on how changes in \( x \) affect the function, treating \( y \) as a constant. Using the chain rule, we get:

• \( f_x = \frac{\partial}{\partial x}((xy + 1)^2) = 2(xy + 1) \cdot y \)
• \( f_y = \frac{\partial}{\partial y}((xy + 1)^2) = 2(xy + 1) \cdot x \)

These derivatives are fundamental for calculating the gradient vector, which we will discuss next.

Gradient Vector

The gradient vector of a function of two variables is a vector that combines all of the function's partial derivatives. For our function \( f(x, y) = (xy + 1)^2 \), the gradient vector at any point \( (x, y) \) is denoted by \( abla f(x, y) \).

It represents the direction of the steepest ascent of the function, providing critical information about how the function behaves near a point. The components of this vector are the partial derivatives we calculated earlier:

• \( abla f = (f_x, f_y) \)

Substituting the point \( (3, 2) \), we computed the gradient as:

• \( abla f(3, 2) = (28, 42) \)

This vector points in the direction of greatest increase in \( f \) around the point \( (3, 2) \).

Unit Vectors

A unit vector is one that has a magnitude of 1, and it is often used to indicate a direction without scaling. In the context of directional derivatives, we convert any given direction vector into a unit vector.

For the vector given in the problem, \( (5, 3) \), we compute the magnitude as follows:

• The magnitude is \( \sqrt{5^2 + 3^2} = \sqrt{34} \)

We then normalize the vector by dividing each component by this magnitude. The resulting unit vector is:

• \( \mathbf{u} = \left( \frac{5}{\sqrt{34}}, \frac{3}{\sqrt{34}} \right) \)

Now, \( \mathbf{u} \) is a unit vector pointing in the same direction as \( (5, 3) \), prepared to be used in our directional derivative calculation.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. This operation is crucial when computing directional derivatives as it combines both the gradient vector and the unit vector.

For our problem, the dot product of the gradient \( abla f(3, 2) \) and the direction unit vector \( \mathbf{u} \) is computed as follows:

• \( abla f(3, 2) \cdot \mathbf{u} = (28, 42) \cdot \left( \frac{5}{\sqrt{34}}, \frac{3}{\sqrt{34}} \right) \)
• This calculates to \( \frac{28 \cdot 5 + 42 \cdot 3}{\sqrt{34}} \)
• Ultimately yielding \( \frac{266}{\sqrt{34}} \)

This result, \( \frac{266}{\sqrt{34}} \), is the directional derivative, indicating the rate of change of the function \( f \) at the point \( (3, 2) \) in the given direction.